Ramanujan's Notebooks
Part II

Bust of Ramanujan by Paul Granlund

Bruce C. Berndt
Ramanujan's Notebooks
Part II
Springer-Verlag
New York Berlin Heidelberg
London Paris Tokyo

Bruce C. Berndt
Department of Mathematics
University of Illinois
Urbana, IL 61801
USA
The following journals have published earlier versions of
chapters in this book:
L'Enseignement Mathematique 26 A980), 1-65.
Journal of the Indian Mathematical Society 46 A982), 31-76.
Bulletin London Mathematical Society 15 A983), 273-320.
Expositiones Mathematicae 2 A984), 289-347.
Journal fur die reine und angewandte Mathematik 361 A985), 118-134.
Rocky Mountain Journal of Mathematics 15 A985), 235-310.
Ada Arithmetica 47 A986), 123-142.
Mathematics Subject Classification A980): 11-03,11P99
Library of Congress Cataloging-in-Publication Data
(Revised for volume 2)
Ramanujan Aiyangar, Srinivasa, 1887-1920.
Ramanujan's notebooks.
Includes bibliographies and index.
1. Mathematics. I. Berndt, Bruce C, 1939-
II. Title.
QA3.R33 1985 510 84-20201
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© 1989 by Springer-Verlag New York Inc.
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987654321
ISBN 0-387-96794-X Springer-Verlag New York Berlin Heidelberg
ISBN 3-540-96794-X Springer-Verlag Berlin Heidelberg New York

Dedicated to my mother
Helen
and the memory of my father
Harvey

The relation between Hardy and Ramanujan is unparalleled in scientific
history. Each had enormous respect for the abilities of the other. Mrs.
Ramanujan told the author in 1984 of her husband's deep admiration for
Hardy. Although Ramanujan returned from England with a terminal illness,
he never regretted accepting Hardy's invitation to visit Cambridge.

** Ч- *
Photograph reprinted with permission from Collected Papers of G. H. Hardy, Vol. 1,
Oxford University Press, Oxford, 1969.

Preface
During the years 1903-1914, Ramanujan recorded many of his mathematical
discoveries in notebooks without providing proofs. Although many of his
results were already in the literature, more were not. Almost a decade after
Ramanujan's death in 1920, G. N. Watson and В. М. Wilson began to edit
his notebooks, but never completed the task. A photostat edition, with no
editing, was published by the Tata Institute of Fundamental Research in
Bombay in 1957.
This book is the second of four Volumes devoted to the editing of Ramanu-
Ramanujan's notebooks. Part I, published in 1985, contains an account of Chapters
1-9 in the second notebook as well as a description of Ramanujan's quarterly
reports. In this volume, we examine Chapters 10-15 in Ramanujan's second
notebook. If a result is known, we provide references in the literature where
proofs may be found; if a result is not known, we attempt to prove it. Except
in a few instances when Ramanujan's intent is not clear, we have been able to
establish each result in these six chapters.
Chapters 10-15 are among the most interesting chapters in the notebooks.
Not only are the results fascinating, but for the most part, Ramanujan's
methods remain a mystery. Much work still needs to be done. We hope readers
will strive to discover Ramanujan's thoughts and further develop his beautiful
ideas.
Urbana, Illinois Bruce С Berndt
November 1987

Contents
Preface ix
Introduction 1
CHAPTER 10
Hypergeometric Series, I 7
CHAPTER 11
Hypergeometric Series, II 48
CHAPTER 12
Continued Fractions 103
CHAPTER 13
Integrals and Asymptotic Expansions 185
CHAPTER 14
Infinite Series 240
CHAPTER 15
Asymptotic Expansions and Modular Forms 300
References 339
Index 355

Introduction
We take up something—we know it is finite; but as soon as we begin to analyze it, it
leads us beyond our reason, and we never find an end to all its qualities, its possibilities,
its powers, its relations. It has become infinite.
Vivekananda
In a certain sense, mathematics has been advanced most by those who are distinguished
more for intuition than for rigorous methods of proof.
Felix Klein
For now we see through a glass, darkly; but then face to face: now I know in part; but
then shall I know even as also I am known.
First Corinthians 13:12
The quoted passages of Vivekananda, Klein, and St. Paul each point to a
certain facet of Ramanujan's work. First, on June 1-5, 1987, the centenary of
Ramanujan's birth was celebrated at the University of Illinois with a series of
28 expository lectures and several contributed papers that traced Ramanujan's
influence to many areas of current research; see the conference Proceedings
edited by Andrews et al. [1]. Thus, Ramanujan's mathematics continues to
generate a vast amount of research in a variety of areas. Second, in the sequel,
we shall see many instances where Ramanujan made profound contributions
but for which he probably did not have rigorous proofs; for example, see Entry
10 of Chapter 13. Third, although St. Paul's passage is eschatological in nature,
it points to the great need to learn how Ramanujan reasoned and made his
discoveries. Perhaps we can prove Ramanujan's claims, but we may not know
the well from which they sprung. These three aspects of Ramanujan's work
will frequently be made manifest in the pages that follow.

2 Introduction
In this book, we examine Chapters 10-15 in Ramanujan's second note-
notebook. In many respects, these chapters contain some of Ramanujan's most
fascinating and enigmatic discoveries. Our goal has been to prove each claim
made by Ramanujan. With a few possible exceptions where the meaning is
obscure, we either give a proof or indicate where in the literature proofs can
be found. We emphasize that many (perhaps most) of our proofs are un-
undoubtedly different from those found by Ramanujan. In particular, we have
often employed the theory of functions of a complex variable, a subject with
which Ramanujan had no familiarity. In no way should our proofs, or this
book, be regarded as definitive. In many instances, more transparent proofs,
especially those that might give insight into Ramanujan's reasoning, should
be sought.
Each of Chapters 10 13 and 15 contains 12 pages, while Chapter 14
encompasses 14 pages in Ramanujan's second notebook. The number of
theorems, corollaries, and examples in each chapter is listed in the following
table.
Chapter
10
11
12
13
14
15
Total
Number of Results
116
103
113
92
87
94
605
In the sequel, we have employed Ramanujan's designations of corollary,
example, and so on, although the appellations may not be optimal. Generally,
we have adhered to Ramanujan's notation so that the reader following our
account with a copy of Ramanujan's notebooks at hand will have an easier
task. At times, for clarity, we have changed notation, especially in Chapter 14
where we make heavy use of complex function theory. Except for some minor
alterations, especially in Chapter 15, we have also preserved Ramanujan's
order of presentation.
Many of the theorems communicated by Ramanujan in his famous letters
to G. H. Hardy on January 16, 1913 and February 27, 1913 may be found in
Chapters 10-15. In the table below, we list these results.

Introduction
Location in Collected Papers
p. xxvi, V, B)
p. xxvi, V, C)
p. xxvi, V, D)
p. xxvi, V, E)
p. xxvi, V, F)
p. xxvi, VI, C)
p. xxvi, VII, B)
p. xxvi, VII, C)
p. xxvii, VII, G)
p. xxvii, IX, A)
p. xxviii, C)
p. xxviii, A0)
p. xxix, A4)
p. 349, V, G)
p. 349, V, (8)
p. 350, VI, D)
p. 350, VI, E)
p. 350, IX, B)
p. 351, last formula in first letter
p. 352, penultimate paragraph
of3
p. 352, last paragraph of 3
p. 353, A6)
3
Location in Notebooks
Chapter 10, Section 7, Example 15
Chapter 10, Section 7, Example 14
Chapter 14, Section 13, Corollary (iii)
Chapter 14, Entry 25(ii)
Chapter 14, Entry 25(vii)
Chapter 11, Section 20, Example 2
Chapter 12, Entry 48, Corollary of Entry 48
Chapter 13, Entry 6
Chapter 13, Corollary (ii) of Entry 10
Chapter 15, Section 2, Example (iv)
Chapter 12, Section 25, Corollary 1
Chapter 10, Equation C1.1)
Chapter 11, Entry 29(i)
Chapter 12, Entry 27
Chapter 14, Entry 25(xi)
Chapter 14, Entry 25(xii)
Chapter 13, Corollary of Entry 21
Chapter 13, Example for Corollary of Entry 21
Chapter 12, Entry 34
Chapter 10, Entry 29(b)
Chapter 15, Section 2, Example (ii)
Chapter 15, Section 2, Example (iv)
Chapter 12, Corollary to Entry 34
Several of Ramanujan's published papers and problems posed in the Journal
of the Indian Mathematical Society have their origins in the notebooks. In
most cases, only a small portion of the published paper is actually found in
the notebooks. We list below those papers with their geneses in Chapters
10-15, together with the respective locations in the notebooks.
Paper
Location in Notebooks
On question 330 of Prof. Sanjana
Modular equations and approximations
to n
\3~
On the product [] 1 + -
n = 0 |_
Some definite integrals
a + nd
Some definite integrals connected with
Gauss's sums
Chapter 10, Section 13
Chapter 14, Section 8, Example
Chapter 13, Section 27
Chapter 13, Entries 14, 15, 16(iii),
Corollary of Entry 19, Entry 21,
Corollary of Entry 21, Entry 22
Chapter 14, Section 6
Chapter 14, Entry 22(ii)

Introduction
Paper
Location in Notebooks
On certain arithmetical functions
On certain trigonometrical sums and their
applications in the theory of numbers
Asymptotic formulae in combinatory
analysis (with G. H. Hardy)
A class of definite integrals
Question 289
Question 294
Question 296
Question 358
Question 387
Question 769
Chapter 15, Sections 9, 10,12, 13,
and 14
Chapter 14, Entry 13
Chapter 15, Section 2, Example (iv)
Chapter 13, Sections 23-25
Chapter 12, Section 4, Examples
(i), (ii)
Chapter 12, Section 48
Chapter 13, Entry 6
Chapter 13, Section 21, Example
Chapter 14, Corollary of Entry 14
Chapter 14, Section 8, Example
Chapter 13, Entry 11 (Hi)
We now provide brief summaries for each of Chapters 10-15. More de-
detailed descriptions may be found at the beginning of each chapter.
Of all the topics examined by Ramanujan in his notebooks, only modular
equations received more attention than hypergeometric series. Chapter 10 is
the first of two chapters devoted almost entirely to the latter subject. In 1923,
Hardy [1], [7, pp. 505-516] published a brief overview of the corresponding
chapter in the first notebook. Ramanujan rediscovered most of the classical
formulas in the subject, including those attached to the names of Gauss,
Kummer, Dougall, Dixon, and Saalschiitz. Ramanujan possessed the uncanny
ability for finding the most important examples of theorems, and Chapter 10
contains many elegant examples of infinite series summed in closed form.
Ramanujan was the first to discover identities for certain partial sums of
hypergeometric series, and these may be found in the latter parts of Chapter
10. Ramanujan continues his study of hypergeometric series in Chapter 11.
Two topics dominate the chapter. The first concerns products of hypergeo-
hypergeometric series, and most of these results are original with Ramanujan. Second,
Ramanujan offers several beautiful asymptotic formulas for hypergeometric
functions. By far, the most interesting is Corollary 2 in Section 24. Quadratic
transformations of hypergeometric series are also featured in Chapter 11.
Chapter 12 is almost entirely devoted to continued fractions and is one
of the most fascinating chapters in the notebooks. Ramanujan's published
papers contain only one continued fraction! However, Ramanujan submitted
some continued fractions as problems to the Journal of the Indian Mathe-
Mathematical Society, and his letters to Hardy contain some of his most beautiful
theorems on continued fractions. Nonetheless, the great majority of the results
in Chapter 12 are new. Perhaps the most exquisite theorems are the many

Introduction 5
continued fraction expansions for products and quotients of gamma functions.
We have no idea how Ramanujan discovered these formulas. Especially awe
inspiring is Entry 40 involving several parameters.
Equally astonishing is Chapter 13. In the first 11 sections, one finds se-
several beautiful, deep asymptotic expansions for integrals and series. Entries 7
and 10 are perhaps highlights. Ramanujan left us no clues of how he discov-
discovered these fascinating theorems. Are these results prototypes for further yet un-
undiscovered theorems? Although we have given proofs, we do not have a
firm understanding of how these wonderful theorems fit with the rest of
mathematics.
Those readers who are fascinated by elegant series evaluations and identi-
identities will take great pleasure in reading Chapter 14. Here, one can find several
series identities that have a symmetry that one often associates with certain
applications of the Poisson summation formula, which, however, does not
seem to be applicable in most cases here. Several closed form evaluations
of series involving hyperbolic functions are given. Some of the results in
this chapter can be established by employing partial fraction decomposi-
decompositions. We have utilized two additional primary tools: contour integration and
some theorems of the author on transformations of Eisenstein series. Since
neither of these techniques was in Ramanujan's arsenal, we do not know how
Ramanujan discovered most of the results in Chapter 14.
Chapter 15 is the most unorganized of all the chapters in the second
notebook. The first seven sections are primarily devoted to interesting asymp-
asymptotic expansions of several series. Entry 8 offers an elegant transformation
formula for a modified theta-function.
In the sequel, equation numbers refer to equations in the same chapter,
unless another chapter is indicated. Unless otherwise stated, page numbers
refer to pages in Ramanujan's second notebook [15] in the pagination of the
Tata Institute. Part I refers to the author's account [9] of Chapters 1-9, and
Part III refers to his account [11] of Chapters 16-21.
In what follows, the principal value of the logarithm is always denoted by
Log. The set of all (finite) complex numbers is denoted by <€. The residue of a
function / at an isolated singularity a will be denoted by R(a). (The identity
of/ will always be clear.)
A small portion of this book has been aided by notes left by G. N. Watson
and В. М. Wilson in their efforts to edit Ramanujan's notebooks. We are
grateful to the Master and Fellows of Trinity College, Cambridge, for pro-
providing a copy of these notes and for permission to use this material in this
book.
We sincerely appreciate the collaboration of Robert L. Lamphere on
Chapter 12 and Ronald J. Evans on Chapters 13 and 15. Because of their
efforts, our accounts of these chapters are decidedly better than what we would
have accomplished without their help. Most of the material in this book
appeared in previously published versions of these chapters. We are grateful

6 Introduction
for the cooperation shown by each of the journals publishing our earlier
accounts. A table below indicates the bibliographic data for the original
publications. (Portions of Chapter 15 were published in two parts.)
Chapter
10
11
12
13
14
15
15
R. L.
R.J.
R.J.
R.J.
Coauthors
Lamphere, В. М. Wilson
Evans
Evans
Evans
Publication
J. Indian Math. Soc. 46 A982), 31-76
Bull. London Math. Soc. 15 A983),
273-320
Rocky Mt. J. Math. 15 A985), 235-310
Expos. Math. 2 A984), 289-347
UEnseign. Math. 26A980), 1-65
J. Reine Angew. Math. 361 A985),
118-134
Ada Arith. 47 A986), 123-142
Although only one author is listed on the cover of this book, several
mathematicians have made valuable contributions. We are very grateful to
George Andrews, Richard Askey, Henri Cohen, Ronald Evans, Jerry Fields,
P. Flajolet, M. L. Glasser, Mourad Ismail, Lisa Jacobsen, Robert Lamphere,
David Masser, F. W. J. Olver, R. Sitaramachandrarao, and Don Zagier for the
many proofs and suggestions that they have contributed. In particular, Askey,
Evans, and Jacobsen have each supplied several proofs and offered many
helpful comments, and we are especially indebted to them. Others, not named,
have made helpful comments, and we publicly offer them our thanks as well.
The author bears the responsibility for all errors and would like to be
notified of such, whether they be minor or serious.
The manuscript was typed by the three best technical typists in Champaign-
Urbana—Melody Armstrong, Hilda Britt, and Dee Wrather. We thank them
for the superb quality of their typing.
Lastly, we express our deep gratitude to James Vaughn and the Vaughn
Foundation for the generous funding that they have given the author during
summers. This book could not have been completed without the support of
the Vaughn Foundation.

CHAPTER 10
Hypergeometric Series, I
In 1923, Hardy published a paper [1], [7, pp. 505-516] providing an overview
of the contents of Chapter 12 of the first notebook. This chapter, which
corresponds to Chapter 10 of the second notebook, is concerned primarily
with hypergeometric series. It should be emphasized that Hardy gave only a
brief survey of Chapter 12; this chapter contains many interesting results not
mentioned by Hardy, and Chapter 10 of the second notebook possesses
material not found in the first. Quite remarkably, Ramanujan independently
discovered a great number of the primary classical theorems in the theory of
hypergeometric series. In particular, he rediscovered well-known theorems of
Gauss, Kummer, Dougall, Dixon, Saalschiitz, and Thomae, as well as special
cases of Whipple's transformation. Unfortunately, Ramanujan left us little
knowledge as to how he made his beautiful discoveries about hypergeometric
series. The first notebook contains a few brief sketches of proofs, but the only
sketch in the second notebook is found after Entry 8, which is Gauss's
theorem. We shall present this argument of Ramanujan in the sequel.
As the reader will see, this chapter contains a wealth of beautiful evaluations
of hypergeometric functions, usually at the argument +1 or — 1. In this
connection, we mention the recent work of R. Wm. Gosper, I. Gessel, and
D. Stanton. By employing "splitting functions" and the computer algebra
system MACSYMA, Gosper discovered many new hypergeometric function
evaluations. Most of these, in the terminating cases, were ingeniously proved
by Gessel and Stanton [1]. Two conjectures of Gosper were established by
P. W. Karlsson [1].
Many elegant and useful binomial coefficient sums can be evaluated,
usually quite simply, by employing the theorems of Gauss, Dixon, Saalschiitz,
Kummer, and others. See the paper by R. Roy [2] for many illustrations.

10. Hypergeometric Series, I
We now offer several remarks about notation. As usual, we put
Г(а + к)
{a)
where к is any complex number. The generalized hypergeometric series pFq is
defined by
1_ (ai)t(a2)t(«A*
where p and 4 are nonnegative integers and al,a1,...,ap and &, j82,..., j8,
are complex numbers. If the number of parameters is "small," we may some-
sometimes use the notation pFq(a1, ce2,..., cep; J31; |?2,..., /Jg; x) in place of the nota-
notation on the left side of @.1). In this chapter, we are concerned only with
the cases when p = q + 1. In these instances, the series defining pFq con-
converges when |x| < 1 for all choices of the parameters <x;, fa, 1 < i < q + 1,
1 < j < q. However, q+iFq can be continued analytically into the complex
plane cut at [1, oo). If x = 1, the series converges for Re^ + • • • + ccq+1) <
?! + • • • + j?,); if x = - 1, there is convergence for Re^ + • • • + a,,+1) <
?! + ¦•• + fiq) + 1. In all the theorems and examples that follow, when
x = +1, we state the conditions for convergence, but without further com-
comment. Moreover, as is customary, if x = 1, we omit the argument in the
notation @.1). It should be remarked that Ramanujan has no notation for
hypergeometric series. All formulas are stated by writing out the first few terms
in each series. This practice has one distinct advantage in that the elegance of
formulas involving series is often more easily discerned. Frequently, a compact
notation obscures the aesthetic beauty of a series relation. For brevity, we
usually use a compact notation, but, at times, in particularly elegant instances,
we follow Ramanujan's practice. To aid readers examining this chapter in con-
conjunction with the second notebook, we have usually adhered to Ramanujan's
notations for the parameters.
For the most part, we refer only to primary sources. For example, we give
a reference to Dougall's paper wherein his famous theorem is initially proved,
but we do not usually offer further references to other proofs, applications,
and so on. The classical texts of Appell and Kampe de Feriet [1], Klein [1],
Bailey [4], and Slater [1] contain excellent bibliographies on which it would
be difficult to elaborate. In the sequel, Bailey's well-known tract [4] will be
our basic reference. We also indicate which formulas have been discussed by
Hardy [1] in his overview. For those readers wishing to learn more about the
history of hypergeometric functions, we recommend the papers of Askey [1],
Dutka [3], and Buhler [1].
In the sequel, always, \j/(z) = P(z)/r(z). Frequent use is made of the classical
representation (e.g., see Luke's text [1, p. 12])
ф{2 +1) = -у + ? (l - _L_\ @.2)

10. Hypergeometric Series, I
where у denotes Euler's constant. We also often employ the simple differentia-
differentiation formulas
= ±{k- 1)!, fc> 1. @.3)
u = 0
Entry 1. Suppose that at least one of the quantities x, y, z, u, or — x — у — z —
и — 2n — 1 is a positive integer. Then
Г n, \n + 1, -.x, —y, -z, —u, x + y + z + u + 2n + l 1
7 6\_\n, x + n+l,y + n+l,z + n + l, u + n + 1,— x ~ у — z — и — n]
Г(х + n + \)Г(у + п + 1)Г(г + и + 1)Г(м + и + 1)Г(х +y + z + n + l)
Г (у + z + и + п + 1)Г(х + и + z + п + 1)Г(х +у + м + п+1)
Г(х + z + п + 1)Г(у + и + п + 1)Г(х + y + z + u + n + l)
A.1)
Ramanuj an did not indicate that A.1) holds when — x — у — z — и — In —
1 is a positive integer.
Entry 1 is originally due to Dougall [1] in 1907, which is probably less
than three years before Ramanujan discovered the theorem. Hardy [1, Eq.
B.1)] has thoroughly discussed Entry 1 and gives Dougall's proof, as does
Bailey [4, p. 34].
Entry 2. // either x, y, or z is a positive integer, then
Г Y v 7 1
3 2\_n + 1, -x - y- z-n]
_ Г(п + 1)Г(х + y + n + l)T(y + z + n+ l)T(z + x + n + 1)
~ Г(х + n + l)T(y + n+ l)T(z + n+ 1)Г(х + у + z + n + 1)'
Entry 2 is known as Saalschiitz's theorem [1], [2], although according to
Jacobi [1], [2] and Askey [1], the result was first established by Pfaff [1] in
1797. In Hardy's paper [1], Entry 2 corresponds to Eq. E.1) there. It should
be mentioned that Hardy's formulation is incorrect. For a proof of Entry 2,
see Bailey's tract [4, p. 9].
Entry 3. // x, y, z,or — x — у — z — 2n is a positive integer, then
Г \n + 1, 1, -x, -y, -z, x + y + z + 2n 1
5 \_jn, x + n + 1, у + n + 1, z + n + 1, -x — у — z - n + lj
6 5 \
_ (x + n)(y + n)(z + n)(x + у + z + n)
n(x + у + n){y + z + n)(x + z + n) '

10 10. Hypergeometric Series, I
Proof. Set и = -1 in Entry 1. ?
Entry 4. //either x, y, z, or —x - у - z — 2n — 1 is a positive integer, then
f (n + 2k)(-x)k(-y)k(-z)k(x + y + z + 2n+l)k
*ti k(n + k)(x + n + l)k(y + n+ l)k(z + n + l)t(-x -y-z-n)k
= ф{х + n + 1) + ф(у + n + 1) + ф{г + n + 1) + ф(х + у + z + n + 1)
- ф(п + 1) — ф(х + у + n + 1) - ф(у + z + n + 1) — ф(г + x + n + 1).
Proof. Logarithmically differentiate both sides of A.1) with respect to и and
then set и = 0. Using @.3), we complete the proof after a little simplification.
?
Example (i). // x is a positive integer, then
x + 1) 4x - 3 \(x + l)(x + 2O Dx - 3)Dx - 4)
_ Г4(х + 1)Г4(Зх - 1)
Г6Bх)ГDх - 2) '
Proof. In Entry 1, put и = 1, replace x by x — 1, and set у = z = и = x — 1.
After some simplification, the desired equality follows. ?
Example (ii). // x is an odd, positive integer, then
(x - lKCx -1I /(x - l)(x - 3)VCx - l)Cx + 1)
x + lKCx - 3) 2 \{x + l)(x + 3O Cx - 3)Cx - 5)
Proof. In Entry 4, put n = 0, replace x by j(x — 1) and set у = z = |(x — 1).
The proposed equality now readily follows. ?
Example (iii). // x is a positive integer, then
/x-lV3x-l /(x-l)(x-2)V Cx-l)Cx)
\x + \) 3x - 3 \(х + l)(x + 2O Cx - 3)Cx - 4)
x3Cx - 2)
Proof. In Entry 3, set n = 1, replace x by x — 1, and let у = z = x - 1. The
displayed equality now easily follows. ?

10. Hypergeometric Series, 1 11
Example (iv). // x is a nonnegative integer, then
3x V 2! /3xCx-l) Г3(х + 1)Г(Зх+ 1)'
Proof. In Entry 2, set n = 0 and x = у = z to achieve the desired result. ?
In the notebooks (p. 118), Ramanujan has mistakenly put ГCх + 1) in the
numerator instead of the denominator in Example (iv).
Example (v). If x is a positive integer, then
x x — 1 x x(x — 1) (x — l)(x — 2) x(x — 1)
1! x + 1 4x - 1 2! (x + l)(x + 2) Dx - l)Dx - 2)
_ 8Г3(Зх + 1)Г(х + 1)
= 9Г3Bх + 1)ГDх + 1)'
Proof. In Entry 2, put n = z = x and у = x — 1. The proposed equality
readily follows. ?
Entry 5. // Re(x + у + z + n+ l)>0, then
Г |n + l,n, -x, -y, -z
Un x + n+1 y + n+1 z + n+1
_ Г(х + n + 1)Г(у + n + l)T(z +n+ 1)Г(х + у + z + n + 1)
~ Г{п + 1)Г(х + у + n + 1)Г(у + z + n+ 1)Г(х + z + n + 1)' ( ' '
Entry 5 is again due to Dougall [1]. Hardy [1] discusses Entry 5 (C.1) in
his paper) and gives a proof based on a theorem of Carlson. For another proof,
see Bailey's monograph [4, p. 27]. It is interesting that a ^-analogue of Entry
5 was established by L. J. Rogers [1] in 1895, twelve years before Dougall's
discovery.
Wilson [1] has shown that Dougall's theorem is intimately connected with
the orthogonality of certain orthogonal polynomials. Moreover [1, p. 694],
Г
Jo
Г (a + ix)T{b + ix)T(c + ix)T(d + ix) 2
ax
FBix)
is a continuous analogue of the sum in Entry 5. The special case с = 0, d = j
was, in fact, evaluated by Ramanujan [8], [16, p. 57]. For further related
comments, see Section 22 of Chapter 13.
For brevity, let
[«!,...,«p+1 1
P+lrp\ о a m \
denote the sum of the first m + 1 terms of p+iFp(a1,..., ap+1; ft,..., /fp; 1).

12 10. Hypergeometric Series, I
Entry 6. // a + P + у + 1 = n, then
T(n + 2)Г(а
m
Г(п - а + 1)Г(п - ft + 1)Г(п - у + 1)
\{п + 3), п + 1, а + 1, Р + 1, у + 1
(п + 1), п - а + 1, п - Р + 1, п - у + 1
~ 2 Log m - ф(а + 1) - <КД + 1) - ф(у + 1) - С,
as m fends to oo, vvnere С denotes Euler's constant.
In our originally published account of Chapter 10 (see the reference in the
Introduction), we gave a proof of Entry 6 supplied to us by J. L. Fields based
on his paper [1]. R. J. Evans [1] has since found a much simpler proof of a
slightly stronger result. We reformulate this stronger version of Entry 6 and
give Evans's proof.
Entry 6 (Second Version). // a, b, c, and a + b + с are not nonpositive integers,
then as m tends to oo,
т
T(a + b + c)T(a)T(b)T(c) R(a + b + с + 1), a + b + с - 1, a, b, с
Г{Ь + с)Г(а + с)Г(а + Ъ) 5 4|_ \(а + Ь Ч- с - 1), b + с, а + с, а + Ь
= 2 Log т — у — ф(а) — ф(Ь) — ф(с) + ОI
\ т
where у denotes Euler's constant.
Proof. Recall Whipple's transformation [1] (Bailey [4, p. 25]),
Г a,\ +$a,b,c,d,e, -m 1
7 6 |_ia, l+a-b, l + a-c, l + a-d,l+a-e, I + a + m]
_ A + a)m(l + а - d - e)m г Г l + a-b-c,d,e,-m
A + а - d)m(l +a-e)m4 3[1+а-Ь, l+a-c, d + e-a-
F.1)
where m is a nonnegative integer. Replacing a, d, and e by a + b + с — 1, а,
and a + b + c + m + e, respectively, in F.1), where e > 0, we find that
+ b + с — 1, ^(а + b + с + 1), b, с, a, a + b + с + m + в, — т
(a + b + с — 1), a + c, a + b, b + c, —m — s, a + b + с + т
= (q + fe + с)т(-а-т-е)„
(b + cU-m-e)m
Letting ? tend to 0, we deduce that
[a, a, a + b + с + m + в, —т~\
a + c,a + b,a + l+e J

10. Hypergeometric Series, I
13
5*4
г(а + b + с + 1), a + b + с — 1, a, b, с
${a + b + c- l),a + c,a + b,b + c
a + b + c)Ja + 1),
m
(b
[a,a,a + b + c + m, — ml
a + c, a + b, a + 1 J'
Thus, the left side of Entry 6 is equal to
T(a + b + c)T(a)T(b)T(c) (a + b + cU«
c)T(a + c)T(a + b) (b + с)иA)„
— m
[a,a,a + b + c + m, —
a + c,a + b,a+l
F.2)
We now apply a transformation for 1-balanced terminating 4F3 series
found in Bailey's tract [4, p. 56]. \fu + v + w = x + y + z — m + 1, then
x, y, z,-m\ (v - z)m(w - z)n
U, V, W
и — х, и — у, z, —т
1 — v + z — т, 1 — w + z — т, и_
F.3)
Letting х = а + b + с + т,у = a,z = а,и = а + b,v = а + с, and w — а + 1,
we find that
Га, a, a + b + с + m, —ml
43|_ a + c, a + b, a + \ J
3^2
a, b, —с — т
m .
(a + c)m(a+l)m~ -\_a + b,l-c-m
Using this equality in F.2), we find that the left side in Entry 6 equals
T{a + b + с + m)T{c + m)
where:
T(b + с + m)T(a + с + m)R"
Т(а)Г(Ь) „ Г a,b,-c-m
F.4)
_ ™ Г{а + к)Г{Ь + k)(c + m)
~~ fc=o Г(а + Ь + ^)ГA + к){с + т-к)'
By Stirling's formula, the coefficient of Rm in F.4) equals 1 + O(l/m). Ex-
Examining Entry 6, we see that it remains to show that
Rm = 2 Log m - у - ф(а) - ф(Ь) - ф(с) +
т
m
F.5)
Let

14 10. Hypergeometric Series, I
where
- T(a + к)Г(Ь + к)
m h
kh T(a + b + к)ГA + к)
and
Г(а + к)Г(Ь + к)
From Luke's book [1, p. 110, Eq. C5)],
Um = Log m - у - ф(а) - ф(Ь) + O(l/m), F.6)
as m tends to oo. (A slightly weaker version is given in Entry 15 below. See
also B4.5) of Chapter 11 for F.6).) By Stirling's formula and @.2),
1 / 1 m f 1 1
| O\ У I |
\m + с t \m + с к к
t=i с + к - 1 \m + с n=i \m + с - к kj
= ф(т + с) - ф(с) + О(\/т)
= Log m - ф(с) + ОA/т), F.7)
by Stirling's formula for ф(г) (Luke [1, p. 33]). Combining F.6) and F.7), we
deduce F.5) to complete the proof. ?
Corollary. Let 0 < x < 1. Then as x tends to 0,
Proof. Let n = a. = p = y= — ^ in Entry 6 to obtain the formula
as m tends to oo, where on the right side above у now denotes Euler's constant.
Since ф({) = -2 Log 2 - у (see Luke's book [1, p. 13]), we find that
5 1 1 1 1
4 2> 2> 2> 2
4> *> ^ x
as m tends to oo. It follows that
n2 Z
-, т л т
Hence,
-^-^x*U3Log2.

10. H) pergeometric Series, I 15
Therefore, as x tends to 1 —,
The corollary now follows. ?
For further expansions of hypergeometric functions in the neighborhoods
of logarithmic singularities, see Section 15 of this chapter and Sections 24-26
in Chapter 11. В. С Carlson [1] has established expansions about logarithmic
branch points for several classes of related functions.
Entry 7. // Re(x + у + \n + 1) > 0, then
_T(x + n + l)T(y + n+ 1)Г(^п + 1)Г(х + у + \n + 1)
~ Г(и + 1)Г(х +у + п+ 1)Г(х + \п + 1)Т(у + \п + 1)'
Proof. Set г = — \п in Entry 5. ?
Entry 7 is a famous theorem of Dixon [1]. In Hardy's paper [1], see C.2).
A terminating version of Dixon's theorem can be used to evaluate Selberg's
integral in two dimensions (Andrews [3]). The case n = 3 of the Dyson -
Gunson-Wilson identity can also be established from a terminating case of
Dixon's theorem (Andrews [1]). Gessel and Stanton [2] have found new short
proofs; of both Saalschutz's theorem (Entry 2) and Dixon's theorem by com-
computing the constant terms in certain Laurent series in two variables.
Corollary 1. // Re(x + у + n + 1) > 0, then
S \k + ^TfcJ(x"+7+iyy + n + l)t
= ф(х + n + 1) + ф(у + n + 1) - ф(п + 1) - ф(х + у + n + 1). G.1)
Proof. Logarithmically differentiate both sides of E.1) with respect to z and
then set z = 0. With the aid of @.3), we obtain the identity above after a little
simplification. ?
Corollary 2. // Re(x + у + 1) > 0, then
Г \n + 1, n, n, -x, -y 1
4 [}n, x + n + 1, у + n + 1, 1J
Г(х + n + 1)Г(у + n + 1)Г(х + у +
~ Г> + 1)Г(х + у + п+ 1)Г(х + 1)Г(у

16 10. Hypergeometric Series, I
Proof. Set z = - n in Entry 5. ?
Corollary 3. // Re(x + у + n) > 0, then
Г in + 1, -x, -y, 1 I = (x + n)(y + n)
4 3 [_|n, x + n + 1, у + n + 1J n(x + у + n) '
Proof. Put z = -1 in Entry 5. ?
Corollary 4. // Re(x + у + i(n + 1)) > 0,
\n + 1, n, -x, -y
, x + n+ \,y + n +
Г(х + n + 1)Г(у + n + l)r(j(w + 1))Г(х + у + $(и + 1))
~ Г(в + 1)Г(х + у + n + 1)Г(х + i(n + 1))Г(у + i(n + 1))'
Proof. Set z = -|(n + 1) in Entry 5. ?
Corollary 5. For ReBx + 2y + n + 2) > 0,
|n+l,n,-x,-y 1 Г(х+ n+1)Г(у+ n
4 3 [in, x + n + 1, у + n + 1 ; ~ J ~ Г(и + Г)Г(х + у + и + 1)' '
Proof. Corollary 5 follows from Entry 5 by letting z tend to oo. The details
are easily justified by using Stirling's formula. П
Bailey [4, p. 28] gives a proof of Corollary 5 based on Whipple's trans-
transformation F.1).
Corollary 6. // Re(x + n + 1) > 0, then
f Л 1 \ (-x)t(fc-l)! = « 1 _ S _J
kk \k n + k)(x + n + l)k(n + l)k h (k + x + nJ tti (fc + nJ'
G.3)
Proof. Differentiate both sides of G.1) with respect to у and then set у - 0.
With the use of @.2) and @.3), we complete the proof. ?
On the left side of G.3), Ramanujan (p. 119) has written k\ instead oi(k - 1)!.
Corollary 7. // Re(x - n + 1) > 0, then
sinGtn)r(x + n + 1)Г(х - и + 1)
Rn + 1, n, n, n, -xl _ si
5 4[_in,x + n+l, 1, lj"
тспГ2(х+1) ¦
Proof. Set у = z = — и in Entry 5. ?

10. Hypergeometric Series, I 17
Corollary 8. // Re(x - in 4 1) > 0, then
[n,n,—x ~| Г(х 4 n 4 l)F(in 4 1)Г(х — in 4 1)
x 4 и 4 1, 1J ~ Г(л 4 1)Г(х 4 1)ГA - ?л)Г(х 4 in 4 1)'
Proof. Puty = —и in Entry 7. ?
Corollary 9. // Re(x - in 4 i) > 0, then
fin 4 1, n, n, -xl _ Г(х 4 n 4 l)r(in 4 j)T(x - \n 4 i)
4'3[in,x4n4 1, 1 J Г(п4 1)Г(х4 l)F(i - in)F(x 4 in 4 i)'
Proof. In Entry 5, set у —- —n and z = —i(n 4 1). ?
Corollary 10. // ReBx - n 4 2) > 0, then
in4l,n, n,-x ~| Г(х 4 и 4 1)
Г(п 4 1)Г(х 4 1)
Prooj-. Let у = — n in Corollary 5. ?
Corollary 11. If Re x > - -1, then
Г(х 4 n 4 I)r2(in 4 1)Г(х 4 1)
Г \n,n,-x 1 =
3 2[in4 1, x4n4 lj
Г(и 4 1)Г2(х 4 in 4 1)
Proop. Put у = -in in Entry 7. ?
Corollary 12. If Rex > - i, tnen
, , 1Ч Г(х 4 и 4 1)ГBх 4 1)
Raimanujan probably deduced Corollary 12 from Entry 7 by setting
у = - i(n 4 1) and then using Legendre's duplication formula to simplify the
resulting evaluation. However, in fact, Corollary 12 is a special case of Gauss's
theorem, which is given by Ramanujan in complete generality in Entry 8
below. See Bailey's monograph [4, pp. 2, 3] for a proof.
Corollary 13. // Re x > - 1, then
Г(х 4 n 4 l)F(in 4 1)
2F1(n, — x; x 4 n 4 1; — 1) =
Г(х 4 in 4 1)Г(п 4 1)'
Corollary 13 is known as Kummer's theorem [1], [2, pp. 75-166] and is
most commonly proved by using a quadratic transformation also due to
Kumrner. See Bailey's tract [4, pp. 9, 10] for details. Ramanujan evidently
derived Corollary 13 by letting у tend to oo in Entry 7.

18 10. Hypergeometric Series, I
Corollary 14. // Re x > —\, then
fin + l,n, -x _ Л _ Г(х + n + l)F(jn + j)
3 2 [in, x + n + 1 ' J ~ Г(и + 1)Г(х + in + i) '
Proof. Put у = -\{n + 1) in Corollary 5. Q
Corollary 15. If Ren < i, then
I sin(Ttn) tan(Ttn)
fin + 1> n, "> "> n~| Г2(п) si
5 4|_ in, 1,1,1 J= n2
TBn + 1)
Proof. Let x = y = z= —n in Entry 5 and use the reflection formula to
simplify the resulting evaluation. ?
Corollary 15 is Eq. C.33) in Hardy's paper [1].
Corollary 16. If Re n < \, then
[~in + 1, n, n, n] sin(nn)r(jn + j)r(j
4 3L in, 1,1 J ™r2(i-in)
-fn)
Proof. Put x = у = — n in Corollary 4. ?
Corollary 16 is C.31) in Hardy's paper [1] and can also be found in Bailey's
text [4, p. 96].
Corollary 17. If Ren < §, then
sin(nn)
fin + 1, n, n, n 1 :
L 2n, 1, 1 J
nn
Proof. Set x = у = - n in Corollary 5. ?
Corollary 17 is Eq. C.32) in Hardy's paper [1] and is recorded by Bailey
[4, p. 96].
Corollary 18. // Re n < 1, then
Г in, n, n 1 _ 2 tan(j
32Lin+l,lJ" nn
ппГ2{п + 1)
Proof. Put x = — n in Corollary 11 and use the reflection principle to sim-
simplify the resulting equality. ?
Corollary 19. // Re n < 2, then
in, \n,n "I _ %nT2{\n + 1)
2~sin(i7rn)r(n + 1)'

10. Hypergeometric Series, I 19
Proof. Put x = —\n in Corollary 11. ?
For the evaluation of certain other classes of 3F2 and 4F3 series at the
argument 1, see the papers by Lavoie [1], [2].
Corollary 20. // ReBx + n + 2) > 0, then
n + kj (x + n
Proof. Take the logarithmic derivative of both sides of G.2) with respect to
у and then set у = 0. Simplifying with the aid of @.3), we achieve the desired
equality. ?
Corollary 21. If Rex > -1, then
( — x)ftA °° / 1 1 1 1
k
n k + x k + x + n к
Proof. In Entry 5, set z = - n, logarithmically differentiate both sides of E.1)
with respect to y, and then set у = 0. Using @.2) and @.3), we deduce the
desired result. ?
Ramanujan (p. 120) neglected to record the summands - 1/fc, 1 < к < oo,
in Corollary 21.
Corollary 22. If Ren > 0, then
oo
У
?0
3 •
+ \I к (к + п)
Proof. In G.3), replace n by n — 1, differentiate both sides with respect to x,
and then set x = 0. Use @.3) in completing the proof. ?
Corollary 23. If Re n > -2, then
Дl
^Wk h (к + w *-i № + "J¦
Proof. In Corollary 6, set x = —\n. After a little simplification, the desired
result follows. ?
Corollary 24. // Re n < 1, then

20 10. Hypergeometric Series, I
Proof. Differentiate both sides of G.4) in Corollary 7 with respect to x
and then set x = 0. Using @.2) and @.3) and simplifying, we complete the
proof. ?
Example 1. // Re x > %, then
V пь ^ ,^-*)? Г3(х + 1)Г(Зх - 1)
Proof. In Entry 5, let и = 1, replace x by x - 1, and set у = z = x — 1. ?
Example 1 has been given by both Hardy [1, Eq. C.45)] and Bailey [4,
p. 96]. The following example is also recorded by Hardy [1, Eq. C.43)].
Example 2. // Re x > j, then
oo i
I Bk + 1)
x2
х\ь 2x 1
Proof. In Corollary 2, let n = 1, replace x by x — 1, and set у = x — 1. ?
Example 3. // Re x > \, then
- 2)V 2хГ4(х + 1)ГDх + 1)
(x + l)(x + 2)/ Dx - 1)Г4Bх + 1) '
Proof. In Entry 7, put и = 1, replace x by x — 1, and let у = x — 1. After
using Legendre's duplication formula to simplify, we obtain the proposed
formula. ?
Example 3 is found in Hardy's paper [1, Eq. C.49)] and Bailey's book [4,
p. 96]. The next example is equality C.44) in Hardy's paper [1].
Example 4. // Re x > j, then
(I — x)? Г2(х + 1)
Proof. In Corollary 5, let и = 1, replace x by x — 1, and set у = x — 1. ?
Example 5. // Re x > j, then
1 13х ,5(X-1)(X-2)|- ::

10. Hypergeometric Series, I 21
Proof. Put n = 1 and replace x by x — 1 in Corollary 14. ?
Example 5 is given by both Hardy [1, Eq. C.41)] and Bailey [4, p. 96].
Example 6 is also given by Hardy [1, Eq. C.46)].
Example 6. // Re x > 0, then
x - 1 (x - l)(x - 2) 22д:Г2(х + 1)
x+ 1 + (x + l)(x + 2) + = f Bx
Proof. In Corollary 13, put и = 1, replace x by x — 1, and use Legendre's
duplication formula to simplify. ?
Example 7. // Re x > ^, then
x - 1 (x - l)(x - 2) x
+
~
x + 1 (x + l)(x + 2) 2x - 1
Proof. In Corollary 12, set n = 1 and replace x by x — 1. ?
Examples 7 and 8 are given by Hardy [1, Eqs. C.47), C.42)]. See also
Bailey's tract [4, p. 96] for Example 8.
Example 8. // Re x > 1, then
xTT + 5(x + l)(x + 2)~ ' = '
Proof. Put n = 1 and replace x by x — 1 in Corollary 9. ?
Example 9. // Re x > 0, then
1
k% (k + 1)A + x)k ГBх) 2 tti
Proof. Replace x by x — 1 in Rummer's formula, Corollary 13. Then loga-
logarithmically differentiate both sides with respect to n and set n = 0. Using @.2)
and @.3), we find that
G.7)
Zk^l \К. К -t- X - 1/
Now,

22 10. Hypergeometric Series, I
-х), = i_ » (-!)*(!-x^fc-
x ?i fc(l + x)
*ti fc(l + X)^i X kti A + Х),_!
» (-!)"(! -x)k 22*~1Г2(х + 1)
-^(fc + lXl+x)»" хГBх+1) ' G'8)
by Example 6. Combining G.7) and G.8), we deduce the desired result. ?
Example 10. // Re x > 0, then
ttb(fc + 1)A +x)t 2x &i\k + x k + 2xj'
Proof. By Entry 9 below and @.2),
k^o (k + 1)A + x),
i _ JL V ( l _ !
x 2x ,^i I/c + x k + 2xj'
and the proof is complete. ?
The next example is in Hardy's paper [1, Eq. C.48)] and Bailey's book
[4, p. 96].
Example 11. // Re x > 0, then
x - 1 (x - l)(x - 2) 24дТ4(х + 1)
~
3(x + 1) + 5(x + l)(x + 2) ~ ~4хГ2Bх+
Proof. In Corollary 11, put n = 1 and replace x by x — 1. After using the
Legendre duplication formula, we easily obtain the proposed equality. ?
Example 12. If x is a positive integer, then
» A-х), = 1 * 1 1 f
k% (k + 1JA +x)k xhk + x 2 h
Proof. Consider Dixon's formula, Entry 7, and logarithmically differentiate
both sides with respect to y. Setting у = 0 and using @.2) and @.3), we find
that

10. Hypergeometric Series, I 23
(n)k(-x)k
-z
% k{x + n + \)k(n
1 1 1 1
~й\/с + х + п к + n) + ttl \k + \n кл-хЛ-\п)
Next, replace x by x — 1 and differentiate both sides of G.9) with respect to n.
Setting n = 0 and using @.3), we deduce that
\( + X —
On the other hand, by Example 10,
h k\x)k h k2(l+x)k_lX
f
x ,4 (fc + 1)A + x)k h (k + 1JA + x),
J
x [2x + ii Vfc + x fcT2x
By combining G.10) and G.11) and using the fact that x is a positive integer,
we complete the proof. ?
Example 13. // Re x > f, then
з(*-1)(*-2) ,
p + З + 5
x + 1 (x + l)(x + 2)
Proof. We shall apply Entry 31 below with n = 1, у = -1, z = и = -|, and
x replaced by x — 1. Accordingly, we find that
i,i|,f, 1-х, l
1
; J~
Г(х)
= x{l+'
1,1,-x, П
i,i J
= xDx - 3). ?
Example 14

24 10. Hypergeometric Series, I
Proof. Let n = -x = —y = j in Corollary 5 to obtain
1ЛЛЛ
i,i,i'; J = r(f)r(i)'
which is equivalent to the proposed formula. ?
Examples 14 and 15 were communicated by Ramanujan in his first letter
to Hardy [16, pp. xxvi, xxv, respectively]. Hardy [2], [7, pp. 517, 518] has
observed the simple proofs that we offer here. Evidently, Example 14 was first
established in 1859 by Bauer [1]. Examples 14 and 15 may also be found in
Bailey's tract [4, p. 96] and Hardy's book [9, p. 7].
Example 15
Proof. Set x = у = z = ~n = —$ in Entry 5, and the proposed equality
follows forthwith. ?
Example 16
Proof. In Dixon's theorem, Entry 7, let x =¦ —\,y— — i, and n = \. ?
Example 17
Proof. In Dixon's theorem, Entry 7, put n = \ and x = у = — \. ?
Example 18
2-4
Proof. Set — x = — у = n = \ in Dixon's theorem, Entry 7. ?
Example 19
2-4
Proof. In Rummer's theorem, Corollary 13, set n = -x = \. ?

10. Hypergeometric Series, I 25
Example 20. If Ren < f, then
/n\3 (n(n + 1)Y 6 sm(+7)
1 + tt + —^ +•••=-
V.J \ 21 J я2в2A+2со8(яп))Г(|и+1)"
Proof. In Dixon's theorem, Entry 7, set x = у = — n. After several applica-
applications of the reflection principle and some simplification, we deduce the desired
formula. ?
Example 20 is due to Morley [1] in 1902. See Bailey's tract [4, p. 13] for
further references.
Entry 8. If Re(x + у + n + 1) > 0, then
п ^п Г(п + 1)Г(х + у + п + 1)
Л(-х, -у, п + 1) = г(х + п + 1)г(у + п + 1)- (8-1)
As mentioned earlier, Entry 8 is Gauss's theorem [1]. Following Entry 8,
Ramanujan indicates, in one sentence, how he deduced Entry 8. This is the
only clue to the methods used by Ramanujan in his derivations of the several
theorems in Chapter 10.
Assume that n and x are integers with n > 0 and n + x > 0. Expanding
A + u)y+n and A + l/u)x in their formal binomial series and taking their
product, we find that, if an is the coefficient of it",
= У (У + "Y*^ - Г<У + и + 1) у (x)t(-A
" % \k + n) \k) Г(и + 1)Г(у + 1) % (n + Щ1) " l " '
On the other hand, expanding A + u)x+y+n in its binomial series and dividing
by ux, we find that
(x + у + n\ T(x + y + n + 1)
1)" ( ' -1
Comparing (8.2) and (8.3), we deduce (8.1).
Entry 9. // Re(a ~P)>0, then
Proof. In Gauss's theorem, Entry 8, put ft = — x and a = n + 1. Take the
logarithmic derivative of both sides of (8.1) with respect to у and set у = 0.
Using @.3), we complete the proof. ?
Entry 10. // Re x > -1, then
f (~х)к Г(и)Г(х
A0.1)

26 10. Hypergeometric Series, I
Proof. In Gauss's theorem, Entry 8, let у = — n. ?
Example 1. If Ren > — 1, then
» (k - 1)! » 1
Proof. Differentiate both sides of (9.1) with respect to fi and set fi = 0 and
a = n + 1. Using @.3), we complete the proof. ?
Example 2. // Re n < 1, then
1 n n(n + 1) n
n (n + 1I! (n + 2J! sin(Ttn)'
Proof. Set x = -n in Entry 10. ?
Example 3. // n is arbitrary, then
Proof. Let x = — \ and replace n by n + 1 in Entry 10. ?
Example 4. // Re n > — 1, then
n n(n - 1) _ _ y/nT(n + 1)
~37ТУ+ 5-2! 2Г(п+|) "
Proof. In Entry 10, replace x by n and n by j. ?
Example 5. // Re x > — 1, then
» (-x)t = Г(п)Г(х + 1) » / 1 1
Proof. Differentiate both sides of A0.1) with respect to n and use @.2). ?
Example 6. // n is arbitrary, then
(„ + 1J (n + 2J\2
И"
^Г(п + 1) » f 1 1
Г(и + |)
Proof. Let x = — \ and replace n by n + 1 in Example 5. ?

10. Hypergeometric Series, I 27
Example 7. // Re n < 1, then
1 n n(n + 1) _ я », z' 1 1
n2 (n + 1J1! (n + 2J2! sin(Trn) »ti \fc + n - 1 ky
Proof. Let x = — n in Example 5. ?
Entry 11. Let n > 0 and suppose that Re(a - j? - 1) > 0. Гйеи
X {(a + kT - {p + 1 + /c)«} ^tv1 = «"•
((a + l)J
Proof. Observe that, for each positive integer m,
Thus, it suffices to show that
Since Re(j5 + 1) < Re a, the statement above is true by Stirling's formula. ?
Corollary 1. // Re(a - /9 - 1) > 0, then
f (/0* /?
Proof. If n = 1, Entry 11 yields
Multiplying both sides by J?/{a(a - /5 — 1)}, we obtain the desired formula.
Alternatively, in Entry 8, set n + 1 = «, x = — 1, and у = -/?, and the
formula of Corollary 1 readily follows. ?
Corollary 2. // Re(a - /9 - 1) > 0, then
Proof. Apply Entry 11 with n = 2. Since
(a + kf - (/? + 1 + kJ = (a - p - l)(a + /? + 2k + 1),
we find that

28 10. Hypergeometric Series, I
(a.-P- 1) X {<x + P + 2k+ 1)-
k=0 l« + l)k
Multiplying both sides by /?2/{a2(a — /? — 1)}, we complete the proof. ?
An alternate proof can be obtained by letting x = у = — p and n = a +
P — 1 in Corollary 3 of Section 7.
Entry 12(a). Suppose that f(x) = ?"=1 (Akxk/k) in some neighborhood of the
origin. Define Pk, 0 < к < оо, by
e/<*> = ? ptX*. A2.1)
ГЛеи Po = 1 a"^> /or и > 1,
«ft = Z ^*ft-*-
k=l
Proof. It is clear that Po = 1. Differentiating both sides of A2.1) with respect
to x, we find that
00 GO
X j ? * X ft
j=0 k=l n=l
Equating coefficients of x" on both sides, we deduce the required recursion
formula. ?
Entry 12(b) is an instance of the inclusion-exclusion principle, but
Ramanujan cleverly deduces Entry 12(b) from Entry 12(a). According to
Macmahon [1, p. 6], Entry 12(b) is due to Newton.
Entry 12(b). For positive integers n and r, define
Sr = Sr(n) = t a{
and
0>r = 0>r{n)= ? aklak2---akr, r < n,
i<k,<n
кх<кг<--- <kr
where alta2, ...,an are arbitrary nonzero complex numbers. Then, if r > 1,
A2.2)
where 0>o= I.
Proof. In Entry 12(a), let

10. Hypergeometric Series, I 29
Let я = m
axi<t<j<
exp(
jk\. Then
'Д AjXJ
Vj=i J .
if|x|< I/a,
/ expU=i
= exp( I
n
- (-1У+1(акхУ
j
\
Log(l + akx)
/
Hence, in the notation of Entry 12(a), Pr = ^r, and A2.2) follows immediately
from the conclusion of Entry 12(a). ?
In preparation for Entry 13, we need to make two definitions and prove
one lemma. For each positive integer r, define
Let <p{0) = 1, and define <p(n, x, r) = <p(r),r > 1, recursively by
Mr) = X Sk<p(r - к). A3.2)
t=i
Lemma. If r is a positive integer, then
^W-i^^-4. A3.3)
an k=i
Proof. We proceed by induction on r. If r = 1, equality A3.3) implies that
d d
<pW j Si = -s2,
an an
which is easily verified from the definition A3.1).
Now assume that
t к), l<j<r- 1. A3.4)
Hence, by A3.2), A3.1), and A3.4),
= -(- t kSk+Mr -k)-?sk "f SJ+1<p(r ~ к -j)
r\ *i ti i

30 10. Hypergeometric Series, I
( ? kSk+Mr -k)+t S]+l I Sk<p(r - к - j)
У \k=l j=l «c=l
t kSk+1<p(r -k)+t W - ЛФ - j)
k=i j=i
which completes the proof. ?
Entry 13. // Re x > — 1 and r is any positive integer, then
ttt, (n + k)r+ik\ = Г(и + x + 1) ^ (
Proof. Now by Example 5 in Section 10,
» (-*)» Г(п)Г(х+1) Г(п)Г(х + 1)
kU) (и + kJk\ Г(п + x + 1) l T{n
Thus, A3.5) is valid for r = 1.
Proceeding by induction, we assume that A3.5) holds for any fixed positive
integer r and show that A3.5) is true with r replaced by r + 1. Differentiating
both sides of A3.5) with respect to n and using the foregoing lemma, we find
that
-fr + 1> jo(n|^!
Г(п)Г(х +
Г(и + x + 1)
Г(и)Г(х + 1)
Г(и + x + 1)
Г(п)Г(х +
Г(и + х +
Г(п)Г(х +
(г + \)ф + 1),
Г(и + х+
from which A3.5), with r replaced by r + 1, follows. ?
Corollary 1. Let Sr(n, x) and <p(n, x, r) be defined by A3.1) and A3.2), respec-
respectively. lfn = \andx=-\, then Sx = 2 Log 2, Sr = Br - 2)?(r), r > 2, and

10. H ypergeometric Series, I 31
Proof. The proposed formulas for Sr, r > 1, are easily determined from A3.1)
after brief calculations. Setting и = \ and x = — \ in Entry 13 yields
1 1 /1\ 1 /l-3
Ы + h
(!ГЧ2-4,
from which A3.6) trivially follows. ?
In the notebooks (p. 124), Ramanujan redefines Sr for Corollary 1. We
emphasize that his formulation of Corollary 1 is correct, however. Likewise, in
Corollary 2, Ramanujan has redefined Sr in the notebooks. In fact, Ramanujan
has proved Corollary 1 in his second published paper [1], [16, pp. 15-17] by
another method. Entry 13 and Example 1 below are also given in [1].
Corollary 2. Let Sr and <p(r) be defined by A3.1) and A3.2), respectively, with
n = \andx= -%. Then Sv = 2 - 2 Log 2, Sr = B - 2r)?(r) + 2r,r> 2, and
1
Proof. Let и = 1 andx = —j in Entry 13. The proof is completely analogous
to that of Corollary 1. ?
Exam pie 1
. n3
Proof. Letting S denote the infinite series above, we find from Corollary 1
and A3.2) that
D
lot J J
Example 2
в cot в Log(sin в) dd = Log2 2 .
,o 4 ° 48
Proof. Letting и = sin в and integrating by parts, we first find that
f*/2 1 f1 Log2«
0 cot 0 Log(sin 0) rf0 = -- - ё du. A3.7)
Jo 2 Jo y/l — u2

32 10. Hypergeometric Series, I
Next, for each nonnegative integer k, an elementary calculation shows that
1 A3.8)
A3.9)
(fc + 1K
Lastly, recall that
2-4
Now substitute A3.9) into A3.7) and integrate termwise with the help of
A3.8) to obtain
f1 1 /1\ 1 /1 -3\
-I в cot в Log(sin в) йв = 1 + --у (-J + ^з (—J + ¦ • •.
Using Example 1, we complete the proof. ?
In preparation for Entry 14, define
oo
Cm,n(X) — Z_,
ft-l
k% (k + x)m jf0 (j + xf
and
where m and и are positive integers with m > 2.
Entry 14. Let и fee аи integer with и > 2. 77ien
li = l (K + X) j=i J r=i
x 1 к 1
-2 У — У-
Л (к + X)" /f0 j + X
Proof. Consider the decomposition from Nielsen's book [1, p. 48]
1 " l{
lh _1_ Y-WI- i\ '—' IV Л- v\n~ri i' J_
\ly ~~r~ Л/ v.«v — J) r=0 V^ ' X) \J i -
Summing on j, 0 < j < к — 1, we find that
1 It 1 n-l 1 *-l 1 k-l
1 у _ = _ у у ^_i L. у
Next, sum on fc, 1 < к < oo, to obtain

10. Hypergeometric Series, I 33
oo 1 It 1 n-2
Z ~tu , v\n Z 7 = "" Z Cn-r,r+i - QiW
x *-l I ( I 1 \
+ Z Zttt^t JT~/-rr^ • A4Л)
fc=l j=0 (J + X) \li — J К + XJ
Observe that
Sm{x)Sn(x) = Sm+n(x) + CmJx) + Си,т(х), т,п>2.
Thus, A4.1) may be written in the form
2 Z ТГГТп Z - = (« - 2)Sn+1(x) - "f 5r+1(x)Sn_r(x)
t=i (к + x) j=i 7 r=i
CO 1 * 1
- 2 Z jj—j; Z :—- + 2Sn+1W
+ 2? У-i
^Д(
This completes the proof. ?
Ramanujan's formulation of Entry 14 (p. 124) is somewhat imprecise. For
several other results of this type, see Chapter 9 and the relevant references
mentioned in Part I [9].
Entry 15. // a. and ft are arbitrary complex numbers, then
"-1 Г(« + к + 1)Г(р + к + 1)
Z —г, , й , , , Os,, Log n - ф(а. + 1) - ф($ + 1) - у,
t=o Г(а + р + к 4- 2)к!
as n tenrfs to oo.
Proof. From a theorem in Luke's book [1, p. 110, Eq. C5)],
~4 ШЬ)к
Г(а + Ъ) к% (а + Ъ)кк\
as n tends to oo. Putting а = а + 1 and b = р + 1, we deduce Entry 15. ?
Corollary. Let 0 < x < 1. Then as x tends to 0,
^2^1 (h i; 1; 1 - x) ~ Log x + 4 Log 2.
Proof. From a general theorem in Luke's text [1, p. 87, Eq. A1)],
2Fx{a, b;a + b;l-x)~ -^.^(Log x + tfr(e) + ф(Ъ) + 2у), A5.1)
Г(а)Г(й)
as x tends to 0, 0 < x < 1. The corollary now follows by putting a = b = \
and using the fact that if/{\) = -y - 2 Log 2 (Luke [1, p. 13]). ?

34 10. Hypergeometric Series, I
It follows from Entry 15 that
"-1 Г(а + к +
Г(а + р + к + 2)к\
Log и,
as n tends to со. This weaker result is due to Hill [1], [2]. See also Copson's
book [2, p. 266]. According to Copson [2, p. 267], Gauss showed that
г 2Fda, b; a + b; x) _Г(а + Ь)
Г(а)Г(ЬУ
which is a consequence of A5.1). See also Whittaker and Watson's text [1,
p. 299].
Entry 16. // Ao, A1,..., Arare any complex numbers and
pr= ? л(-1)'(Т)' r^°>
then
r t r\
r>0.
A proof of this well-known inversion formula can be found in Riordan's
book [1, pp. 43, 44].
Entry 17. Suppose that
is analytic for \x\ > R. For \x\ > sup(K, \h\), write
Then
Bk ~ fa Ajh \j
Proof. For |x| > R,\h\,
Now equate coefficients of x ' " in A7.1) and A7.3) to deduce that

10. Hypergeometric Series, I 35
n\ k% k\ \n~
After a straightforward calculation, the foregoing equality yields, for h ф 0,
= ? Bk(-h)k("), n>0.
k=o W
ft"
Applying the inversion formula of Entry 16 and simplifying, we conclude that
BJT"= ? Akh-k("X n>0,
t=o \k/
which implies the desired conclusion. Q
Entry 18(i). Suppose that A7.1) TioMs. Assume also that
/or |jc| > sup(R, 1). Furthermore, assume that ^л=о(^кхк/к\) is analytic for
\x\ <R*. Then for \x\ < R*,
k=o
SO
=
^f. Apply Entry 17 with h = -1. Comparing A7.2) and A8.1), we find
'that
fc>0. A8.3)
On the other hand, for |x| < R*, by the Cauchy multiplication of power series,
e^A^/L=l^, A8.4)
where:
Q=?(-l)M/fcY fc>0. A8.5)
i=o V/
By A8.3) and A8.5), Ak = Ck,k> 0. Thus, A8.4) becomes the equality that we
sought to prove. ?
In Entry 18(ii), Ramanujan claims that if A7.1) amd A8.1) hold, then
1 f (r)kAkfr(x)-<p(-x)\k
<p'(x)k% k\ I <p(x)
is always an even function of x. This is clearly false. For example, letting
<p(x) = x and r = 1 provides a counterexample.

36 10. Hypergeometric Series, I
Entry 18(iii). Suppose that A7.1) and A8.1) hold. Then, if n is an even integer,
n>2 (n\
\nK-i = Z ( )B2к- 1)В2^п-2ь «>2, A8.6)
where B, denotes the jth Bernoulli number.
Proof. From the generating function (Abramowitz and Stegun [1, p. 804]),
ir^r = Z -^t-. W < 2n>
we find that, for |x| < n,
x x 2x " ?t(l - 2*)x*
x _ i p2x _ 1 Zj Ь|
A8.7)
We now use the representation for ex given by A8.2) on the left side of
A8.7). After some manipulation and simplification, we deduce that, for |x| <
minGt, R*),
» Aj(-xY = » Bk(\ - 2k)xk » Л^'
¦= /! = fc! •= B/)!
If we equate coefficients of x", with и even, on both sides above, we readily
deduce A8.6). ?
Entry 19. Suppose that \x\, \x - 1| > 1. Then
x~r 2F1{r, m; n; 1/x) = (x - l)"r 2F1(r, n-m;n; - l/(x - 1)).
This transformation is well known (Bailey [4, p. 10]) and is generally
attributed to Gauss or Kummer. However, Askey [1] has indicated that it
was originally discovered by Pfaff [1]. We shall give what was evidently
Ramanujan's argument.
Proof. Apply Entry 17 with Ak = {m)k/(n)k and h = 1. We then see that it
suffices to show that
fc>0. A9.1)
But this is simply Vandermonde's theorem (Bailey [4, p. 3]), which is a special
case of Gauss's theorem, Entry 8. ?
Entry 20. Let
00 U3(r)d)
be analytic for \x — 1| < R, where R > 1. Suppose that m and n are complex

10. Hypergeometric Series, I 37
parameters such that the order of summation in
" (m)k « yW(i)(-iy(-r)t
h(n)kk\h r!
maI be inverted. Then
"' "№)/m °° ' '*'" rai ~№>—• B0.2)
k% (n)kk\ h (n)kk\
Proof. Using B0.1) to calculate q>m@), 0 < к < oo, and inverting the order
of summation by hypothesis, we find that
f
r!
r=o г!(и)г
by Vandermonde's theorem A9.1). D
Note that if q>{x) = (x — l)r, where r is a nonnegative integer, then B0.2)
yields
V \т>ь\~г'к _ (n ~ m)r
k% \n\k\ {n\ ¦
Hence, in this case, B0.2) reduces to Vandermonde's theorem.
Entry 21. For any complex numbers m, n, and x,
x^(-l)k(n-m)kxk « (m\x'
Proof. Now,
x »
e k%
The coefficient of xr on the right side is
' (-lftn-m)k= (m)r
&о (n)kkl(r - k)! (n)rr\'
by Vandermonde's theorem A9.1). This completes the proof. ?
Entry 21 is due to Kummer [1]. An alternate proof can be obtained from
Entry 19 by replacing x by r/x and letting r tend to со.

38 10. Hypergeometric Series, I
Entry 22. Suppose that |x|, |x + 1| > 1. Then
(x + I)"' 2F1{r, m; 2m; l/(x + 1)) = x~r 2F1(r, m; 2m; - 1/x).
Proof. Set n = 2m and replace x by x + 1 in Entry 19. ?
Entry 23. Let m and x be any complex numbers. Then
x » (-mm)kxk = ^ (m)rxr
6 ttb Bm)kk\ t%{2m\r\'
Proof. Set и = 2m in Entry 21. ?
Corollary 1. If x is any complex number, then
ел 1 - - —
Proof. Let m = \ in Entry 23. ?
Corollary 2. // |x| < 1 and Re x < \, then
Proof. Id Entry 22, let r = m = \ and replace x by - 1/x. ?
The function 2Fi(h h U x) is a constant multiple of the complete elliptic
integral of the first kind and is central to the theory of elliptic functions. See
Part III of our account [11] of Ramanujan's notebooks.
T. Matala-Aho and K. Vaananen [1] have studied the arithmetic properties
°f 2^1B> h 1; 0) when в is algebraic.
Entry 24. Let |x|, |x — 11 > 1 and suppose that m is arbitrary and that Re n > 0.
Then
V ( = f (-l)*(n+l-Bi)t
k% (n + k)k\xn+k k%
/c)/c!x"+t k% (n + k)k\(x - \)n+k'
Proof. In Entry 19, replace и by n + 1 and set r = и + 1 to obtain
у (m)k _ у (-#+ 1 -m)k
k = @ Ix.X fc = 0 /v:^X — 1.)
Integrate both sides over [x, со) to achieve the desired result. О
Entry 25. Let |x|, |x - 1| > 1 and suppose that n is arbitrary. Then
Й k\ = » (-1)"

10. Hypergeometric Series, I 39
Proof. Put r = m = 1 and replace n by n + 1 in Entry 19, and multiply both
sides by 1/n. ?
Entry 26. lf\x\ < 1 and a., j8, and у are arbitrary, then
A - xf+" 2f,(a, P; y; x) = A - x)" 2F,(y - а, у - 0; >S *)•
Entry 26 is elementary and well known; see Bailey's tract [4, p. 2].
Entry 27. //Re(n + 1) > -Re(x + y), -Re(p + q), then
Г(х + y + n+ 1)
Ч х'+п+1,уУ+п+1 J
Г(р + n + 1)Г(<? + и+1K
F-,
-х, -у,
Entry 27 is a famous theorem of Thomae [1] and can be derived from Entry
26. Hardy [1, p. 499], [7, p. 512] has extensively discussed Entry 27 and has
given references to other proofs. In Bailey's book [4, p. 14], Entry 27 is
equivalent to formula A).
Entry 28. IfRe(n + 1) > -Re(x + y), -Re(p - 1), then
n)T(x + у + n + 1)
Г-х, -у, р + nl (p + п)Г(п
21_ n, p + n + 1 J ~ Г(х + n -
+ l)F{y + n + 1)
n + l, у + n + :
Proof. Set q = — 1 in Entry 27. ?
Entry 29(a). // Re и > -1, then
Vj, \, n + П 4(n + 1) Г — n, 1, 1~|
3-^2 . 2 = ^T 3 2 3. 1
Proof. In Entry 28, put x = у = — \, n = 1, and p = n. ?
Entry 29(b). // n is a nonnegative integer, then
l Tjjrc+n r2(n + i) » (jJ
n+13 2L l.n + 2 J Г2(я+|ьй(/с!J- ( ' ^
This extremely interesting result was communicated in Ramanujan's [16,
p. 351] first letter to Hardy and was first established in print by Watson [4]
in 1929. A flurry of papers was written on this form ula and certain generaliza-
generalizations in the years 1929-1931. References may be found in Bailey's book [4,
pp. 92-95]. Related results are given in Entry 32 and Section 35 below. A

40 10. Hypergeometric Series, I
more recent proof of B9.1) has been given by Dutka [1]. Further identities
for partial sums of hypergeometric series have been established by Lamm and
Szabo [1], [2] in their work on Coulomb approximations. The finite sum on
the right side of B9.1) arises in the theory of functions of one complex variable
and is called Landau's constant. For details of this connection, see Watson's
paper [5].
Entry 29 (с). // и is any complex number, then
2 .ribn+f
Proof. In Entry 27, let p = — и — 1, q = —\, x = — n — §, and у = —j, and
replace и by и + f. ?
Entry 29(d). // Re и > -f, then
1,! J-
Proof. In Entry 27, put p = —j, q = —l,x = n,y= —j, and и = 1. ?
Corollary 1. // G denotes Catalan's constant, that is,
then
f3^(ii,i;i,l) = G. B9.3)
Proof. Putting и = — ^ in Entry 29(a), we find that
On the other hand, from Example (i) in Section 32 of Chapter 9 (see Part I [9]),
3F2& 1,1; f,|)== 2G.
Combining these two equalities, we deduce B9.3). ?
Corollary 2. As n tends to oo,
n 3^2B* 2> и; 1, n + 1) ~ Log и + 4 Log 2 + y.
Watson [5] has established an asymptotic formula for the finite sum on
the right side of B9.1) as n tends to 00. Thus, Corollary 2 follows from Entry
29(b), Watson's theorem, and Stirling's formula. We shall not relate any more
details, because Entry 35(i) below gives a very closely related, fuller asymptotic
expansion. R. J. Evans [1, Theorem 21] has generalized Corollary 2 by

10. Hypergeometric Series, I 41
showing that
Г(а)Г(Ь)
Г(й + b) 3'z la + b,,
as real с tends to oo.
Entry 30. // Re и > - Re x, - Re y, then
Г-х, 1, у + n~| _ y + n Г-у, 1, x + n~|
^ln,y + n + l\-x + n3*2ln,x + n+l\
Proof. Let у = — 1 and p = у in Entry 28. ?
Entry 31. // Re(x + у + n + 1) > 0 and ReBx + 2y + 2z + 2и + Зи + 4) >
0, then
Г i«+ 1, и, -х, -у, -z, -и
6 5 [К х + и+1, y + n + l,z + n+l, и + и + 1 '
и + 1) Г-х, -у,.
~ Г(и + 1)Г(х + у + п + 1) 3 2|_ z + w+1, и + и+1
Entry 31 is an immediate consequence of Whipple's transformation F.1).
See Bailey's tract [4, p. 28] for details.
It is interesting to note that although Ramanujan did not discover Whipple's
transformation, he did find this important special case approximately 20 years
before Whipple's proof [1] in 1926. An enlightening discussion of Whipple's
theorem can be found in Askey's paper [3].
Suppose that we set —и = x = y = z = u= —j in Entry 31. Then
:nfj' C1Л)
by Example 18 in Section 7. This result may be found in Ramanujan's [16,
p. xxviii] first letter to Hardy as well as in Hardy's book [9, p. 7, Eq. A.4)].
Equality C1.1) was established by Watson [6], who gave the same proof that
we have given. Another proof was given by Hardy [2], [7, pp. 517, 518].
Entry 32. // x + у + z = 0 and x is a positive integer, then
\n, -x, -Л Г(п + 1)Г(х + 1) * Щу + z\
Л L n + 1, z J = Г(п+х+1) к (z)kk\ ¦ C2Л)
Proof. Consider the following result

42 10. Hypergeometric Series, I
"-i (a)k(b)k T(a + n)T(b + n)
Va,b,f+n-ll
llf,a + b + n ]'
*2Г,.Л. ' C2)
due to Bailey [2], [4, p. 93]. Set a = n, b = у + z, f = z, and n = x + 1 in
C2.2) and use the fact that x + y + z = 0to complete the proof. ?
In fact, Entry 29(b) is not a special case of Entry 32. However, C2.2) does
generalize Entry 29(b). The hypothesis x + у + z = 0 is not mentioned by
Ramanujan. If x + у + z ф 0, C2.1) is false in general. For example, if x = 2
and у = z = —I, then C2.1) is erroneous, as can be seen by a comparison with
the correct formula C2.2) with the proper parameters. For Entry 33 below,
Ramanujan does provide the hypothesis x + у + z = 0.
Entry 33. // x + у + z — 0 and x + у + nis a positive integer, then
Ги, -x, - Л = Г(и + 1)Г(х + у + и + 1) х+у+« (-х)к(-у)к
3 2 [_ и + 1, z J Г(х + и + 1)Г(у + и + 1) ?„ (z)»*!
Proof. In C2.2), set а = —x,b= —у, and f = z, and replace и by x +
n+ 1.
у +
?
Entry 34. // x and у are arbitrary, then
Entry 34 is due to Gauss [1]. In Bailey's text [4, p. 11], Entry 34 is Eq. B).
The following result is due to Kummer [1] and can be found in Bailey's
monograph [4, p. 11, Eq. C)].
Corollary. // x and n are arbitrary, then
F(x iY i , iY.i,,.i.M
ГA{и _ х + 2})гA{„ + х + 2})'
We refrain from explicitly stating Examples 1 and 2 which are merely the
special cases x = 0 and x = \, respectively, of the previous corollary.
In Entry 35(i), Ramanujan defines
n-l fl'|
and then states an asymptotic formula for <p({n + l}/4) as и tends to oo. More
properly, <p(n) should be defined by B9.1). Thus, for all complex n, define
<p{n)= 3F2\ . C5.1)
Г(и)Г(и + 1) |_1, n + 1J

10. Ffypergeometric Series, I 43
Entry 35 (i) is thus an extension of Corollary 2 in Section 29. Watson [5]
and Dutka [1] have each derived asymptotic expansions for <p(n). How-
However, the expansions of Watson, Dutka, and Ramanujan are all of different
forms. We shall employ Dutka's asymptotic series to establish Ramanujan's
formulation.
Entry 35(i). Let (p(n) be defined by C5.1). Then as n tends to oo,
(n + l\ (n + l\ 3 99 999
Proof. According to Dutka [1], as и tends to oo,
na> I I ~ \l/1 I + 4 Log 2+7
V 4 / \ 4 /
where
1-3
3(n/2 + f)
From Legendre's duplication formula, it is easy to show that
2 Log 2-
(
Thus, as n tends to 00,
4 /
- Log 2 - Un.
Using Stirling's formula for Log i//(x) (Luke [1, p. 33]),
where x tends to 00 and Bn, 0 < и < oo, denotes the nth Bernoulli number,
we find that, as n tends to 00,
n + \\ ,/n+l\ „т л
^ + 3 Log 2 + 7 +
Recalling that Un is defined by C5.2), we now express the terms of l/(n + 1) -

44 10. Hypergeometric Series, I
Un in terms of quotients of gamma functions. Thus, as и tends to oo,
fn + 1\ . fn + Г
и Zn^
2kBk + l)Bk + 2) 2kBk + l)Bk + 2)Bk + 3)
бй3 + 24Й*
F(in+i) 12-32 V 4 / 12-32-52
242!2 г/п+ 11\ 263!3 /и + 15
t2.32.52.72.92 ry 4
284!4 ^fn+l9\ ' 21O5!5 (n + 23
12-32-52-72-92-112 4 4,
2126!6 /и + 27ч ' l j
For each quotient of gamma functions displayed above, we use a general
asymptotic formula for Г(х + а)/Г(х + b) due to Tricomi and Erdelyi [1] and
reproduced in Luke's book [1, p. 33]. Omitting the numerical calculations,
we find that, as и tends to 00,
2 ) 22f. 4 13 40 12Г, _. _„. ,_
C5.4)
2
n + 3'
' =^j1_12 + Z|_5_^_,llliUO(n-7), C5.5)
4 / 26 f 21 310 3990] .
15\ 3 ( 2 3 j
n + 15\ и3 ( и n2 n3 j
C5-6

10. Hypergeometric Series, I 45
(n + 19\ n4 { n n2
,3,7,
10
and
fn
Г
Substituting C5.4)-C5.9) into C5.3), we now calculate the coefficients of и к,
2 < к < 6. After some lengthy calculations, we find that all the coefficients
agree with what Ramanujan has claimed in Entry 35(i). ?
Entry 35(ii). Let <p(n) be defined by C5.1). Then for each nonnegative integer n,
— cp{n + %) = "X ^- + 2G, C5.10)
where G is defined by B9.2).
Entry 35(iii). Let (p(n) be defined by C5.1). Then for each nonnegative
integer n,
Entry 35(iv). // <p(n) is defined by C5.1), then
/14 8 /14 1
We shall first prove Entry 35(iv) and then prove Entries 35(ii) and 35(iii)
by induction.
Proof of Entry 35(iv). By C5.1) and Corollary 1 in Section 29,

46 10. Hypergeometric Series, I
By C5.1) and Dixon's theorem, Entry 7, with и = \, x = -\, and у = — i,
we find that
*LF2(hhhhl) = h a
Proof of Entry 35(ii). We proceed by induction on n. For и = 0, formula
C5.10) is valid by Entry 35(iv). Assume now that C5.10) holds for any fixed
nonnegative integer n. Thus, it remains to prove C5.10) with n replaced by
и + 1.
We first establish the recursion formula
rtn + l) = rtn) + nr*li++iy C5Л2)
where и is any complex number, or, by C5.1),
In the course of proving Entry 29(b), Darling [1, p. 9, line 1] proved precisely
the formula C5.13).
Hence, by C5.12) and C5.10),
л2 з _ л2 лГ2(п + 1)
which completes the proof. ?
Proof of Entry 35(iii). We induct on и. If и = 0, then C5.11) holds by Entry
35(iv). Assume now that C5.11) holds for any fixed nonnegative integer n, and
so it suffices to prove C5.11) with n replaced by и + 1.
By C5.12) and C5.11),
_

10. Hypergeometric Series, I 47
= 1 , 16ГФ У (ft* I l
* ?(i)?
and the desired result follows. ?
In the first notebook (p. 239), Entry 35(iv) is listed before C5.10) and C5.11).
Furthermore, prior to the latter two formulas, Ramanujan states the recursion
formula C5.12). Thus, it seems clear that Ramanujan also used induction to
establish C5.10) and C5.11).
At the beginning of Darling's paper [1], in conjunction with Entry 29(b),
he remarks, "His (Watson's) own proof is by transformation of series, and it
seems probable that Ramanujan obtained the theorem in a similar manner;
but the following two proofs by induction, which will perhaps appeal more
to the average analyst, may be of interest." It appears that Darling's specula-
speculation is incorrect, and that he, in fact, had likely found Ramanujan's proof.
Dutka [1] has found a different proof of Entry 3:5(ii).

CHAPTER 11
Hypergeometric Series, II
Much of Chapter 11 is contained in Chapters 13 and 15 of the first notebook,
while some formulas from Chapter 11 may be found scattered among the
"working pages" of the first notebook.
In Chapter 11, Ramanujan gives many results on quadratic transforma-
transformations of hypergeometric series. Several of these results can be traced back to
Kummer [1], [2]. Ramanujan also offers many theorems on products of
hypergeometric series. Although some of these results were established in the
19th century, most are originally due to Ramanujan. Entry 34(iii) is a parti-
particularly elegant formula which combines a product formula and a quadratic
transformation. Much of Bailey's work in the 1930s on products of hyper-
hypergeometric series was motivated by Ramanujan's discoveries.
Corollary 2 in Section 24 offers a certain asymptotic formula for zero-
balanced 3F2 series. Such formulas in the literature have previously been
established only for zero-balanced 2Fy series. It is interesting that this elegant
formula had been overlooked for 60 years after Ramanujan's death. We
provide here an elegant proof of this asymptotic formula by R. J. Evans and
D. Stanton [1]. However, their proof depends on knowing the formula in
advance. It would be interesting to have a more direct proof that might shed
some light on Ramanujan's approach.
There are two additional formulas in Chapter 11 which are amazing indeed.
The first is Entry 22, which involves a remarkable recursively defined sequence
An and which leads to two intriguing binomial coefficient identities B2.22) and
B2.23). The second is Entry 31 (ii), which we were only able to prove by using
the theory of second-order inhomogeneous linear differential equations and
equating coefficients in 75 power series. Unfortunately, we have no idea how
Ramanujan discovered these two extraordinary formulas (as well as most of
the results in this chapter). Our proofs of these two theorems are certainly

11. Hypergeometric Series, II 49
not those found by Ramanujan; he must have derived these formulas more
naturally. Although differential equations have traditionally played a strong
role in the theory of hypergeometric series, there is no evidence that Ramanu-
Ramanujan significantly utilized this connection. The hypergeometric differential
equation does appear in somewhat disguised form in Entry 31 (i). The formulas
in Sections 30 and 31 of Chapter 11 are the only ones in Chapters 10 and 11
with links to differential equations.
A few formulas in Chapter 11 are apparently without meaning. Entry 24
is such an example; we have not been able to find any functions for which the
proposed formula is valid.
We use the notation that was set forth in the introduction to Chapter 10.
In that chapter, we considered the case p = q + 1. Since in this chapter, we
establish theorems for p ф q + 1, we offer further remarks about convergence.
If p < q + 1, then pFq converges for all finite values of x; if p > q + 1, then pFq
converges for only x = 0 unless the series terminates. For most of the theorems
and examples in the sequel, we shall not state the region of validity because
it can readily be ascertained from the general remarks we have made about
convergence.
In the sequel, we shall frequently appeal to the treatises of Erdelyi [1] and
Bailey [4].
Entry 1. Let <p be any function. Then, provided the series converges,
<p'{x) k% Bm)kk\ { <p{x)
is an even function of x.
Proof. Consider the quadratic transformation found in Erdelyi's work [1,
p. 112, formula B6)] and due to Kummer [1, p. 78], [2, p. 114],
2Fi(r, m; 2m; z) = A - z) r/2 2FX \$r,m- \r\ m + %;
z2
4(z - 1),
Setting z = 1 — (p( — x)/<p(x), we find after some simplification that
-r-T2*i r, m; 2m; 1 —
<p(x) \ Ф)
, {ф)-<р(-х)У
jr, m — jr; m + j, — -
which clearly is an even function of x. ?
Entry 2
2F, (r, m; 2m; -^-) = A + xf 2F^r, Цг + 1); \{2т + 1); x2).

50 11. Hypergeometric Series, II
Entry 2 is a well-known quadratic transformation (see Erdelyi's book [1,
p. Ill, Eq. D)]) that is due to Kummer [1, p. 78], [2, p. 114].
Entry 3
2F1 U m; 2m; ** j = A + xJr 2Fi(r, r-m + ±;m + b x2).
Entry 3 is precisely Eq. E) of Erdelyi's treatise [1, p. Ill] and is due to
Gauss [1]. This formula is also mentioned by Hardy [1, p. 502], [7, p. 515].
Entry 4
2F, (V, i(r + 1); iBm + 1); ^^г) = tt + *)' 2F1{r, r - m + \; m + \; x).
Proof. In Entry 2, replace x by 24/x/(l + x) to find that
2Fa (r, m; 2m;
= (x + If 2Fl(r, r -m + \;m + \; x),
by Entry 3. П
Entry 5
4x
Proof. Put m = \ in Entry 3. ?
Entry 6
2F, ^r, i(r + 1); 1; ^^i) = U + х)Г 2Fx(r, r; 1; x).
Proof. Put m = \ in Entry 4. П
Entry 7
^(m; 2m; 2x) = ex 0^(т + |; x2/4). G.1)
Entry 7 is due to Kummer [1, p. 140], [2, p. 134] and was recorded by
Hardy [1, p. 502], [7, p. 515]. Entry 7 follows from Entry 2 by replacing x by
x/r there and then letting r tend to oo.

11. Hypergeometric Series, II 51
Corollary, ^(fc 1; x) = е*12 0^A; (x/4J).
Proof. In Entry 7, put m = \ and replace x by x/2. П
Entry 8. Lef <p(x) be analytic for \x — 1| < R, where R > 1. Suppose that m
and q> are such that the order of summation in
у
may be inverted. Then
tb Bm)/c! „tb
Proof. Since <p is analytic for |x - 11 < R, R > 1, we readily find that
№)@) = у 9 ЧЛ *H Hfe fc > 0
n=k П\
Hence, inverting the order of summation, by hypothesis, we find that
« д>тфJ?(т)к = » 2\m\ » (p<"»(l)(-l)"(-n),
I ! »tb h
tb Bт),/с! h n\
= « (?(n)(l)(-l)" y 2"(m),(-n),
Now multiply both sides of G.1) by e~x and then equate coefficients of x",
n > 0, on both sides to obtain the evaluation
if»is even,
n\ ?0 Bm)kk\ _ .
if и is odd.
If we substitute (8.2) into (8.1), we complete the proof. ?
Entry 9. // n is an integer, then
1
У^х" I V ' ' 2x
+ cos(n7i)e~* 2F01 n, 1 - n; ——
\ ix
Observe that
where /v is the Bessel function of imaginary argument usually so denoted (see
Watson's treatise [9, p. 77]). Thus, Entry 9 is a well-known result in the theory
of Bessel functions (ibid. [9, p. 80, formulas A0), A1)]).

52 11. Нуpergeometric Series, II
Corollary. .4s x tends to oo,
l2 12-32 12-32-52 \
Л (9-2)
Proof. Undoubtedly, Ramanujan formally deduced this formula from Entry
9 by setting n = j there.
However, by (9.1), which holds for any complex number n,
Remembering that x is positive, we observe that (9.2) is precisely the asymp-
asymptotic expansion of/0(x) given by Watson [9, p. 203]. ?
It is possible that Ramanujan did not restrict n to be an integer in Entry
9. In such a case, the right side of Entry 9 is an asymptotic expansion for the
left side as |x| tends to oo when |arg x| < § n (Watson [9, p. 203]), provided
that cos(njr) is replaced by ехр(штг).
Entry 10. If n is an integer, then
.8(±ИЯ - X) ? -
I
Proof. Replace x by ix in Entry 9 and equate real parts on both sides. After
some simplification, we achieve the desired equality. ?
Corollary. Suppose that n is an integer. Let x0 be a root of
Let /л be an odd integer chosen so that |x0 — \п{ц + n)\ is minimal. Then if x0
is "large"
п{ц + n) n{\ - n) иA - и){7иA - ri) - 6}
X°K—2~ + rt^) ^bVf +'"- (ШЛ)
Proof. By Entry 10, we want to approximate large roots of
- x) ? {2k + i)!BxJt+1 = 0. A0.2)
We shall use a method of successive approximations.

11. Hypergeometric Series, II 53
From A0.2), it is clear that we should take as a first approximation
x = ^n{fi + n).
For our second approximation, consider \п(ц + n) + у, where, by A0.2), у
should satisfy the equation
cos(ln7r — {\n(u + ri) + y}) — sm(\nn — {hn(u + ri) + v}) = 0.
After a short calculation, we find that
n(l - и)
tan у = — -.
п(ц + ri)
Hence, as our second approximation, we shall take
к(/л + ri) n{\ — ri)
x = 1 .
2 п(/л + n)
For our third approximation, consider
п(/л + n) n{\ - n)
2 + п{ц + n) + Z'
where, by A0.2), z is to satisfy the equation
. /иA -и) \ f n(n + 1)A - и) B - ri)
— Sin .-in
п(ц + n) ){ 2п\ц + nJ
n(n + 1)(и + 2)A - и)B - и)C - и) _
6п3(ц + nK
Hence,
t (n(\ -ri) \ fиA - ri) / 2иA - ri)
tan — + z « <— -I 1 =- -r
\7r(jU + ri) ) \n\\i + ri)\ n (ц + ri)
n(n + \){n + 2)A - и)B - и)C - и)] Г, n{n + 1)A - и)B - и)
иA - n) n{\ - n) ( (и + l)(n + 2)B - и)C - и)
+ < zn(i —n)
7г(уЦ + И) Л3{Ц + ИK
+ и(и+1)A-и)B-и)) (ШЗ)
Now,

54 11. Hypergeometric Series, II
(n{\ -n) \ иA - и) n3(l - nf
\( ri) ) ( ) З3( )
(n{\ n) \ иA и) n(l nf
tan i-j '- + z\x-/ L + z + —± '-,. A0.4)
Thus, from A0.3) and A0.4),
, (" + W* + 2)B - и)C - и)
\ц + иK { v ' 6
n(n + 1)A - n)B - n) n2(l - nJ>
Hence, our third-order approximation is precisely that claimed by Ramanujan
in A0.1). ?
Entry 11. If
Cx sin и n
du = r cos(x — в) (HI)
Jo м 2
and
11 - cos и , лч
аи = у + Log x — г sin(x — У), A1.2)
о "
where у denotes Euler's constant, then
k = O
rsin0~^ 2k "> A1-4)
and
r2 ~ V , A1-5)
asx tends to oo.
Proof. By A1.1),
= — — r cos x cos в — r sin x sin 0. Ш-6)
By successively integrating by parts, we easily find that

11. Нуpergeometric Series, II 55
(l)B/c)! . - (-l)k+1B/c - 1)!
du ~ cos x ? y2t+1 + sin x X 3* . (П.7)
as x tends to oo. Thus, A1.3) and A1.4) follow from A1.6) and A1.7).
We next show that A1.2) is consistent with A1.3) and A1.4). From A1.2)
and the tables of Gradshteyn and Ryzhik [1, p. 928],
1
x 1 — cos и f °° cos и
аи = у + Log x + аи
о » Jx "
= у + Log x — г sin x cos в + r cos x sin в. A1.8)
On the other hand, by successively integrating by parts,
cost* . » (-1)кBкУ. » (-l)t+1Bfe-1)!
аи ~ —sin x 2, 2T+1 1~ cos x ^L 2k '
A1.9)
as x tends to 00. Using A1.8) and A1.9), we again deduce A1.3) and A1.4).
From A1.3) and A1.4),
2 1 X" \ / \^('*l" I 1 т~" \ / -fa/v 1 I.
k=0 X ) [k=\ X
as x tends to 00. The coefficient of x 2", n > 1, above is equal to
(_ iy+i ? Bfe)!B« - 2 - 2fe)! + (-1)" X B/c - 1)!Bи -2к- 1)!
+i ? Bfe)!B« - 2 - 2fe)! + (-1)
0 k=\
= (-1)" Z (-l)fc+1/c!Bn-2-/c)!. A1.10)
Comparing A1.5) and A1.10) and replacing n by n — 1, we see that it suffices
to show that
У (-l)"fc!Bn-fc)! = l—^A n>0. A1.11)
*о n + 1
*=о
n + 1
Let Sn denote the left side of A1.11). Using C2.2) in Chapter 10 with n
replaced by 2n + 1, a — b — 1, and / = — 2n — e, where ? > 0, we find that
2" A)
*й(-2и)к
= Bи)! hm
ot=o(-2« ~?)t/c!
Г2Bп + 2)
' ГBп + 1)ГBи + 3) ?™
1 + lim
¦[-й'-^ + з]
ГBп + 3) V е-о *=2Й+1 (- 2п ~ е)кBп + 3)к

56 11. Hypergeometric Series, II
Bn + 1)! / (lJ.+i(-eJ,+i ? Bи + 2)tBn +l-
- 4Dn + 4)k
Bи + 1)! / Bи + 1)!Bи + 2)! Г2п + 1, 2п + 2
= ^2пТ~2~1 + Dи + 3)! 2Fl'
Bп + *)! Л Bп + 1)!Bн + 2)! ГDп
+
2п + 2 \ + Dп + 3)! ГBп + 3)ГBп + 2)
2и+1)! п = Bи+1)!
1 + ' п + 1 '
2п + 2
where we have employed Gauss's theorem, which is Entry 8 of Chapter 10.
This completes the proof of A1.11). ?
For results similar to Entry 11, see the author's [9] account of Chapter 4
of Ramanujan's second notebook.
Example 1. Jjf cos(;r sin2 в) d9 = 0.
Proof. Letting
sin2 в = ?A - cos 29) A1.12)
and replacing в by тг/2 — 9, we find that
f»/2 f*/2
cos(rc sin2 9) dB = sin(^n cos 26) d9
Jo Jo
f*/2
= — sin(jn cos 29) d9,
Jo
from which the desired result follows. ?
Example 2. j^2 cosBrc sin2 9) d9 = -|J/2 cos(rc sin 9) d9.
Proof. As above, the proof is quite elementary. First, use the identity A1.12)
and then replace 29 by тг/2 — д. After simplifying, we obtain the desired
equality. ?
fW2 /2Л \ f*/2 fn \
i. cos -— sin2 в Ыв = И cos I - sin 9\ dd.
Jo \3 / Jo \3 /
Example 3,
Proof. The steps are exactly the same as in the previous proof. ?
Entry 12. If x + у + z = j, then
2^1 (-x, -y, z; p) = 2Л(-2л:, -2y; z; f(l - ^1 - p)).

11. Hypergeometric Series, II 57
With obvious changes in the parameters, Entry 12 is the same as equality
A0) of Erdelyi's book [1, p. 111]. A formula equivalent to Entry 12 was given
by Hardy [1, p. 502], [7, p. 515] in his overview. Entry 12 is due to Gauss [1].
Corollary
l2 + n (l2 + n)E2 + n) , (l2 + n)E:!+n)(92 + n) з
1+* +X +X +
l + n/l,/lx\
22 V 2 /' 22-42
+ •••¦
Proof. In Entry 12, set x = (-1 + iy/n)/4, у = (-l - iy/n)/4, z = 1, and
p = x. П
Example 1
Bk-2f)Bk
xk
Proof. In the corollary above set n = 3. For к > 2, we are led to examine
(I2 + 3)E2 + 3)(92 + 3)---(Dfc - 3J + 3)
42-82---Dfc-4JDfcJ
(I2 + 3)E2 + 3)(92 + 3)¦ • • (Dk - 3J + 3)
" 22-42---Bk-2JBkJ4k
B2 + 2 + 1) D2 + 4 + 1) ({2k - 2J + Bk - 2) + 1) 1
22 42 Bk-2J BkJ
2-4---Bfc-2)lB2 + 2+ 1KD2 + 4+1)
1 ¦ 3 ¦ • • Bfc — 3) 2~ъ 41
Bk - 3)(Bfc - 2J + Bk - 2) + 1) 1
Bk - 2K BkJ
where in the middle expression above we used the equality 4(BnJ + In + 1) =
Dи + IJ + 3.
We are also led to examine, for к > 1,

58 11. Hypergeometric Series, II
(I2 + 3)C2 + 3)---(Bfc - IJ + 3)
22-42---BfcJ
_B2-2+ l)---(fc2-fc + l)
2A2 -1 + 1) 3B2 -2+1) (k + l)(k2 - fc + 1) 1
I3 23 к3 /c+1
1
it + iV т1зд 23; v *V'
where in the second expression above we used the equality 4(n2 — n + 1) =
{In - IJ + 3.
Using A2.1) and A2.2) in the previous corollary, we obtain the desired
result. ?
Example 2. // a + ft = 1, then
Example 2 is well-known (e.g., see Erdelyi's compendium [1, p. 112, formula
B2)]).
Entry 13. Ifa + P + y = 0, then
2F?(-a, -p;y+bx)= 3F2(-2a, -20, у; у + i 2y; x).
Entry 13 is a famous result of Clausen [1]. Other results on products of
hypergeometric series are given in the sequel. See also Bailey's tract [4,
Chapter 10]. Entry 13 was mentioned by Hardy [1, p. 503], [7, p. 516].
Corollary 1
42-82
Proof. Put a = (-1 + iy/n)/A, $ = (-1 - 1\А)/4' and 7 = 2 m Entrv 13-
П
Corollary 2. // Jo denotes the ordinary Bessel function of order 0, then for all x,
Jo (*>/*) = 1^2A; 1, i; x).

11. Hypergeometric Series, II 59
Proof. In Watson's text [9, formula F), p. 147], set v = 0 and z = i^/x to find
that
Since Bк)\/(к\22к) = (|)fe, the desired result follows. ?
According to Watson [9, p. 145], Corollary 2 is originally due to Schlafli
[1].
Entry 14. // a + P + 1 = у + 5, then
2FM P; y; i(l - JT^)) 2Fx(a, j8; 8; i< 1 - JT^c))
This equality can be found in Bailey's monograph [4, p. 88, formula C)],
where x = 4z(l — z). The first published proof of Entry 14 is due to Bailey [3]
in 1935.
Entry 15. For any x,
F{y;X)F{S;x) l
A short calculation shows that
a? + ^ + , ,,,,2» _
(y + s- 1)*
Thus, Entry 15 can be written in terms of hypergeometric series,
M S), X.Y + 8-l):y,8,y + 8-U 4x).
In fact, Entry 15 gives a formula for Jy_lBi^G,)Ji.lBi^/x), where Jv
denotes the ordinary Bessel function of order v. This result is due to Schlafli
[1] and thus represents a generalization of Corollary 2 in the previous section.
Entry 15 is also given by Watson [9, p. 147, formula E)], Hardy [1, p. 503],
[7, p. 516], and Erdelyi [1, p. 185, formula B)]. Bailey [1] has also established
Entry 15 as well as generalizations.
Entry 16. // x is arbitrary, then
0F2{m + 1, и + 1; x) 0F2(m + 1, n + 1; -x)
_» (-l)*(m + n + 2fc+l)tx2*
k% (m + Щи + l)k(m + \)u{n + \Jkk\ ¦ ¦

60
11. Нуpergeometric Series, II
A brief calculation shows that
),(±(n + 2))k26
(m + n + 2k + l)k
Thus, Entry 16 may be written in the form
0F2{m + 1, n + 1; x) 0F2{m + 1, n + 1; -x)
2]
The first published proof of Entry 16 is evidently due to Hardy [1, p. 503],
[7, p. 516] who stated Entry 16 in the latter form. See also Erdelyi's treatise
[l,p. 186, formula G)].
Entry 17
0F2(m + n + 1, n + 1; x) 0F2(m +1,1 — n; — x)
п + к + 2Щ2х)к
=
(m + n+ \)k(m + l)kkl '
( ¦ '
where, for к > 1,
4 =
(и2- 12)(и2-32)---(и2-
1
(n2-22)(n2-42)---(n2-k2)
, if к is odd,
, if к is even.
A7.2)
Proof. For r > 0, the coefficient of xr on the left side of A7.1) is equal to
k% (m + n+ \\.k{n
_t(r - k)\{m + 1),A - n\k\
{-m -n- r)k{-n - r)k(-r\
(m + n + l)r(n + l)Pr! k% (m + l)t(l - n)kk\
where we have used the elementary relation
with a = m + n + 1, n + I, and 1.
We now apply Dixon's theorem, Entry 7 of Chapter 10, with x = m + n + r,
у = r, and n replaced by — n — r. Accordingly, we find that

11. Hypergeometric Series, II 61
ГЩ2 - n - r))T(m + 1)ГA - п)ГЩ2 + 2m + n + 3r))
(m + n + l)r(n + 1)гг!ГA - n - г)Г(±B + 2m + л + г))Г(|B - n + г))ГA + m + r)
(|Bm + и + r + 2))гГ(п + 1)ГA - и)Г(±B - и - г))
~(т + п+ l)r(m + 1)гг!ГA + п + г)ГA - и - г)Г(±B - и + г))'
A7.4)
after a considerable amount of simplification. Comparing A7.4) with A7.1),
we find that it suffices to show that
Г(и + 1)ГA - п)Щ{2 - и - г)) _
24 ^° A75)
After using the functional equation of the gamma function, we readily establish
A7.5), and therefore Entry 17 is proved. ?
Entry 18
JA-P\ r, -x) M-P; 7i x) = 2f3(-P, P + г- у, b, kb + i); x2/4).
Evidently, the first published proof of Entry 18 was given by Hardy
[1, p. 503], [7, p. 516]. (There is a misprint in Hardy's formulation; read x2/4
instead of -x2/4.) See also Erdelyi's book [1, p. 186, formula E)]. For exten-
extensions and ^-analogues of Entries 16 and 18, see the paper by Srivastava [1].
Ramanujan (p. 133) has an extra factor of (y + 4) in the denominator of the
coefficient of x4 on the right side above.
If we replace x by — x/fi in Entry 18 and let fi tend to сю, we find that
0F,{y; -x) 0Fx{y- x) = 0F3(y, b, i(y + 1); ~x2/4). A8.1)
Entry 19. If a or P is a nonnegative integer,
2F0(-a, -p;xJF0(-a, ~P;-x)
= 4Fi(-a, -p, -i(a + P), -i(a + P - 1); -a - j8; 4x2).
Entry 19 may be proved by multiplying term wise the two series on the left
side and applying Dixon's theorem. Entry 19 may be found in Erdelyi's treatise
[l,p. 186, formula D)].
Entry 20. // x is arbitrary and ak, к > 1, is defined by A7.2), then
1F1( — m;n + 1; —xIF1( — m — n; J. — n; x)
= l+f ak(U-2m-n-kM-2x)k
k=i k\

62 11. Hypergeometric Series, II
Proof. Using A7.3), we find that the coefficient of xr, r > 1, on the left side
ofB0.1)isequalto
'" h (n + l),-k(r - k)\(l - n)kk\
(-m-m)r ' {-r)k(-m-n)k{-n-r)k
(n+l)rr\ kk (m - r + Щ1 - n)kk\
Apply Dixon's theorem A7.4) with a = — n — r,b = —r,andc = — m — nand
use A7.6) to get
d = ГA - п)ГЩ2 - n - r)) (-iyr(U2 + 2m + n + r))
' (n + 1)ГГA - n - г)ГЙB - n + r)) T(iB + 2m + n - r))
(-т)гГ(т+ 1 -r)
X ?.Г(т + 1)
This completes the proof. ?
Example 1
» xM » (-xKt = 1 2 » (-3x2K"
^o Cik)! 4lo Cfc)! 3 + 3^О Fik)! "
Proof. In Entry 16, let m = — | and n = —f and replace x by (x/3K. Then
A6.1) becomes
S (-*K
eoC^*4 C*)!
_, + ? (-«№«
- l)!C/cJ• 5 • ¦ ¦ Ffc — 1I • 4¦ ¦ ¦ Ffc — 2)
= 1 -
from which the desired result follows. ?
Example 2
Zj /i,i\3 Zj П,\\Ъ ~ 2-i

11. Hypergeometric Series, II 63
Proof. Putting m — n = 0 in Entry 16 yields
0F2(l, 1; x) 0F2(l, 1; -x) = I
* = 0 (К-
The desired equality readily follows. ?
Example 2 is mentioned by Hardy in his book [9, p. 7] and is found in
Ramanujan's letters [16, p. xxvi] to Hardy.
Example 3
» x3t+1 -(-lfx3t+1 2 » (¦
k% Ck + 1)! k% Cfc + 1)! 3 »^o Ffc + 2)! '
Proof. In Entry 16, set m = ^ and и = —^ and replace x by (x/3K. We then
find that
- x3t
I
Ck + 1I k% Ck + 1)! k% Ck + 1)!4 • 7 • • • Fk + 1J ¦ 5 ¦ ¦ • F/c - 1)
l)B*)!4-7"-F*+
3 k% Fk + 2)! '
On multiplying both sides above by x2 we complete the proof. ?
oo ( — \)kBx2Jk
Example 4. cos x cosh x = ? — .
Proof. In A8.1), set у = j and replace x by x2/4. After some simplification,
the desired result follows. ?
oo (—i)kBx2Jk+l
Example 5. sin x sinh x = У .
k=o Dk + 2)!
Proof. In A8.1), let у = f and replace x by x2/4. ?
GO yK GO
Examples S_X 7^=1 fLi7}
k=o(k\J k% (k\J k^0(k\JBk)\'
Proof. Set у = lin A8.1). ?
Example 7
1* 1\2» J' л' lrlv2' 1» л/ — lr2v2> J! '> л /";•

64 11. Hypergeometric Series, II
Proof. Set ft = -\ and у = 1 in Entry 18. ?
Example 8
2k
Proof. Apply Entry 18 with /J = — 1, у = f, and x replaced by x/2 to obtain
^A; f; x/2) ^A; f; -x/2) = 2F3A, \; |, |, f; x2/16).
An elementary calculation shows that
1 _ 26kBk)\
(!)*(!)* ~ Dfe + 1)!'
The proposed equality now follows. ?
Example 9
!^A; n + 1; x) ^A; и + 1; -x) =
nx2*
t=o (и + k)(n + lJk
= 2F3A, и; и + 1, i(n + 1), i(n + 2); x2/4).
Proof. Set ft = -1 and у = n + 1 in Entry 18. ?
Example 10. If n is a nonnegative integer, then
2F0(-n, 1; x) 2F0(-n, 1; -x) = ? (-и)»(-и + 1 + fc)kx2k.
Proof. In Entry 19, let a = n and /i = —I. The proposed equality easily
follows. ?
Entry 21
x))
Entry 21 is originally due to Kummer [1, p. 82], [2, p. 118]. See also
Erdelyi's compendium [1, p. Ill, formula C)].
Entry 22. Let mbe a nonpositive integer and put
l). B2.1)

11. Hypergeometric Series, II 65
For each nonnegative integer k, let
k k(k-l) t_t k(k - \)(k - 2)Ck - 1) k_2
Ak = P з| P + <jj P
2(k - 1)!B2* - l)B2k
+ + l-3-3-Bfc-l) P'
w/iere B,, 0 <j < oo, denotes thejih Bernoulli number. Then, if \x\ < n,
B2.3)
Before commencing the proof of Entry 22, we make one comment. We have
stated Entry 22 exactly as Ramanujan gives it. Note that, by B2.2), Ak is not
well defined because there apparently is no general formula for the coefficient
of pj, 1 < j < k. However, Ak is well defined by a recursion formula given by
B2.13) below.
Proof. Replacing m by — n and x by ix in B2.3), we rewrite B2.3) in the form
f (-
flx).= e>» f tM^l _ e-2i*f =l+
ш- к (k\J A } \k 2W
We show first that /„(x) = ^„(cos x), n > 0, where Pn denotes the nth Legendre
polynomial. (This fact was first kindly pointed out to us by R. J. Evans.) By
Bailey's book [4, p. 4],
-'"* 2^(-и, i; -и + I; e 2lJC). B2.5)
Г(п +
By using A7.3), we may easily show that
Hence, from B2.5),
игр cos(" -2k)x' B2-6)
where the prime on the summation sign indicates that if к = n/2, this summand
is to be multiplied by \. From a representation for Pn(cos x) in Whittaker
and Watson's text [1, p. 303], it follows that fn(x) = Pn(cos x). Hence, it
remains to show that
00 (— l)kA x2k
Pn(cos x) = 1 + X ' * , |x| < n. B2.7)

66 11. Hypergeometric Series, II
It is well known (e.g., see Copson's text [2, p. 273]) that Pn(cos x) is a
solution of Legendre's differential equation
y" + (cot x)y' + n(n + l)y = 0. B2.8)
Since Р„A) = 1 (Whittaker and Watson [1, p. 302]) and Pn(cos x) is an even
function of x, Pn(cos x) has a power series expansion of the form
Pn(cos x) = ? a2kx2k, ao=l. B2.9)
Our procedure will be as follows. We shall actually assume that B2.7) holds;
that is, we assume that
а2к = Ц_11_ф5 fc>i; B2.10)
and then we show that Ak has the properties evinced by the formula B2.2).
Recall that
cot x = ? t-1^*-, \x\ < n. B2.11)
Substituting B2.9) and B2.11) into B2.8), we find that
X а2к2Ц2к - l)x2*
'п -y2kY2k-l
k=O (Щ'- k=l
+ n(n + 1) ? a2kx2k = 0.
k=O
Equating coefficients of x2r, r > 1, on both sides, we find that
4 | '-^-^. + „(„ + Щ2,_, . 0^ ,22.,2,
Noting, from B2.1), that p = \n(n + 1) and using B2.10), we find, after some
simplification, that the recursion relation B2.12) takes the form
where r > 1 and Ло = 1.
First, letting r = 1 in B2.13), we find that
A1 = p. B2.14)
Second, letting r = 2, using B2.14), and recalling that B2 = ?, we find that
А2 = р2-Ь. B2.15)
Third, letting r = 3, using B2.14) and B2.15), and recalling that B4 = -^,
we find that
Лз=Р3-Р2+|Р- B2.16)

11. Hypergeometric Series, II 67
Observe that the formulas for A x, A 2, and A 3 given by B2.14)-B2.16), respec-
respectively, are in complete agreement with the formula for Ak given by B2.2). In
particular, the coefficient of p in B2.2) is in corroboration with B2.14)-B2.16)
for к = 1, 2, 3.
We now proceed by induction and assume that, for к = 1, 2, ..., r, the
leading three coefficients and the last coefficient of Ak are in agreement with
those prescribed in the formula B2.2). Thus, from B2.13) and the inductive
hypothesis,
4+1 = pAr - X
*tl (r + 1 - k)\(r - k)\Bk)\
= pA, - \rAr + ?r2(r - \)Ar_,
2(r - l)!B2r -
45 v
23-1(r!JB2r
Br)! P'
The coefficient of pr+1 above is equal to 1 in agreement with B2.2). The
coefficient of pr above is equal to
r r(r - 1) _ (r + Y)r
~3 6 ~ 3! '
which also agrees with B2.2). The coefficient of pr-1 above is found to be
r2(r - 1) r(r - l)(r - 2)Cr - 1) 2r2{r - 1) (r + l)r(r - l)Cr + 2)
18 + 5j + 45^ ~ 5J '
which again is what we desire by B2.2).
Lastly, the coefficient of p above is equal to
1(r!JB2t(r - k)\T+2'\r + 1 - k)\{22'+2-2k - l)B2r+2-
2fc
(r + 1 - k)\(r - fc)!Bfc)!Br + 2 -
Recalling the Laurent expansions for cothBx) and tanh x, we have, for
|x| < я/2,

68 11. Hypergeometric Series, II
? 2*l-lBuxa-1 ' 2"B" - 1)в„х"-'
Z g I Л»
The coefficient of x2r, r > 0, on the right side is equal to
*~ 1>кг+2-2к
dr:=2'+i X
\4.Н,у.\4.Г T Z, — АН.)'.
Г ->rp2r+2 _ nR_ 4
B2.18)
Bfe)!Br + 2 - 2fe)!
cr 2rB2r+2 - 1)Д2г+
(r!J Br + 2)!
by B2.17). On the other hand, for |x| < л/2,
cothBx) tanh x = 1 — \ sech2 x
1 d
= 1 — tanh x
2 ax
„о 22fc->B2* - l)Bk - \)B2kx2k2
= 1 - У
Hence, we have also found that, for r > 1,
Equating B2.18) and B2.19) and solving for cr, we find that
r!(r + l)!B2r+2 - \)B2r+2
_
Br
Examining the coefficient of p in B2.2) when к = r + 1, we find that this
coefficient is indeed equal to cr. This completes the inductive proof. ?
Corollary. If p = 1, then
A = Ak(l) = -^-, к > 1; B2.20)
if p = 3, t/ien
Д = XtC) = 3 • 22*-2ЛA), к > 1. B2.21)
Proof. In Entry 22, let m= -1. Then by B2.1), p = 1. From B2.6),
P^cos x) = cos x. Thus, the coefficient of x2k, к > 1, in P^cos x) is equal
to (-l)*/Bfe)!. But from B2.7), the coefficient of x2k is also equal to
(- l)Mt/{B*(fc!J}, к > 1. Equating these two coefficients, we deduce B2.20).
Second, let m = -2 in Entry 22. Then p = 3. From B2.6),
P2(cos x) = | cosBx) + 3.
Thus, the coefficient of x2\ к > 1, in P2(cos x) is equal to 3(- l)k22k~2/Bk)\.
Equating this with the coefficient of x2k given by B2.7), we deduce B2.21).
D

11. Hypergeometric Series, II 69
If we expand cos(n — 2k)x, 0 < к < [и/2], in its Maclaurin series in B2.6)
and, for Pn(cos x), equate the coefficient ofx2J,j > 1, with the coefficient of x2J
in B2.7), we obtain an elegant identity involving binomial coefficients. We
shall further separate this identity into two cases. Replacing n by 2n and then
n by In + 1, we find, respectively, that
where n,j > 1 and p = nBn + 1), and that
where n>0,j> 1, and p = (n + l)Bn + 1). These identities are apparently
new and cannot be found in the tables of Gould [1] or Hansen [1], for
example.
Entry 23 is apparently meaningless. Ramanujan claims that if
<p(x) = Cj + УГ = c2 + ф. = •••== с + y/n
and "if cb c2, c3, ..., с„ appear to be similar," then they are all identically
equal to с. Не then concludes that
The intent of this entry shall perhaps always remain a mystery.
Entry 24. Let
and
Then
fc=0
<24I)
Corollary 1. Let <p(r) be defined as above and let
T(m + 1) » (m + l)k
Q' - Ц»+1-г)Г(г+-1) Jo (^Т
Tnen

70 11. Hypergeometric Series, II
X ср(т + k)(l - x)m+k = ? Q'k(-xf
k=O k=0
Log
— I Pk( Log—- .
t=o \ 1-х/
1 -x
Entry 24 and Corollary 1 are enigmatic. It seems likely that there are no
functions q> for which either of the proposed identities holds. For most choices
of (p, the series for Qr and Q'r diverge. Employing the definition of Qj, we
formally find that
?0 oooo I i Л- \\
J = 0 j=0 k=0 k]
GO П
„
= I <p(n)I M-xY
n=O j=O\J/
00
= Z <p(n)(l - x)".
Comparing the formula above with B4.1), we find that the logarithmic series
does not appear!
Corollary 2. Let я + Р + у+\=д + Е with y>—\. Then as x tends toO + ,
Г(а + 1)Г(Р
<5 + i,? + i ' J
where i//(z) = I~"(z)/r(z) and С denotes Euler's constant.
We cannot see how Corollary 2 would follow from Entry 24. Corollary 2
should be compared with the more precise formula for 2F1 in Entry 26
below. Corollary 2 is a very beautiful and significant formula, for it is the
only asymptotic formula for zero-balanced series besides that which can be
obtained from Entry 26. R. J. Evans and D. Stanton [1] have recently found
an elegant proof of Corollary 2 as well as of a q-analogue. They provide a
complete proof of the q-analogue and sketch a proof of Corollary 2. In fact,
they establish a slightly stronger version of Corollary 2. We follow Evans and
Stanton in our development below. It will be convenient to trivially alter the
notation of Corollary 2 above.
Theorem 1. Ifa + b + c = d + e and Re с > 0, then
- [Г(а + k)T(b + k)T(c + k) __ _1_Д =
h \ k)T(e + k)T(l + k) Л + IJ ' ( ''

11. Hypergeometric Series, II 71
where
L=-2y- ф(а) - ф{Ь) + f ^=^?=Д B4-3)
k = l Wk(D)kK
where у denotes Eulefs constant. Furthermore, as m tends to со,
^ (a)k(b)k(c)k r(d)r(e)
l
{22A)
where the implied constant depends on a, b, c, d, and e but not on m.
If с = <?, then B4.4) reduces to the following asymptotic expansion for a
partial sum of a zero-balance 2Ft series (e.g., see Luke's book [1, p. 109,
Eq. C4)]):
as m tends to со. A slightly less precise version of B4.5) is given by Ramanujan
in Entry 15 of Chapter 10. It would be interesting if there existed a theorem
for zero-balanced q+lFq series that included B4.4) and B4.5) as special cases.
Theorem 2 below is a slightly more precise theorem than Ramanujan's
Corollary 2 given above.
Theorem 2. Ifa + b + c = d + e and Re с > 0, then as x tends to 1 with
0<x< 1,
Г(а)Г(Ь)Г(с) ^ Ub, с . 1 Log(i x) + L + m_x) LogA _
[a,b,c 1
B4.6)
where L is defined by B4.3).
In the sequal, we shall deduce Theorem 2 from Theorem 1.
In order to establish Theorems 1 and 2, we shall need four lemmas.
Lemma 1. // Re О 0, S = D + E - A - В - С, and Re S > 0, then
,F,
i В, С] =
T(D)r(?)r(S)
321 D E J Г(С)Г(А + S)r(B + S) 3 2
Г(С)Г(А + S)r(B + S)
D - C, ? - C,,
Л + S, В + S
Lemma 1 is a reformulation of Entry 27 in Chapter 10.
Lemma 2. // a and d are bounded, then as z tends to со with Re z > 0,
= zfl-d(l -
F(d + z)
Lemma 2, of course, is an easy consequence of Stirling's formula for the
gamma function, which can be found in Entry 23 of Chapter 7.

72
11. Hypergeometric Series, II
Lemma 3. Let e > 0 be fixed and let a complex number E be fixed. Let Re z > e
and suppose that к is any positive integer. Then there exists a constant N > 0
such that
' z\E /zN\
1+- -1=0 - , B4.7)
k) \fc,
where the implied constant is independent of z and k.
Proof. Let F = Re E. If F > 0, then, since Re z > e,
-E
гГ*'Ч)'-Ь(ИУ-».
Hence, it suffices to consider the case F > 0. Let N = F + 1. First, suppose
that к < \z\. Then
Thus, B4.7) easily follows. Finally, suppose that к > \z\. Then
m
m=l
m
where the last series does indeed converge because F > 0. This completes the
proof. ?
Lemma 4. Let Re D be fixed, where D is not a nonpositive integer. Let к be
any positive integer, and suppose that Re z > 0. Then
(D — z)k _, э-ы/зч
(D)k ч
where the implied constant is independent of z and k.
Proof. For some constant N > 0 that is independent of z and k,
- z)k
(D)k
= n
1=0
k-l
-д
D+j-z
1 -
D+j
k-i
f Г]
l=o
D+j
k-l
|z|f П
\D+j
D+j
2\l/2

11. Hypergeometric Series, II
73
fc —X
г п
2\ 1/2
|z|
\ m
1/2
e2lt|z|/3
Proof of Theorem 1. By Lemma 2,
\r(d + к)Г(е + /с)ГA + к) к + 1
as m tends to со. Also, from Ayoub's text [1, p. 43],
m-l 1 /
Z 7 r = Logm + y + 0
as ш tends to со. Thus, it is readily seen that B4.4) follows from B4.2). It
remains to prove B4.2).
We first prove B4.2) for с = 1. Then, inducting on c, we prove B4.2) for
each positive integer с Lastly, we establish B4.2) for all с with Re с > 0.
For each e > 0, write
where
and
i = 3F2
a, b, 1 1
i, с + ej
j_w
2 ~
М .''
п, а + т
(d)m(e + Е)т 3 |_а -г т, е -г 6
upon a change of index of summation. By Lemma 1,
Г(Л)Г(е + ?)Г(?) Vd-l,e + E-l,
1 Г(а + ?)Г(Ь + ?) 3 2 L а + ?, Ь + ?
and
_
2 ~
+ е)Г(е)Г(Ь + от) Td - a, e - a + s, e
Г(а)Г(Ь)ГA + ?)Г(Ь + ш + ?K 2 L 1 + ?, b + m + e
Thus, we may write
where
G3,
B4.9)

74 11. Нуpergeometric Series, II
/Г(Л)Г(е + е)Г(е) Г(й)Г(е + е)Г(е)Г(Ы-m) \
1 ~ Е™ \Г(й + е)Г(Ь + е) ~ Г(а)Г(Ь)ГA + е)Г(Ь + m + г))
lim I <- + Г'A) + ••• п^ггт ~ 1-2/- ч? + '
?-о V L? J (,^(а) * (а)
1 Г'(Ь) , 1 If, r'(ft + m)
Г2(Ь) J Г(в)Г(Ь)е
T{d)Y{e) ( Г(«) Г'(Ь) Г'(Ь + т)
Г(а)Г(Ь){ У Г(а) Т(Ь) + Г(Ь + m) j'
1 ' ;
G = lim Г(^)Г(е + Е)Г(е) « (d - l)t(e + е -
2 ™ Г(в + е)Г(Ь + е) Д
Г(в + е)Г(Ь + е) Д (а + е)к(Ь + е)кк\
(d - Ще - 1).
Г(а)Г(Ь)^ (а)к(Ь),/с 1 " >
and
G Ит Г(<0Г(е + е)Г(е)Г(Ь + от) » (d - a)t(e - а + e)t(e)t
е) Д
= Ит
3 ,^о Г(в)Г(Ь)ГA + е)Г(& + от + е) Д A + е\{Ь + т + г)кк\
--^ V ^—^—?1* B4 12)
Г(а)Г(Ь),4 (ЩЫ-mLfc l ;
Since (Luke [I, p. 33, Eq. (8)]),
Гф + m)
Гф + от) =
as ги tends to oo, we find from B4.10) and B4.12) that, respectively,
and
G3 = oQY B4.14)
as m tends to oo.
Putting B4.11), B4.13), and B4.14) in B4.9) and then B4.9) into B4.8), we
conclude that we have established B4.4) for с = 1.
Assuming that B4.4) holds with с replaced by 1, 2,..., с — 1, we examine
m^ {а)кФШ\
h (d\(e\k\
(ft - l)(c - l)ttb (d -
1

11. Hypergeometric Series, II 75
= (d - l)(e - 1) - (b - \\(c - 1), |(fl)t _ (a - \\
(b - l)(c - 1) к (d - \\(e - \\{k - 1)! { к к
(d- l)(e - 1) » (a),(fo - l)t(c - 1).
(b - l)(c - 1) h (d - Ще - l)kkl
(d - l)(e - 1) » (a - l)t(fr - l)t(c - 1),
(b - l)(c - 1) tU (d - Ще - l)tfe!
Г(в)Г(Ь)Г(с
(d - l)(g - 1) », (а - 1)к(& - l)t(c - 1д + Q / ±
(b - l)(c - lhtb (d -
as m tends to oo. Using again Lemma 1, we deduce that
к~ШАп
-у ~ф(а)-ф(Ь)+
Г(а)Г(Ь)Г(с)
у (d - 4t(g - c)t 1__ » (d - 4(g ~ с).
»ti (a)t(b-l)kfe b-b4 (а)»(Ь)к
T{d)Y{e) [
Log ш — у ¦
Г(в)Г(Ь)Г(с)
(a)k \(b-\\k
Thus, B4.4) has been established for each positive integer с Letting m tend to
oo in B4.4) and recalling the opening paragraph of this proof, we conclude
that B4.2) holds for each positive integer с
To prove that B4.2) is valid for all с with Re с > 0, it suffices by Carlson's
theorem (Bailey [4, p. 39]) to prove that, for a, b, d, and s > 0 fixed, both sides
of B4.2) are analytic in с and equal to 0(е2*|с|/3) for Re с > е.
Let D = Re(d — e), with d adjusted, if necessary, so that D is not a non-
positive integer. Let z = с + D — d. Thus,
S ¦=
(a)k(b)kk к " (D)k '
where
AkL

76 11. Hypergeometric Series, II
By Lemma 2, Ak = 0(k~l~e), while by Lemma 4, (?> - z)J(D)k = O(e2n|z|/3).
Thus, S is analytic in z and equals 0(е2я|г|/3) for Re z > 0. It follows that S is
analytic in с and equal to О(е2я|с|/3) for Re с > e.
It remains to prove that
, Г(а + к)Г(Ь + к)Г(с + к)
к)Г(а + b-d + c + k) k+l
is analytic in с and equal to O(e2jt|c|/3) for Re с > ?. Let E = d — a — b. By
Lemma 2, since Re с > e,
kf(l + fc ^
/C + 1
} l 0A),
where the expressions 0A) are bounded analytic functions of с for Re с > ?.
By Lemma 3, A + c/kf - 1 = O(cN/k) for some positive constant N. Thus, T
is analytic in с and equals 0(cN) for Re с > е. This then completes the proof
of Theorem 1. ?
Proof of Theorem 2. Define
f = Па + к)Г(Ь + к)Г(с + к)
J ' T(d + к)Г(е + /с)ГA + к)
and
V(x) = f /(fe)x4 + Log(l - x) - L,
k = 0
where 0 < x < 1 and L is defined by B4.3). We must show that
V(x) = 0((\-x)Log(l-x)), B4.15)
as x tends to 1. By B4.2),
1 \ oo xk _ *1
/(fc) - (* l) E
oo / 1 \
= Z /(fe) - 7ГГТ
fe=0 \ K+l/
Now, by Lemma 2,
T
1 \ v _ 1
00
I
k=l
1
+ 1
-1)
00
«I
= A
1 -
k2
-X)
xk
00
I
k~2 "f x"
n = 0

11. Hypergeometric Series, II 77
00 00
= A -x) ? x" X k~2
n=0 k=n+l
»
« A - x) Log(l - x).
Using this in B4.16), we complete the proof of B4.15) and so also that of
Theorem 2. ?
The special case с = e of Theorem 2 gives an asymptotic expansion of a
zero-balanced 2Ft as x tends to 1 —. This special case is also an easy con-
consequence of Entry 26 below. Moreover, it is equivalent to B4.5).
For further remarks on Theorems 1 and 2 as well as ^-analogues, consult
the paper of Evans and Stanton [1]. A generalization of Theorem 2 has
recently been established by Biihring [1] who uses the differential equation
satisfied by 3F2. His proof has the advantage that the form of the asymptotic
formula does not have to be known in advance. Because Ramanujan showed
little interest in differential equations, he likely had yet a different proof.
Entry 25. Suppose that n is not an integer. Then
+ n+\,b + n+\ л
Г (а + b + n + 2)Г(-и) Га + n + \,b + n+ I
Г(а+ 1)Г(Ь+ 1) z 4 и+ 1
Г(й + b + п + 2)Г(п)х-п Га + 1, Ъ + 1
Entry 25 is a basic formula for the analytic continuation of hypergeometric
series and can be found in the treatises of Bailey [4, p. 4] and Erdelyi [1,
p. 108, formula A)].
Corollary 1. // n is a nonnegative integer, then
+ n+l, b + n+ [
1
b + n + 2)Г(я)х-» "-1 (a + \)k(b + l)kxk
п + 1)?ь {-п+1)кк\
(-l)T(a + b + n + 2) ™ {a + n+ \)k{b + n
Г(в + 1)Г(Ь + 1)Г(и + 1) »ti (и
x {ф(а + n + к + I) + ф(Ь + n + к + I)
- ф(п + к + I) - ф(к + 1) + Log x}x\

78 11. Hypergeometric Series, II
where i/>(z) = F(z)/r(z). Ifn = 0, the first expression on the right side above is
understood to be equal to 0.
Corollary 1 can be found in Erdelyi's synopsis [1, p. 110, formula A4)].
Corollary 2. // n is a nonpositive integer, then
Га + и+l, b + n+1 t
;1
Г (а + b + n + 2)Г(-и) ~^ (а + n + \)k(b + n + \)kxk
Г(а + 1)Г(Ь + 1) k% (n + l)kk\
Г(а + Ь + п + 2)(-х)-" » (а + l)k(b
T(a + n+ 1)Г(Ь +п+ 1)ГA - n) кЬ0 A - n)kk\
x {ф{а + к + 1) + ф(Ь + к + 1)
- ф(к - п + 1) - ф(к + 1) + Log x}xk.
If п = 0, we employ the same convention as in Corollary 1.
Corollary 2 is a reformulation of another formula in Erdelyi's treatise
[1, p. 110, formula A2)].
Entry 26. We have
+ Logx2F1(a+ l,b+ 1; 1; x)
= О.
Entry 26 is simply the case n = 0 of either Corollary 1 or Corollary 2
above. Ramanujan has given a less precise version of Entry 26 in Chapter 10
(Section 15).
Corollary
n 2РгA, i; 1; 1 - x) = Log №\ 2Ft(i i; 1; x)
A\2 к I
M
Proof. Putting a = b = — \ in Entry 26 and using familiar formulas for
ij/(k + 1) and ф(к + \) (Gradshteyn and Ryzhik [1, p. 945]), we find that

11. Hypergeometric Series, II 79
(X\2
k=0
- Log (?) Л<«; l; „> - 21 § ? --^x., ,26.1,
which completes the proof. ?
Example. I/O < x < 1, then
rn/2 p/2 tan(<p/2) d9 dcp _n Cn'2 dq>
Jo Jo ^1 — x cos2 в cos2 <p 4 Jo y/T^~ A — x) sin2 <p
+ iLogx Ф ¦ 2 •
Jo ^1 — x sin ф
B6.2)
Proof. First, for |x| < 1,
Jo л/1 — X Sin ф Jo *=0 К!
ttb fc! X 2r(fe
Second, for |1 — х| < 1,
dУ= = | 2fl(i 2; 1;
Third, using an integral evaluation in Gradshteyn and Ryzhik's tables
[1, p. 376] and the calculation B6.1), we find that, for |x| < 1,
11/2 f*/2 tan(y/2) rfg dy
о Jo x/l — x cos2 в cos2
oo A\ pi/2 /•»
= J] TT^*1 tan((jp/2) со82кф й(ф
*=o к! Jo Jo
ir/2
cos2t0 dQ

80 11. Hypergeometric Series, II
k%(k\J
JFdhh l;x). B6.5)
Using B6.3)-B6.5), we find that B6.2) is equivalent to the identity
РЛ i 1 1 ) ^
у Лъ i; 1; 1 - x) + ^(Log x) 2^(|, i; 1; x),
where 0 < x < 1. This last identity follows from the foregoing corollary, and
so the proof is complete. ?
The integral in B6.3) is the complete elliptic integral of the first kind, and
the formula B6.3) is a basic, well-known result in the theory of elliptic func-
functions. For further ramifications, see Section 6 of Chapter 17 in Part III [11].
Entry 27. For |x| < 1,
Гх" = ~i 2*"i(i> i; 1; 3c) Log(l - x). B7.1)
,4* (/с!J М 2/ -
Proof. For n > 1, the coefficient of x" on the right side of B7.1) is equal to
4 „tl {(и - fe)!}4 4(n!J „ti (I - n)t2fc'
where we have employed A7.3). It thus suffices to show that
я t — n)} " 1
*=i (l — "л fc j=i -4/ — x
Let Sn denote the left side of B7.2) and rewrite Sn in the form
- nJ+1 (i - nJ h (f - nJB),fc! • (Z'-i}
The right side of the equality above is a balanced 4F3 and so can be trans-
transformed by F.3) in Chapter 10. Let _y = z=l,x= — n, и = v = j — n, w = 2,
and m = n. Then
1, 1,-и,-и1 (-!-»)„(!)„
j A-4B^, 4 3
2я+1 - (&(-и-&

11. Hypergeometric Series, II 81
Bи + IJ - 1
и+1 k%Bk+l)Bn+\-2k)
= Bя + IJ « / 1
~ 2(и + IJ *tb \2k + 1
1
2n + 1 - 2k
= Bя + IJ " 1
~ (и + IJ *tb 2k + Г
Replacing n by и — 1 above and using the result in B7.3), we complete the
proof of B7.2). ?
The expression on the left side below is fundamental in the theory of elliptic
functions. See Section 6 of Chapter 17 in Part III [11].
Example 1
2F&A;i; i - x)\ _ x / i 21 2
Proof. By the corollary in Section 26,
/ 2^i(ii;i;i-
eXP 'К Г-/1 1.1..Л
1 21 \// 1 9 2
1 13 2
1 13 , 1/1 13 , V
Proof. Putting a = —% and b = — § in Entry 26, we find that
F(i i-M
Г 2rH3> 3> J>
= Log x 2^(|, f; 1; x) + ? %J|* {^(/c + f)
fc=o (л!)
from which the sought result follows. ?
Example 2

82 11. Hypergeometric Series, II
= Logfe) 2F1(hh 1; x) + ? ЩФ^{Зф(Щ - ф(к) - 2ф(к + 1)}х"
\27/ /c=i (к!J
where we have used the facts (Gradshteyn and Ryzhik [1, p. 945]), (/>(§) +
ф(^)-2ф{\)=-ЪЬо%Ъ and ф(к + %) + ф(к + ^) = Зф(Зк)-ф{к)-3 Log 3,
for к > 1. Hence,
2n 2F1(hhUl-
eXP|
5 2rU3> 3> l' x)
Example 3
/ д|Д ( 5
exp V271 z7/i з i7 I m\ ! + o
Proof. Putting a = — 5 and fr = —f in Entry 26, we find that
-s/2n2F1&i,l;l-x)
— T П(Г V F ("— —* 1 * v^
(f)k(z)k
- ф(к) -ф(к + $)-2 Log 2 - 2ф(к + 1)}хк
where we have used the facts (Gradshteyn and Ryzhik [1, p. 945]),
Ф(г) + ФИ) - 2^A) = —6 Log 2 and ф(к + ±) + ф{к + f) = 4^Dfc) - ф(к) -
ф(к + i) - 8 Log 2. The proposed formula now easily follows. ?
Example 4
13
Proof. In Entry 26, put a = — \ and Ь = — f to find that

11. Hypergeometric Series, II 83
-2n 2F1(lh U I ~ x)
Ё ~T^rWk + б) + Ф(к + I) - Щк
k=o (k\Y
S ^7 + У ~
As in Examples 1-3, we have employed familiar properties of \j/{z) (Gradshteyn
and Ryzhik [1, p. 945]). We also have used the fact that t//(?) + ^(f) -
2i//(l) = —4 Log 2 — 3 Log 3, which can be deduced from results in Chapter
8 of the second notebook. (See the author's book [9, Chap. 8, Eq. E.2) and
Corollary 3 in Sec. 6]. See also Gradshteyn and Ryzhik's tables [1, p. 944,
formula G)].) The desired formula now readily follows. ?
We do not know Ramanujan's intention in giving Examples 1-4.
Entry 28. Let <p denote a polynomial of degree m. Suppose that n is not an
integer and that Re(a + b + m + n + l)<0. Then
n + 1)Г(а + 1)Г(Ь + 1) « (a + \)k(b
~ f(a + b + n + 2) tt'o (a + b + n + 2)kk\
Proof. Since 1, x, x(x — 1),..., x(x — 1) • • • (x — m + 1) form a basis for the
set of all polynomials of degree m over the field of complex numbers, it suffices
to prove the result for cp(x) — (pm{x) := x(x — l)---(x — m + 1). We first
observe that
ГО, к < т,
cpm(k) = \m\, к = т,
[()()m, k>m.
Next, since
j=o

84 11. Hypergeometric Series, II
where r is an integer, we find that
/k\ f°' k<m'
A>m@) = X (-lW . ШЛ = < (-l)mm!, к = т,
W [o, k>m.
Thus, for <p(x) = <pm(x), the proposed identity may be written as
(„, ? <¦ + '">+ 'М-'П-Ц.
k^ A - n)/c!
+ T(a + п + 1)Г(Ь + п
у (а + п+ ЩЬ + п + Щ- 1Н-и - fcb.
Х ttb (и + 1)кк1
_ Г(а + и + 1)Г(Ь + и + 1)Г(а + 1)Г(Ь + 1)(а + l)m(b + l)m(- l)mm!
Г(д + b + п + 2) (а + Ь + и + 2)тт\
B8.1)
Let St denote the first sum on the left side of B8.1). Replacing к by к + т,
employing Gauss's theorem, Entry 8 of Chapter 10, and simplifying, we find
that
00 tr, _L_ \\ fh I 1\ ( \\m( _ 1, vm\
Sx = Г(в -
- n)
+m(
Г(а + 1)Г(Ь + 1)Г(я)(а + l)m(fo + l)m " (a + m + l\(b + m
A - n)m kh A - и +
Г(и)Г(а + m + 1)Г(Ь + m + 1)Г(т - и
A-п)яГ{-а-п)Г(-Ь-п)
Г(а + т + 1)Г(Ь + т + 1)Г(а + и + 1)Г(Ь + и + 1) sin л(а + n) sin л(Ь + п)
Г(а + b + m + n + 2) sin(ren) sin я(а + b + m + п + 1)
B8.2)
Let S3 denote the expression on the right side of B8.1). Then
_ (- l)T(a + n+ l)T(b + n + 1)Г(а + m + 1)Г(Ь + m+ 1)
i3~ Г{а + Ь + т + п + 2) ' {ЖЗ)
If S2 denotes the second series on the left side of B8.1), then, by B8.2) and
B8.3), we must show that
_ Г (a + n + 1)Г(Ь + и + 1)Г(а + m + 1)Г(Ь + m + 1)
2 Г(й + b + m + n + 2)
1Г +
sin(Tm) sin л(а + Ь + m + n + 1) j
We shall prove B8.4) by inducting on m. For m = 0, B8.4) is valid by Entry

11. Hypergeometric Series, II 85
25, since Entry 28 reduces to Entry 25 for x = 1 when cp(x) = 1. Assume then
that B8.4) holds with m replaced by 0, 1, 2,..., m - 1. Observe that
(pm(n + к) = (n + k)(pm_x(n — 1 -(- fe).
Thus, we may write
S2 = T(a+n+ 1)Г(Ь + n + 1)Г(-и)
" (a + n+ l)k(b + n+ \)k(pm-y(n - 1 + k)
= -Y{a + 1 + (и - 1) + 1)Г(Ь + 1 + (и - 1) + 1)Г(-(и - 1))
» (a + 1+ (и - 1) + 1)*(Ь + 1 + (и - 1) + У)к(рт-Х(п - 1 + к)
We now apply the induction hypothesis, but with a, b, and n replaced by a + 1,
b + 1, and n — 1, respectively. Hence,
Г(й + и + 1)Г(Ь + и + 1)Г(а + m + 1)Г(Ь + m + 1)
2 ~ Г {а + Ь + т + п + 2)
sin п(п + a) sin я(п + Ь)
x ^-
sin я(и — 1) sin п(а + b + т + п + 1)J
from which B8.4) follows. This completes the proof. ?
Corollary. Assume the hypotheses of Entry 28. Then
Г(а + 1)Г(Ь +1) " (а + 1)к(Ь + l)t/
" Г(а + Ь + 2) ttb (а + Ь + 2)tfe!
« (а + l)t(b + l)t<p'(fc; S (а
h (k\J h
(k\J h (к!J
х {ф(а +\ + к) + ф(Ь+1 + к)- 2ф{к + 1)} = 0.
Proof. After some manipulation, we write Entry 28 in the form
Г(д + 1 + k)T(b + 1 + fc)y(fe)
ГA - и + к)к\
Г(д + и + 1
Г(и + 1 + к)к\
S Г(а + 1 + к)Г{Ь + 1 + fe)AV(O)
* к% Г {а + Ъ + п + 2 + к)к\
Differentiating both sides with respect to n and then setting n = 0, we find that

86 11. Hypergeometric Series, II
S Г(д + 1 + к)Гф + 1 + к)ф(к + l)y(fc)
L (/с!J
« Г'(а + 1 + к)Гф + 1 + к)(р{к)
" Г(а + 1 + к)Г'ф + 1 + fc)<p(fc)
» Г(а + 1 + к)Гф + 1 + к)<р'(к)
*=о (к\)
„г» +1, i Г±±1±Ш±1±!^ . 0.
^ Г(й + Ь + 2 + fe)fe!
After some manipulation and simplification, the formula above reduces to the
proposed formula. ?
Entry 29(i). // Re(a + P + у - 5 - e), Re(<5 - у - 1) < 0, then
Га, р, у] Г(8)Г(8 -а-р) Г a, j8, e - у 1
|_ <5, ? J Г(д — а)Г(<5 — P) |_a + P — 5 + 1, ej
ГE)Г(е)Г(а
Г(а)Г(/?)Г(е - у)Г(<5 + e - a -
S — ot, д — Р, 5 + e — a — P~)
8-a-P+l,5 + s-a-P
Entry 29(i) was communicated by Ramanujan in his second letter to Hardy
[16, p. xxviii]. For a proof of Entry 29(i) and an illuminating discussion of
this formula, see Hardy's paper [1, pp. 498, 499], [7, pp. 511, 512]. Another
proof can be found in Bailey's tract [4, p. 21].
Entry 29(ii). If a, p, or у is a nonnegative integer,
-a,-p,-y,y
B9Л)
Proof. R. Askey and J. Wilson [1] have recently given a short proof of Entry
29(ii) when either a or /? is a nonnegative integer. Now suppose that у is a
nonnegative integer. If we multiply both sides of B9.1) by ( — а — p + j)y,
then on each side we obtain a polynomial in a of degree y. These two
polynomials agree for each nonnegative integer a. Hence, they must be identi-
identically equal, and this completes the proof. ?

11. Hypergeometric Series, П 87
If n is a nonnegative integer, define
n,n + a + y-^,y+ ix,y-ix
These polynomials in x arise from the right side of B9.1) by a renaming of the
parameters. Askey and Wilson [1] have shown that {P2n{x)}, 0 < n < со,
is an orthogonal set on (—oo, oo) with respect to the weight function
|Г(а + ix)F(y + ix)\2. As we pointed out in Chapter 10, the integral over
(—oo, oo) of this weight function was first evaluated by Ramanujan [8],
[16, p. 57]. There also exists a set of similarly defined polynomials P2n+l(x) of
odd degree 2n + 1 so that {Pn(x)}, 0 < n < со, forms a complete orthogonal
set on (—oo, oo) with respect to the aforementioned measure [1].
Entry 30. Leta + p+ \ =y + 8,c = Г(а)Г(Р)/{Г(у)Г(8)}, and
_c2Fl(a,P;8;l-x)
2^(а, Р;у,х)
Then
x'y{l - x)~d
y = ~~
Proof. From Entry 25,
cA1 2Fl(a, P; y; x) + cA2x1~y 2Fl(8 — a, 8 — /?; 2 — y; x)
2Ft(a, P;y;x)
where Al and A2 are constants with cA2 = l/(y — 1). Thus,
1 d (xl~y 2FA8 - a, 8 - Й; 2 - y; x)
У' = г i
— x1 y 2F1(<5 — a, «5 — jS; 2 — y; *)— 2-^1 (a> i^; y; x)
1
^(a, jS; y; x), x1 y 2Ft(8 — a, 8 — P; 2 — y; x))
C0.1)
(y - 1) 2FHa, p; y- x)
1
(У - 1) 2^2(«,/?; y; x) " v""
where W(f, g) = Ж(х) denotes the Wronskian of /(x) and #(x). Now these two
functions are linearly independent solutions of the hypergeometric differential
equation (Bailey [4, p. 1])
x(l - x)y" + {y - (« + p + l)x}y' - ару = 0. C0.2)

11. Hypergeometric Series, И
By Abel's formula (e.g., see the text of Coddington [1, p. 113]),
——. ;—-dx
where С is a particular constant. Suppose that we write W(x) = x~yF(x). Then
С = F@). If we perform the differentiation in C0.1), we readily find that
С = 1 - у. Thus,
W(x) = -(y ~ l)*~r(l - x)~\ C0.3)
and, by C0.1), the proposed formula for y' follows. П
Corollary. Let
n 2F1(n, 1 - и; 1; 1 - x)
у -— .
sin(Ttn) 2-F\(n> 1 — и; 1; x)
Then
У x{\-xJFl(n,\-n;\\x)'
Proof. Apply Entry 30 with a = n, p - 1 - n, and у - S = 1. П
Entry 31 (i). Let у = 2Fl(ix, 0; y; x). Then
у dx - x(l - x)y' = (y- l)(y -1)-(ol + P- \)xy.
Proof. Upon differentiation, it is found that the proposed formula is equi-
equivalent to the formula
- \)y - A - x)y' + xy' - x(l - x)y"
= {y- l)y' -(a + p-l)y-(a + p- \)Xy'.
Upon simplification, this formula reduces to C0.2), the hypergeometric differ-
differential equation satisfied by 2-f\(«, P\ У, *)¦ ?
Entry 31 (ii). Let a. + P + 1 = у + д. Assume that n > 1 and that n>Rey. If
У = У(х) = 2Fl(tx, P; y; x), then
X Г Г" 2 ¦ ¦ I du
1 л
y(x) 1 t"-2y(t) dt
uy(\ - uNy2(u)
n,n -y

11. Hypergeometric Series, II 89
The conditions that we have imposed on n are needed only for the conver-
convergence of the integrals on the left side of C1.1).
Entry 31 (ii) is somewhat imprecisely stated by Ramanujan.
Our method of proof will be as follows. We first show that the left side of
C1.1) is a solution of the inhomogeneous hypergeometric differential equation
x(l - x)z" + {y - (a + P + l)x}z' - a/?z = x^^l - xI^".
Then, with considerably more difficulty, we show that the right side of C1.1)
is a solution of the same differential equation. The difference of these two
solutions is, of course, a solution of the associated homogeneous hyper-
hypergeometric differential equation C0.2). Now y± := у =- 2^i(°c> P\ y\ x) and y2 :—
x1'7 2Fi(<5 — а, ё — P; 2 — y; x) are a pair of linearly independent solutions of
C0.2). By examining the power series expansions of both sides of C1.1), we
easily see that the difference of these two functions cannot possibly involve yt
or y2', that is, their difference is identically equal to zero. This then completes
the proof.
Proof. Letting w = w(x) denote the left side of C1.1), we find trivially that
dx\yj xy(l-x)V(
and
lx^W) » *->,(»). C1.2)
?
On the other hand, since y + d = a + p+ 1,
d
= — x"(l - xfy-г ~ x"(l - xf w^
dx \ dx dx
= x^(l - xf-ly{x{l - x)w" + {y-(a + P+ l)x}w')
- x^^l - x)d-lw(x(l - x)y" + {y - (a + p + l)x}y')
= x^!(l - x)s-ly(x(l - x)w" + {y - (a + p + l)x}w' - afiw),
C1.3)
where we have used the fact that у is a solution of the hypergeometric equation
C0.2). Combining C1.2) and C1.3), we deduce that
x(l - x)w" + {y - (a + p + l)x}w' - apw = x^^l - xI"*. C1.4)
It remains to show that the right side of C1.1) satisfies the differential
equation C1.4).
Let

90 11. Hypergeometric Series, II
i=o {n)k(
Then, by C1.4), we must show that
x(l - x)Y" + {y - (a + p + l)x} Г
rn+k~yt
US - Щ1 - X)-' f
i у (я - <х)к{п - Р)к(п + к - у)(п + к-у
П _ u-« у (» ~
- с +, + оо -
- у
(»-
Ж ) *?о Ш» - у + 1Ь
= (и - у)(и - ^x'-^Hl - xI^. C1.5)
We cancel the factor of A — х)~г in the last equality and show that the
coefficients of like powers of x on both sides in C1.5) are equal.
We first examine the coefficients of x". On the left side of C1.5), this
coefficient is equal to
(n - y)(n - у - 1) + y(n - y) = (и - у)(и - 1),
which is in agreement with the right side of C1.5).
Next, the coefficient of x"~y on the left side of C1.5) is equal to
2E - 1)(и - у) - 2(и - y)(n - у - 1) + Un - а)(п - P)(n - у)
+ у(д - 1) - у(и - у) + -у(и - а)(я - Д) - (а + р + 1)(и - у) - а/7.
C1.6)

11. Hypergeometric Series, II 91
Now it is easy to see that C1.6) may be written in the form
-{n - n^in - n2),
where n1 and n2 are the two roots of the quadratic polynomial C1.6). By a
direct verification, it can readily be shown that 1 and у are the roots of C1.6),
although the case n = 1 is moderately tedious. In both computations, the
hypothesis a + /?+l=y + <5is used. Thus, the coefficients of x"~y on both
sides of C1.5) agree.
Lastly, we must show that the coefficient of xn+k~\ к > 1, on the left side
of C1.5) is equal to 0. This coefficient is equal to
<5(<5 — 1) 1- 2F — 1)
(«)t-i(n - 7 + l)k-i (n)k(n - у + l)k
+ 2A - 5){П ~ a)t~l(w ~ P)k-l{n + k ~ * ~ 7)
(n)t-i(n-y + l)t-!
«)t+i(» - i?)t+i(w + fc + 1 - у)(и + к - у)
2(и ~ <х)ц(и ~ j?)t(H + fc - у)(н + fc - 1 - у)
(«)*(« - У + 1)*
| (н - а\-у{п - Р)к^(п + к - 1 - у)(н + /с - 2 - у)
+ у(д — 1)-— -^ + у — —
(и - сс)к(п - р\(п + к - у) . , . о . 1W1 ЕЛи - a^-^n - Д)^!
+
__ (я + /, + !)Ln - а^п ~ № + к-у)
+ (а + р + 1)
-аря~ли'
(п\{п - у + l)t
и - a)k-i(i - Р)к-Лп + fc — 1 — у)
(и)к_!(и - у + 1)к_!
(n)k(n - у + l)k
Next, remove the factor
(n)k.1{n-y+
from each of the 14 expressions above. Then let и = u(k) = n + к — 1. There-
Therefore, it suffices to show that

92 11. Hypergeometric Series, II
ax и 2(д - Щи - a)(u - P)
6{5 - 1) + \- 2A - S)(u - y)
и
| (и + 1 - а)(и - а) (и + 1 - P){u - P) 2{u - <x)(u - p)(u - y)
(м + 1)м м
+ (и - y){u - 1 - у) +
u(u + 1 - у)
у(ц + 1 - а)(м - а)(м + 1 - /?)(ц - /?) у (и - а) (и - Р)
(м + 1)м(м + 1 - у) м
— у)
01.7,
If we multiply both sides of C1.7) by u(u + 1)(м + 1 — у), the left side becomes
a polynomial of degree 5. In order to show that this quintic polynomial is
identically equal to 0, we shall show that the coefficient of м5 is equal to 0 and
that the polynomial vanishes at five distinct points. It is easy to check that the
coefficient of м5 is equal to 0. One can verify that this polynomial vanishes at
u = 0, — 1, у — 1, a, and p. We sympathetically suppress the details. This
completes the proof. ?
Corollary. // n is arbitrary and у = 2-Ми, 1 — и; 1; x), then
fx
x(x — \)y' = n(n — 1) у dx.
Proof. In Entry 31(i), let а = n, p - 1 - n, and у = 1. ?
Entry 32(i). // (p is any function, then
M
у ML
(р(х)
is always an even function of x, provided the series converges.
Proof. Set r = m = \ in Entry 1. ?
Entry 32(ii). lf\<x<2, then
Ffi !¦ |'| _1/v)- [Z F f1 J-' 1 • 1 — x\
2l 1V2' 2» x» L 1/л) — V 21 lV2» 2> A» A л/'
This result is a special case of a transformation
2^(а, b; c; z) = A - z)"- 2F1(a, c-b;c; z/(z - 1)) C2.1)
that is generally attributed to Gauss [1] or Kummer [1], [2] but is due to

11. Hypergeometric Series, II 93
Pfaff [1]. Equality C2.1) is also found in Chapter 10 (Entry 19). For a proof
see Chapter 10 or Bailey's tract [4, p. 10, formula A)].
Entry 32(iii)
'1 -x\
¦YWl + xh^&fcljx2).
Proof. Replace x by (A — x)/(l + x)J in Entry 32(ii) and then apply Entry
5 with r = j and x replaced by — x. This yields
„,,!,, /l-x\2\ 1 + x _ /, , . -4x
= A+хJ^(А^;1;х2). П
Entry 32(iv)
P (I 1. |- 1 | ~ X 1 | _ Л i Y\2 p Л i. .. 4л
2^11 2, 2» *' A ~" I , I I - v1 + x) г'^г» 2. A> x )¦
Proof. From the work of Kummer [1, p. 148, Eq. D6)], [2, p. 142],
\i уь/ \
where c2 = 1 — fo2. If we put Ь = (A — x)/(l + x)J, Ramanujan's proposed
formula easily follows. П
The reader should compare Entries 32(iii) and (iv).
Entry 32(v)
V V A
1 + n'
n
+
Proof. In Erdelyi's treatise [1, p. Ill, formula (8)], let a = b = \ and z = n2
to get
C2.2)

94
11. Hypergeometric Series, II
Next, in the same compendium [1, p. Ill, formula (9)], let a = b = J and
z = «2/(l + и2) and obtain
F (* *• a-
44'4>2'
'l+n2
From C2.2), C2.3), and C2.1),
¦
F U з.з.
T'4'2'
r2ti)
C2.3)
where in the last equality we employed Entry 21 with m = n = \ and x
replaced by in. ?
Entry 33(i)
2J Ц 2, 2.
1+x
Proof. Set r = m = % in Entry 2.
П
Entry 33(ii)
1 (i, i; l; i(i - УГ^^)) = 2f&, i; i;
Proof. In Entry 12, set x = у = -\ and z = 1 and replace p by x. ?

11. Hypergeometric Series, II
95
Entry 33(iii)
Proof. Put a = p = —? and у = \ in Entry 13.
Entry 33(iv)
?
* *;
4x
Proof. Set r = m = { in Entry 4.
Entry 33(v)
?
ж;
Proof. Set a = fo = ? and с = 1 in Erdelyi's book [1, p. 105, formula A)].
?
Example (i)
1 3.1.
2J 1 I 4> 4> l>
-4x
F/l 1.1.
2Г1 I 2' 2^ •>
1 +x
Proof. Replacing x by — x in Entry 33 (iv) and then using C2.1), we readily
find the proposed formula. ?
Example (ii)
F U 1.1. ~4X
21|ы"A-хJ
Proof. Apply Entry 33(iv), Example (i), and lastly Entry 33(ii) with
i(l - л/l-x) replaced by x/(x - 1) to find that
l+x2J4*'?'
X
-4x
Ч4'4'"(Г^^х)
П
Example (iii)
F [I .3,1. ~4x
ТПГ2'ча':

96 11. Hypergeometric Series, II
Proof. Apply Entry 33(v) with x replaced by — 4x/(l — xJ and then use
Example (ii). ?
Entry 34
лщ
(a) —3-= 1.08643481121330801457531612.
(b) —)±4= 1.311028777146060.
4./27Г
(c) -^Д-= 1.180340599016092.
1 D)
(d) —^ = 0.269676300594191.
л3'2
яЗ/2
(e) -щ- = 3.708149354602731.
Both parts (a) and (b) are correct. The last recorded digits in (c) and (d)
should be 6 and 0, respectively, and the last two digits in part (e) should read
44. Numerical values for the relevant powers of л may be found in the tables
of Fletcher et al. [1, Chapter 5]. A numerical value for Г(|) was taken from
Fransen and Wrigge's tables [1].
For brevity, set
and n = —^-. C4.1)
Entry 34(i). If\x\ < I, then
Proof. Evaluating 2F\. (j + 2J + h) + U з) below by a formula of Kum-
mer that can be found in Bailey's tract [4, p. 11, formula B)], we find that
jkjl tkjk\(k-j)\2k
Pop- »%
oo AJ j

11. Hypergeometric Series, II 97
v
l22k ' k% Bk + 1)!(|J22'
= л 2Fdh i; i; *2) + ^ 2^(i I; I; x2), C4.2)
after some simplification. ?
Entry 34(ii)
1 +x'
Proof. First apply Entry 21 with m = n = \ and x replaced by 2x/(l + x2).
Then make two applications of Entry 3 with x replaced by x2 and r = m = \
and r = m = |, respectively. Thus,
,ы 4x2 \ /3 3 з 4х2
- n ^ ^ J F ^
x2K'2 2Fx(i i; |; x4). C4.3) П
Ramanujan(p. 141) has mistakenly written A + x2I'2 instead of A + x2K/2
on the right side of Entry 34(ii).
Entry 34(iii)
I 2Fhh Ь 1; *A + x))-j 2F?& h 1; i(l - x))
-x
V fP
x3 41x5 21x7
_ __^c x3 41xs
~ Г^? " 2A - x2J 120A - x2K + ''' • ( '
Proof. We first establish the latter two equalities. The four displayed coeffi-
coefficients on the right side of the second equality are simply numerical calcula-
calculations of the first four coefficients of the left side. Apparently, Ramanujan
does not possess a simple formula for these rational coefficients. Expanding

98 11. Hypergeometric Series, II
A — x2)~k, к = 1, 2, 3, in binomial series on the far right side of C4.4) and
collecting coefficients of x, x3, and x5, we establish the last equality. Evidently,
Ramanujan is not claiming to have found a general formula for the coefficient
x^-ya-x2)*,^ > i.
We now prove the first equality of C4.4). By Entry 21 and C4.1),
b i;
- \{n 2Fdi, i; b x2) - цх 2Fdl I;!; *2)}2
Г1 i !
L2,4-n,
1 i !_„„
ГГ~2 ' 1 C4.5)
where we have employed Erdelyi's work [1, p. 187, formula A4)]. In compar-
comparing C4.4) and C4.5), we find that we must show that
(IJ
V4/n i
Now from Erdelyi's book [1, p. 85, formula B)] we find that
Thus, it remains to show that
However, this last formula is a special case of Entry 29(i). Thus, the proof is
complete. ?
Example (i)
2^1\2~' 2^ ^> 2~\ ' ^)) =~ /^(Д ^ ) 2-* 1 I 4> 4» 2~' ' У 7
\ x — l
Proof. Employing Entry 34(i) and then C2.1), we easily achieve the proposed
formula. ?
In Ramanujan's formulation of Example (i) (p. 142), he has written
A - х2Г5/4 instead of A - x2)'4 on the right side.

11. Hypergeometric Series, II 99
Example (ii)
/i Z* F I 1 1- 1- 1
W 1 — X 2-^1 I 2' 2> X> 2
1 + X'
= H iF, (k, h !;
Proof. To each of the functions on the right side of Entry 34(ii), apply C2.1).
The desired result easily follows. ?
On the right side of Example (ii), Ramanujan (p. 142) has written 2 instead
of 1 + x2.
Entry 35(i). If n is arbitrary, then
cosBn sin^1 x) = 2F\[n, —и; \; x2).
Proof. In Erdelyi's treatise [1, p. 101, formula A1)], let a = 2n and z =
sin x. ?
Entry 35(ii). // n is arbitrary, then
sinBw sin^1 x) = 2nx 2-Fi(i + n, \ — n; |; x2).
Proof. In Erdelyi's book [1, p. 101, formula A2)], put a = 2n and z = sin x.
?
Entries 35 (i) and (ii) are closely related to the Tschebyscheff polynomials.
Entry 35(iii). // n is arbitrary, then
A - xT1/2 cosBw sin x) = 2F1{h + и, з - и; i; x2).
Proof. By Entry 35(i),
A - х2)-1'2 cosBn sin x)=i Щ±х2^ f ("Y~")<[x2t.
j=O Г- k = O B)kk-
Using A7.3), we find that the coefficient of x2r, r > 0, on the right side is equal
to
!(r - /с)! r! *U
where we have utilized Saalschutz's theorem, Entry 2 of Chapter 10. The
proposed formula immediately follows. ?

100
11. Hypergeometric Series, II
Entries 36 (i), (ii). For n real and k>2, let
(и V - 22)(n2 - 42) • • • (n2 - (k - 2J), if к is even,
k(n)~\n(n2- l2)(n2-32)---(n2-(k-2J), if к is odd.
If 2 - 2ф. < x < 2 + lyfi, then
(x + l)n/2 - (x + l)-"/2 = HX + 2 У
2fc+l
and
(x + 1)"/2 + (x + 1)""/2 = 2 + 2
й BЛ)!
2*
C6.1)
C6.2)
We have stated Entries 36(i), (ii) in somewhat different forms than did
Ramanujan.
Proof. From Corollary 2, Section 14 of our description of Chapter 3 [9],
*=2
where \a\ < 1 and n is any real number. Let a — x/{2^/l + x}. Then an
elementary calculation shows that а + ^/l + a2 = ^/l + x. Thus,
(x + 1)и/2 = 1 +
z^/l + x t=2 k\ \2^/i + x,
Replacing и by -и and using the definition of bk(n), we find that
C6.3)
. C6.4)
Subtracting C6.4) from C6.3), we deduce C6.1); adding C6.3) and C6.4), we
deduce C6.2). Finally, an elementary computation shows that \a\ < 1 if and
^ / ?
C6.5)
Entries 36(iii), (iv). Let n be real and suppose that
A + xK'2
33/2 •
Then
l+4x\n
= 1 + nx(l
4-3!
-x3(l
n(n - l)(n - 9)(n - 11)(и - 13) 5
42-5!
X5(l
C6.6)

11. Hypergeometric Series, II
101
and
¦ +
x
f
. - 6) (и - 8) (и - 10)
4М! ;
C6.7)
We have presented Ramanujan's formulations of Entries 36(iii), (iv). As we
shall observe below, a general formula for the coefficient of (x/(l + xK'2)*,
к > 0, can be given in terms of gamma functions.
Proof. We shall apply the Lagrange inversion formula (Whittaker and
Watson [1, p. 133]). Accordingly, we let
and /(x) = A + x)"/2
+x
and define у by у = x + xq>{y). If we solve this equation for y, we find that
у = \{2x - 1 + УГ+ 4x).
It follows that
Also note that
/Ш2х - 1 + УГТ4Х)) =
Л >
Hence, by the Lagrange inversion formula,
f{y) = /(x)
|
we find that
C6.8)
- к
C6.9)
By Stirling's formula, the series above converges for those values of x given
by C6.5). (The radius of convergence of a more general class of power series
has been calculated in our book [9, Sec. 14 of Chap. 3].)

102 11. Hypergeometric Series, II
We now make a second application of the Lagrange inversion formula. Set
ф) = ^=L= and f(x) = A + xf2
'l+x
and define у by у = x - хф(у). It follows that у = 0. Hence, /(y) = /@) = 1.
Hence, by an application of the Lagrange inversion formula C6.8) like that
above,
1 = A + xf12 - inx(l + x)(")/2
C6.10)
Again, by Stirling's formula, the series C6.10) converges for those values of x
given by C6.5). Subtracting C6.10) from C6.9), we deduce C6.6); adding C6.9)
and C6.10), we deduce C6.7). D
Ramanujan had an affinity for the Lagrange inversion formula or, perhaps
more precisely, for the beautiful expansions that can be derived from it.
Ramanujan undoubtedly learned the Lagrange inversion formula from Carr's
Synopsis [1]. The Lagrange inversion formula is also found in the calculus
books of Edwards [1, pp. 450-457] and Williamson [1, pp. 151-153], both
of which were known to Ramanujan. In Chapter 3 of his second notebook
and in his quarterly reports, Ramanujan offers many applications of the
Lagrange inversion formula. Although perhaps Ramanujan first discovered
some of these expansions via the Lagrange inversion formula, his primary
method for deriving these results arose from one of his favorite discoveries,
a type of interpolation formula in the theory of integral transforms. This
theorem has been thoroughly discussed by Hardy [9, Chapter 11] and by the
author [9] in his account of Ramanujan's quarterly reports.
An excellent survey on the g-Lagrange inversion formula has been given
by Stanton [1].

CHAPTER 12
Continued Fractions
In assessing the content of Ramanujan's first letter, dated January 16, 1913,
to him, Hardy [9, p. 9] remarked: "but A.10)—A.12) defeated me completely;
I had never seen anything in the least like them before. A single look at them
is enough to show that they could only be written down by a mathematician
of the highest class. They must be true because, if they were not true, no one
would have had the imagination to invent them." These comments were
directed at three continued fraction representations. Indeed, Ramanujan's
contributions to the continued fraction expansions of analytic functions are
one of his most spectacular achievements. The three formulas that challenged
Hardy's acumen are not found in Chapter 12, but this chapter, which is almost
entirely devoted to the study of continued fractions, contains many other
beautiful and penetrating formulas. Unfortunately, Ramanujan left us no clues
as to how he discovered these elegant continued fraction formulas. Especially
enigmatic are the several representations for products and quotients of gamma
functions. Three of the principal formulas involving gamma functions are
Entries 34,39, and 40. Entries 20 and 22, giving Gauss's and Euler's continued
fractions, respectively, for a quotient of two hypergeometric functions, also
play prominent roles. Several other formulas are dependent on these five
entries, and it may be helpful to schematically indicate these connections
among entries.

104
12. Continued Fractions
20
13 18 21
41
42
47
34
7 9 10 11 12 21
29
43 44(iii)
39
25
26
28 32(i)
32(iii) 33 36 37
18 30 31 38
We shall use the notation for hypergeometric functions that we introduced
at the beginning of Chapter 10.
In the sequel, ф(г) always denotes F(z)/r(z). We shall employ the represen-
representation (Olver [1, p. 39])
several times in this chapter, usually without comment. Here у denotes Euler's
constant.
We shall usually adopt the notation
b1+b2+b3+
@.2)
for the continued fraction
+a2
b2 + a3

12. Continued Fractions 105
The notation @.2) appears to be the most convenient and widely used notation
for continued fractions. For brevity, it will occasionally be convenient to
employ the notation К(а„/Ь„) instead of @.2). We shall refer frequently to the
well-known texts of Perron [3], Wall [1], Khovanskii [1], and Jones and
Thron [1]. Because Perron's book contains several formulas that we shall
employ and that are not found in the other texts, we shall make many
references to this classic work.
In our initial published account of Ramanujan's work on continued frac-
fractions (see the Introduction for a complete reference), the domains of conver-
convergence were often more restrictive than necessary, and, in a few cases, they were
incorrect. The account that follows has been considerably improved because
of the comments and work of L. Jacobsen. In particular, she [3] has employed
analytic continuation and the uniform parabola theorem to extend the
domains of convergence of many of Ramanujan's continued fractions. The
work of Jacobsen [1] and Waadeland [1] on tails of continued fractions has
yielded a simpler, more uniform approach to several of Ramanujan's formulas.
Entry I. Let a1; a2, ..., ar and b1, b2, ¦¦¦, br be complex numbers such that
an ^ 0 for each positive integer n. Define ALj = 0, No = 1, D_t = 0, Do = 1,
, к > 2, A.1)
and
Dk = bkDk_y + akDk_2, k>L
Then, for r> 1,
b,b2+- + br ' Dr к Dk^Dk ¦ ^ '
Proof. The first equality in A.2) is a somewhat unusual formulation of a basic
elementary formula in the theory of continued fractions (Wall [1, p. 15]). For
future reference, we restate the first equality of A.2) in a more familiar fashion.
Let A_t = 1, Ao = 0, B_l = 0, Bo = 1,
Л = M*-i + e*4t-2. k * 1» A-3)
and
Bk = M*-i + a-i. * ^ 1- A-4)
Then, for r > 1,
b,+b2 + --- + br B/ y''
Thus, ax Nk = Ak+1 and Dk = Bk,k> -1. Note that if we define N_2 = 1, then
A.1) is valid for к = 1 as well. Recall that Ak and Bk are the fcth numerator
and denominator of the continued fraction @.2).

106 12. Continued Fractions
The second equality in A.2) is essentially another version of a well-known
fact (Wall [1, p. 18]) due to Euler [1]. The relations A.3) and A.4) were first
established by Wallis [1] and first studied seriously by Euler [5]. ?
Corollary. Let ax,a2,...,arbe nonzero complex numbers such that aj + aj+1 ф
О,; > l,andr> 3. Then
' ax a2 axa3 a2a4 ar.2ar
k=\ 1 — ax + a2 — a2 + a3 — a3 + a4 — • • • — ar_t + ar
This corollary is due to Euler [1], and a proof may be found in Perron's
book [3, p. 17].
Entry 2. Let x,a1,a2,... denote nonzero complex numbers and define, for each
nonnegative integer n,
i
t=o a^a2-¦ ¦ ak+1
If
lim/,(x)=oo, B.1)
then
x = x - fll + —±— —?— . B.2)
x - a2 + x - a3 + ¦ ¦ ¦
Proof. For each nonnegative integer n (Chrystal [1, p. 516, Eq. A4)]),
"i,1 a1a2---akxk axx bta2x b2a3x bnan+1x
kh. biV'A bt -b2 + a2x-b3 + a3x -¦¦¦-bn+1 + an+1x'
If we set uj = 1 and replace bs by —a^j > 1, we find that
where
axx a2x а„х
x-a2+x-a3 + --- + x- an+l
Letting n tend to oo and using B.1), we deduce that ax — A = 0, which is
equivalent to B.2). ?
Of course, we could impose several sets of conditions on x, ax, a2,... in
order to ensure that B.1) holds. For example, if lim,,.,^ \an\ = p, then, by the
ratio test, B.1) is valid if |x| > p.

12. Continued Fractions 107
Quite possibly, Ramanujan attempted to prove B.2) by the following
nonrigorous argument. Trivially,
ak = ^ , k>\. B.3)
x - ak+1 + ak+1
If we successively employ B.3) for к = 1, 2,..., we find that
atx axx a2x
a — =
x — a2 + a2 x — a2 + x — a3 + a3
a1x a2x a3x
x — a2 + x — a3 + x — a4 + ¦ • ¦
which is equivalent to B.2). This type of argument is valid under certain
conditions which will be set forth in the next theorem.
Entry 2 and some entries in the sequel are consequences of the following
result which is due to H. Waadeland [1] and L. Jacobsen [1].
Theorem. Let К(а„/Ь„) have a sequence g("\ 0 < n < oo, of tails; that is,
g^\bn + 0<">) = an, n> 1,
such that g{n) ф oo, 0 < n < oo. {Thus, gin) #0, -bn, if an ф 0, 1 < n < oo.)
Then K(an/bn) converges if and only if
00 * / b + a{n)
Z П -t
k=o n=\ \ g
converges in 4? = <€ и {oo}. In particular, if B.4) has the sum oo, then К(а„/Ь„)
converges to gm. More generally, if B.4) has the sum L e^, then K(ajbn)
converges to gm(L — 1)/L.
Note that a continued fraction К(а„/Ь„) has infinitely many sequences of
tails; define g@> e У? arbitrarily, and define g{n), n> 1, by
We now show that Entry 2 follows readily from Waadeland and Jacobsen's
result.
If g{n) — an+1, n > 0, then g(n) is a sequence of tails for B.2). Inserting this
into the (truncated) sum B.4), we find that
m k / x \
1 П (~ =al/m(x).
k=0 n=l \ «n + l/
Entry 2 now follows from the theorem above.
We shall interpret Entries 3 and 4 formally. We emphasize that the argu-
arguments that we give are not rigorous. There is a slight misprint in the formula-
formulation of Entry 3 (p. 143).

108 12. Continued Fractions
Entry 3. //x, al,a2,... are arbitrary complex numbers, then
x = a1+(x2 + fll(fll - 2a2) - 2a,(x2 + a2(a2 - 2аъ) - 2«2(- ¦ -I/2I/2I/2.
Proof. It is easy to verify that
x-ak = (x2 + ak(ak-2ak+y)-2ak(x-ak+l)fl2, k>l. C.1)
Using C.1) successively, we find that
x - ax = (x2 + ax(ax - 2a2) - 2ax(x - a2))m
= (x2 + aMi ~ 2a2) - 2fll(x2 + a2(a2 - 2a3) - 2a2(x - аз
and so the desired result follows. Q
Entry 4. Let a, n, and x denote arbitrary complex numbers. Then
f(x) := x + n + a
= (ax + (n + aJ + x(a{x + n) + (n + aJ + (x + n)(a(x + 2n)
Proof. By successively substituting, we find that
f{x) = (ax + (n + aJ + xf(x + n)I'2
= (ax + (n + aJ + x(a(x + n) + (n + aJ + (x + n)f(x + 2и)I/2I/2
and therefore we obtain the proposed formula. ?
Examples. We have
(i) 3 = A + 2A + 3A + 4A + ...)l/2)l/2)l/2)l/2
and
(ii) 4 = F + 2G + 3(8 + 4(9 + ...)i/2)i/2)i/2)i/2_
Examples (i) and (ii) were submitted by Ramanujan [5], [16, p. 323] as a
problem in the Journal of the Indian Mathematical Society and solutions were
subsequently given by him. Example (i) appeared as a problem in the William
Lowell Putnam competition in 1966 (J. H. McKay [1]).
T. Vijayaraghavan (Ramanujan [16, p. 348]) has shown that
, а„ > 0,
tends to a limit as n tends to oo if and only if
D.1)

12. Continued Fractions 109
See also Polya and Szego's book [1, pp. 37, 214]. Vijayaraghavan's theorem
can be used to show that the infinite radicals in Examples (i) and (ii) are
convergent (Ramanujan [16, p. 348]).
The literature on infinite radicals is rather scant, and so Herschfeld's paper
[1] is to be particularly recommended. He points out that Ramanujan's proofs
of (i) and (ii) are slightly incomplete, and he gives full rigorous solutions. This
paper contains a good discussion on the convergence of infinite radicals.
Elementary discussions of nested radicals have also been given by W. S. Sizer
[1] and E.J. Allen [1].
We state Entry 5(i) as Ramanujan records it. But, as we shall see, Entry
5(i) is valid only for 9 = 0. We shall separate Entry 5(ii) into two parts. The
first part will be proved rigorously; the second will be regarded as a formal
identity. However, we shall indicate some values of 9 for which the second
part of Entry 5(ii) is rigorously true. We suggest to readers that they attempt
to develop more thoroughly the theory of infinite radicals, so that perhaps
concrete conditions may be imposed on the formal identities in Sections 3-5
to ensure their validity. Jacobsen's paper [4] is one in this direction.
Entry 5(i). We have
2 cos 9 = B + 2 cos 20I/2 = B + B + 2 cos 40I/2I/2
= B + B + B + 2 cos 80I/2I/2I/2
Proof. Repeatedly apply the identity
2 cosB*0) = +B + 2 cosB*+10)I/2, к > 0,
with the plus sign always chosen on the right side. However, unless 9 = 0,
there clearly will be values of к when cosBk9) < 0, and so we must choose the
minus sign in such instances. If 9 = 0, Entry 5(i) implies that
2 = B + B + B + ---I/2I/2I/2,
which is meaningful since D.1) is easily seen to be satisfied. Furthermore, a
direct proof may easily be given. (This last example appears in Zippin's book
[l.P-51].) ?
Entry 5(ii) (First Part). Suppose that either \9\<л/6ог 5л/6 < 9 < 7л/6. Then
2 cos 9 = B cos 39 + 3B cos 39 + 3B cos 39 + ...)i/3)i/3)i/3_
Proof. For n > 1, let
Rn = B cos 39 + 3B cos 39 + 3B cos 39 + ¦ .-)^)Vi)V\
where n cube roots are taken. Observe that
J?n = Bcos30 + 3J?n_1I/3, n>2.
First suppose that |0| < л/6. Clearly, Rn^ < Rn for each n > 2. Thus,

110 12. Continued Fractions
Я„3 = 2 cos 30 + ЗЯ„_! < 2 cos 30 + 3Rn. E.1)
The polynomial x3 — 3x — 2 cos 30 has three real roots, 2 cos 0 and — cos 0 ±
^Isin 9\. For |0| < я/6, -cos 9 ± y^sin 0| < 0. Therefore, {Rn} is a non-
negative, increasing sequence bounded above by the root 2 cos 0. Thus, {#„}
converges and, by E.1), {Rn} converges to a root of x3 — 3x — 2 cos 39. As
we have just seen, this root must be 2 cos 9.
For 5л/6 < 9 < 7я/6, consider a = 0 — я. Thus, |a| < л/6. Using the fore-
foregoing analysis, we complete the proof. ?
We remark that if л/2 < 9 < 5л/6 or 7я/6 < 9 < Зл/2, then {Rn} converges
to -cos 9 + v^lsin 91 while if n/6 < 9 < л/2 or Зя/2 < 9 < 11 л/6, {Rn}
converges to —cos 9 — ^/^sin 6\.
Entry 5(ii) (Second Part). We have
2 cos 9 = F cos 9 + F cos 39 + F cos 90 + ¦ ¦ -)i/3)i/3)i/3 E 2)
Proof. Repeatedly employ the equality
2 cosC*0) = F cosC) + 2 cosC/l+10)I/3
for k = 0,1,2,....
We now indicate some special cases when the second part of Entry 5(ii)
may be established rigorously.
U 9 = 0, then E.2) becomes
E.3)
To prove E.3), define
Я„ = F + F + ---61/3---I/3I/3, п> 1,
where n cube roots are indicated. Observe that
Я3 = 6 + ./?„_! < 6 + Rn, n>2. E.4)
Now x = 2 is the only real root of the equation x3 — x — 6 = 0. It follows
that #„_! < Rn < 2, n > 2. Thus, {Rn} converges, and, by E.4), the limit of
{Rn} equals 2.
If0 = л, then E.2) yields
-2 = (-6 + (-6 + (-6 + ...)V*)i/3)i/3
= -F + F + F + ---I/3I/3I/3, E.5)
which is valid by E.3).
If 0 = л/3, the right side of E.2) becomes
C + (-6 + (-6 + ...)i/3)i/3)i/3 = (з _ 2I/з = 1(
by E.5). Hence, E.2) is valid for 0 = л/3. In fact, by induction, it is easy to
show that E.2) holds for 0 = n/3\ k>l.

12. Continued Fractions 111
It may also be easily checked that E.2) is valid if в = я/2 or 2n/3, for
example. If в = я/4, E.2) holds, but the verification is more difficult. ?
Entry 6. Let a > 0 but а ф 1. Suppose that n is a nonnegative integer. In the
field of formal power series, put
j=o
where ao(n) = 1, a^n) = — a~", and aj(n),j > 2, is defined recursively by
J 2{a' l - 1) k=i
Then for each nonnegative integer n,
a(a - 2) fa(a - 2) _ fa(a - 2) a
F.2)
vv/iere, on the left side, there are n iterated radicals. Furthermore,
(v/anJ (v/a"K
fn(v) = 1 - v/a" +
2(a - 1) 2(a - \)(a2 - 1)
v/a"f{a + 5) {v/a"M{2a2 + 3a + 7)
+ 8(a- l)(a2- l)(a3- 1) 8(a - l)(a2 - l)(a3 - l)(a4 - 1) + '
F.3)
Proof. If n = 0, F.2) is trivial. Thus, assume that n > 0. Proceeding by
induction and squaring both sides of F.2), we find that we must show that
or, in other words,
a + f a{n _ i)p; = af2(vl n > i,
2 J=i 1
Now, by F.1) and induction, a^n — 1) = ajaj{n),j > 0, n > 1, and so it suffices
to show that
aJaj(ri) = - E аЛ"Ц-*(")> У ^ 2.
But the latter equality is equivalent to F.1), and so the proof of F.2) is
complete.
The expansion F.3) is easily determined by employing F.1). ?
Ramanujan's formulation of Entry 6 is slightly incorrect, for he claims that
(p. 143)

112 12. Continued Fractions
which should be compared with F.1).
Entry 7. // x is not a negative integer, then
л x+lx+2x+3
First Proof. We first derive a consequence of Entry 22 that we shall employ
several times in the sequel. In Entry 22, replace x by x/a.. Since the continued
fraction converges uniformly with respect to a in a neighborhood of a = oo,
we may let a tend to oo to deduce that, for /? ф {— 1, — 2, • • •},
1F1(/?+l;y + l;x)= 1 (/? + l)x (/? + 2)x
У^(|8;у;х) yx + y + lx + y + 2x + '
(An equivalent form of G.2) was also found by Perron [3, p. 278, formula (8)].)
By using Corollary 1 of Entry 21, we can show that G.2) is also valid when p
is a negative integer, provided that y$ {ft, ft — 1, /J — 2,...}.
To prove G.1), set x = 1 and /? = у = х in G.2). The result now easily
follows. ?
Second Proof. The continued fraction G.1) has tails gin) = 1. The Nth partial
sum of B.4) is therefore equal to
i(-l)"(x+l)b G-3)
lc=O
which obviously cannot converge to a finite number. However, if x is not a
nonpositive integer,
x+1 x + 2 x + 3
jc+1 x + 2 x + 3 _ x x(x+l) (x+ l)(x + 2)
x +x+l+x + 2+--- 1 + 1 + 1 +•••'
which converges by Worpitzky's theorem (Wall [1, p. 42]). Hence, by the
theorem in Section 2, G.3) tends to oo as N tends to oo. So by the same theorem,
G.1) converges to gi0) =1. ?
It also should be remarked that Entry 7 follows from Entry 11 by setting
a - 1 and n — x + 1.
Ramanujan (p. 143) has written x instead of 1 on the left side of G.1).
Corollary. We have
2 3 4 5
1 + 2 + 3 + 4+ ¦¦¦¦

12. Continued Fractions 113
Proof. Set x = 1 in Entry 7. ?
Entry 8. Let n denote a positive integer and suppose that хф —ка, where к is
a positive integer such that 1 < к < п. Then
2a)---(x + ka)
x + a x + 2a x + (n — \)a
a+x + 2a— 1+х + За—1+---+ x + na-1
. (8.1)
First Proof. Denote the right side of (8.1) by AJBn in the notation of Section
1. Then by A.3),
An = {x + na - \)An-x + {x + (n - 1)а)Л„_2, п > 3,
or, upon iteration,
А„-{х + па)А„^ = -{Ай-1 -(х+(п- \)а)Ап_2}
= (-1Г1, n>3, (8.2)
since Ax = 1 and A2 = x + 2a — 1.
Similarly, by A.4),
В„ - (x + na)Bn_1 = -{?„_! - (x + (n - 1)а)В„_2}
=0, п>3,
since BY = x + a and B2 = (x + a)(x + 2a). Hence,
В„ = (x + a)(x + 2a) • ¦ • (x + na), n > 1. (8.3)
On the other hand, let the left side of (8.1) be denoted by the rational
function PJQn. Clearly,
Qn = (x + a){x + 2a) ¦ • • (x + na), n > 1. (8.4)
Now, for n > 2,
Р„ Р„_! (-1)"+1 {x+ па)Р„_! +(-l)"+1
6» ~ Qn-i. Qn ~ Qn
that is,
Pn = (x + иа)Р„_! + (-1)"+1, n > 2. (8.5)
Hence, by (8.2) and (8.5), An and Р„ satisfy the same recursion formula. Since
Ax = Pl = 1 and A2 = P2 = x + 2a — 1, we conclude that An = Pn, n> 1.
Also, by (8.3) and (8.4), В„ = Qn, n> 1. Thus, the equality (8.1) has been
established. ?

114 12. Continued Fractions
Second Proof. We induct on n. For n = 1, (8.1) is trivially true.
Suppose that we denote the left side of (8.1) by /„(x). Proceeding by
induction, we thus find that
(x + a)fn+l{x) = 1 -/„(x + a)
1 x + 2a x + na
= 1 -
x + 2a + x + 3a - 1 + ¦ ¦ ¦ + x + (n + l)a - 1
Letting
x + 3a x + na
Л=х + За-1+ -,
x + 4a - 1 + • • ¦ + x + (n + \)a - 1
we then deduce that
1
(x + а)/п+1(х) = 1 -
x + 2а + (x + 2a)/A
(x + 2a - 1) + (x + 2a)/A
~ (x + 2a - 1) + (x + 2a)/A + 1
1
+ (x + 2a - 1) + (x + 2a)/A
Upon dividing both sides of the equality above by x + a, we arrive at (8.1),
but with n replaced by n + 1. This completes the induction. ?
Corollary. We have
1 12 3
e- 1 1 +2 + 3 + ---
Proof. Let x = 0 and а = 1 in Entry 8 to obtain the equality
" (-l)t+1 1 1 2 3 n- 1
й~^ ~Т + Т+2+3 + --Ч-п-Г
Letting n tend to oo yields
1 _ l _I i ? 3
~ e ~ I + I + 2 + 3 + • • •'
The desired formula now readily follows by inverting the equality above.
?
The previous corollary is due to Euler [3].
Entry 9. Let a and x be complex numbers such that either x ф —ка for
к e {1, 2,...} and а ф 0, or that a = 0 and \x\ > 1. Then

12. Continued Fractions 115
x + a + 1 x + a x + 2a x + 3a
Proof. We first indicate a formal nonrigorous argument. Observe that for
each positive integer n,
x + na+l x + na _
(n-l)a+\ / <4 ,
x + (n - \)a - 1 +
x + na + 1
By applying this identity successively for n = 1,2,..., we formally derive (9.1).
We now give a rigorous proof based on G.2). We first assume that а Ф0.
Putting x = I/a, /? = x/a, and у = (x — a)/a in G.2), we find that, under the
restriction fi = x/a $ {— 1, — 2,...},
x + a x 1
x — a /x x — a 1
a ay
1 x + a x + 2a x + 3a
x-a-l+x-1 +x + a-l+x + 2a-l
provided that x ф —ка, where Ac is a positive integer. But,
(9.3)
a a) x — a
r(x-
x —
a
a
a)
X
x-t
(x Л
U )
1
a
- a
Substituting (9.4) into (9.3), taking the reciprocal of both sides, and simplifying,
we arrive at (9.1).
Another proof for а ф 0, depending on the theorem in Section 2, can be
given. Equality (9.2) shows that
... x + na + 1
a1"1 = n > 1,
У
У х + (п-1)а+Г
is a sequence of tails for (9.1). Thus, a partial sum of B.4) equals
f. , * {x + (nl)a+l}{x + na}
M x + (n + l)a + 1
which must tend to oo by the same argument that was used in the second
proof of Entry 7. Since gi0) = (x + a + l)/(x + 1), the proof is complete by the
theorem in Section 2.

116 12. Continued Fractions
For a = 0, the continued fraction (9.1) reduces to the periodic continued
fraction K(x/(x — 1)). The convergence behavior of periodic continued frac-
fractions is well established. See, for instance, the text of Jones and Thron [1,
pp. 47,48]. Thus, K(x/(x — 1)) converges if and only if x/(x — IJ does not lie
on (—oo, —1/4). For |x| > 1, the continued fraction in (9.1) converges to 1, as
claimed by Ramanujan. However, if |x| < 1, the continued fraction converges
to -x. ?
Examples. We have
4_3 4 5 6
W 3 +2 + 3 + 4 + •¦¦
and
,.., 5 4 6 8 10
()
Proof. Set x = 2 and a = 1 in Entry 9 to deduce (i); similarly, set x = 2 and
a = 2 to obtain (ii). ?
Entry 10. // n is a positive integer, then
1 2 3 nn+ln+2
n =
1 -n + 2- n + 3 - n + ¦•• + 0+ 1 + 2 +
First Proof. Putting x = 1, p = 0, and у = 1 - n in G.2), we find that
A - n) ^@; 1 - n; 1) 1 2 3
« = ~^~т.—т. r: = — n +
1 -П + 2-И + 3-И + ---'
which completes the proof. ?
Second Proof. In A1.7), set n = 1 and replace a by n to deduce that
1 2 n-1 n-l
= 1 +
1-Л+2-П + --- 3-П + 4-П + ---
We shall be finished if we can show that, for each positive integer n,
^ + ^+ =». (ЮЛ)
2-П + 3-П + --
We prove A0.1) by inducting on n. If n = 1, A0.1) is trivial. Assuming that
A0.1) holds with n replaced by n - 1, n > 1, we see that
n
2-П + 3-П + -" B - и) + (и - 1)
The interpretation of Entry 11 was made difficult because Ramanujan

12. Continued Fractions 117
left most of his notation undefined. Furthermore, some of his notation is
unnecessary and so will not be given.
Entry 11. Suppose that a is a positive integer and that n $ {0, — 1, — 2,...}.
Define Na and Da by
!^A-a;n + 2-a;-1) = A1.1)
11 ; (n + 2-a)(n+ 3-a)---n
and
1F1(l-a;n+l-a;-l) = - —-^ — -, A1.2)
(и + 1 — a)(и + 2 — a)---(n - 1)
where if a = 1, the denominators on the right sides of A1.1) and A1.2) are
understood to be equal to 1. Then
A13)
Ц, n-a + n-a+l+n-a + 2+---
and
Na+, a-I a-2
-^~=n+2-a+ . A1.4)
Na n+3-a + n + 4- a + ¦¦¦
Proof. Since a is a positive integer, both ^A — a; n + 2 — a; — 1) and
jF^l — a; n + 1 — a; — 1) terminate, and so Na and Da are simply the numera-
numerators of the rational functions respectively obtained. In fact, Na and Da are
polynomials in n of degree a — 1.
Setting P = n, у = n + I — a, and x = 1 in G.2), we find that
n jF^n + 1; n + 2 - a; 1) n n+1 n + 2
(n+ 1 - a) iF^n-n + 1 - a; 1) n-a + n-a+l+n-a + 2 + ...'
A1.5)
where n ф {0, — 1, — 2,...}. But by Kummer's theorem (Entry 21 of Chapter
10),
n xF1{n + 1; n + 2 - a; 1) n ^A - а; и + 2 - a; -1)
(n + 1 - a) iF^n; n + 1 - a; 1) ~ (n + 1 - a) ^A - a; n + 1 - a; -1)
Thus, A1.3) follows from A1.5) and A1.6).
From A1.1),
Ц.+1 (и+ 1 -aIFl(-a;n+ 1-a; -1)
Na 1Fi(l-a;n + 2-a;-l)
a-\ a-2
= n + 2-a +

118 12. Continued Fractions
where we have applied G.2) with /? = — а, у = n + I — a, and x = — 1.
This application of G.2) is valid by our remarks following G.2). For if a
is a positive integer, у = n+1 — аф { — a,—a—1,—a— 2,...}, since
n $ {0, -1, - 2,...}. This proves A1.4). П
By generalizing the proof above, we can easily prove that
n n +\ n + 2 a — I a — 2
=1-1-
n + 2 — a + n + 3 — a + ¦¦•
A1.7)
provided that not both a and — n are nonnegative integers.
Corollary 1. // n is not a nonpositive integer, then
n2 + n + 1 n n + 1 n + 2
Proof. Let a = 3 in A1.3). ?
Corollary 2. // n is not a nonpositive integer, then
n3 + 2n + 1 _ n n+\ n + 2
(n - IK + 2(n - 1) +T = n^4 + ~n^3 + n~^2 + ¦¦¦ '
Proof. Let a = 4in( 11.3). ?
Entry 12. If а Ф 0 and x ф —ка, where к is a positive integer,
1=* + q (x + aJ - a2 (x + 2aJ - a2 (x + 3aJ - a2
a + a + a + a +¦••'
First Proof. In Entry 22, put x = 1, a = 0, p = (x - a)/a, and у = (x + a)/a.
After simplification, we find that
x ~ a x - a (x + aJ - a2 (x + 2aJ — a2
x + a a + a + a + ••¦
Multiplying both sides by (x + a)/(x - a), we complete the proof. ?
Second Proof. In Entry 27, let x = 1. Then set у = 1 + 2xja and n = — 4.
We then find that
2x 4(x + aJ/a2-4 Цх + 2аJ/а2-4 4(x + 3aJ/a2 - 4
1+т = 1+ т—+ г—+ 2 +...
2 Ux + aJ -a2 (x + 2aJ -a2 (x + 3aJ - a2
= 1 + -¦
a a + a + a + ¦••

12. Continued Fractions 119
say. Thus, x = X. Lastly,
a + x a + x
a + x a + X'
which is the desired formula. ?
Third Proof. For x = 0, the result is trivial. Thus, assume that x Ф —ka,
0 < к < oo. A sequence of tails for A2.1) is given by
The sum in B.4) is then equal to
tiMV x + (nl)a / % ' x
By the theorem of Section 2, we conclude that the continued fraction in A2.1)
converges to g<0> = 1. ?
Entry 13. Let a, b, and d be complex numbers such that either d / О, Ь ф —kd,
where к is a nonnegative integer, and Re((a — b)/d) > 0, or d Ф 0 and a — b,or
d = 0and\a\ < \b\. Then
a ab (a + d)(b + d) (a + 2d)(b + Id)
a + b + d - a + b+ 3d - a + b + 5d -
First Proof. For this proof, we shall assume the first set of conditions on a,
b, and d. We shall also need to assume that (a + kd)(b + kd) Ф 0, for each
nonnegative integer k. Let pk = b + kd, к > 0. Then
Л.1.Л.О MW (a + (П + l)d)(b + (П + l)d) ^^
р„ = a + о + Bп + l)a , n > 0.
Pn+l
Writing р„ = xjxn+l, n > 0, we may write the preceding formula in the form
х„ = (a + b + {In + l)d)xn+i - (a + (n + \)d)(b + (n + l)d)xn+2.
Setting a + nd = yjyn+1, n > 0, we easily see that the same recurrence formula
is satisfied by yn.
Now if x0 = 1,
_ ^и + 1 Хп
1
(n- \)d)---b
Similarly, if y0 = 1,
Thus, under our assumptions, xn/_vn tends to 0 as n tends to oo.

120 12. Continued Fractions
We now apply a theorem in Perron's text [3, p. 97, Satz 2.46, C] to deduce
that, under our hypotheses,
x0 , , , л [a + d)(b + d) (a + 2d)(b + Id)
(a + 3d)(b + 3d)
- a + b + Id '
Now take the reciprocal of both sides above and then multiply both sides by
ab to obtain the proposed continued fraction representation. ?
Second Proof. As in our first proof, we assume that the first set of hypotheses
holds. In Entry 20, let a = b/Bd), ft = a/{2d), and у = (a + b + d)/{2d). By our
hypotheses, each of the two hypergeometric series in Entry 20 converges at
x = — 1. By the remarks following Entry 20, we may let x = — 1 in Entry 20.
After a slight amount of manipulation, we find that
fa + d a + 2d a + b + 3d
ab \ 2d 2й 2a
a + b + d fa + d a a + b + d
FT'2d'^2d ;
ab (a + d)(b + d) (a + 2d)(b + 2d)
a + b + d- a + b + 3d - a + b + 5d -•¦¦
If we now apply Gauss's theorem (Entry 8 of Chapter 10) to each of the
hypergeometric series above, we find that the left side of A3.2) becomes
ab V 2d
(b + 2d\
2d j \ 2d
which completes the proof. ?
Third Proof. Assume that either of the first two sets of hypotheses is valid.
Assume that a + nd, b + nd Ф 0 for each nonnegative integer n. Then
-{a + nd), ifn>l,
is a sequence of tails for A3.1). The series in B.4) then becomes
« * b + (n - \)d (b a
?пЧ1 1
which is known to diverge to oo if Re((b — a)/d) > 0. Thus, by the aforemen-
aforementioned theorem, the continued fraction in A3.1) converges to g@) = a. ?

12. Continued Fractions 121
Our last proof is due to Jacobsen [3], who proves Entry 13 under all the
given hypotheses on a, b, and d.
Entry 14. // alt a2,..., a2n and x are arbitrary complex numbers, then
T
о4о5 a2
«1
x -
a2
\- 1 +
«i

x -
a4
h 1
+ ¦¦¦ +
a2a3
x + a2 - x + a3 + a4 - x + a5 + a6 - x + a2n_x + a2n
Proof. We shall induct on n. For n = 1, it is easy to verify that the proposed
identity is valid. Now assume that A4.1) is true with n replaced by n — 1 for
any fixed integer n > 1. Let
a3 a4 a§ a^ ^2n
~x + l +x + l +•¦•+ 1
and
i±a* a^a7 #->„_-> а7И_1
B =
x + a5 + a6 - x + a7 + a8
Then, by induction,
a
3 a4 a2n
x + 1 + x + 1 + ••• + 1 x + I + A
a% a2 aj
a2a
2
a7-
2a3
x + a4 — В x + a3 + a4 — В
which completes the proof. ?
Entry 14 is actually a finite form of a special case of a classical result.
Suppose that К(а„/Ь„) has approximants /„, n > 1. Then the even part of
К(а„/Ь„) is a continued fraction with approximants/2n,n > 1. If Ь2„ ф 0, n > 1,
the even part is given by (up to equivalence transformations)
axb2 a2a3b4 a4a5b2b6
a2 + b{b2- a3b4 + Ь2(а4 + b3b4) - asb6 + b4(a6 + b5b6)
a2n+1b2n+2 + Ь2„(а2п+2 + b2n+lb2n+2) -¦¦•
See, for example, the treatise of Jones and Thron [1, pp. 41,42].
Preece [2] established a slight generalization of Entry 14 in infinite form,
that is, a special case of A4.2). Rogers [2] also proved a corollary of A4.2).
We shall establish two renditions of Entry 15. We first regard A5.1) as a
formal identity and provide a proof that is probably similar to that found by

122 12. Continued Fractions
Ramanujan. By a "formal identity" we mean that the two continued fractions
in A5.1) below correspond to the same power series Ya=o ckx~k- In the second
version, we offer conditions under which A5.1) is valid as an identity between
two convergent continued fractions. We are very grateful to L. Jacobsen for
the latter version.
Entry 15 (First Version). As a formal identity,
ar + h a^ a2 + h a2 at a^ + h a2 a2 + h
1 +x+ 1 + x + ¦¦• ~ 1+ x +1+ x +¦¦¦'
A5.1)
Proof. Let
ak + h a* ak+l + h ak+1
1 + x + 1 + x +¦¦¦
Denoting the left side of A5.1) by F, we find that
+h h(F2 + at) + ax(F2 - h)
F =
1 + aJF2 F2 + a,
ax{F2-h)
= h + -^ - = h +
(F2 -h) + (ax + h)
A5.2)
F2 + a, (F2 -h) + (a, + h)
F2-h
Next, for к > 2,
+
Fk+1 ) Fk+i-h
Now use A5.3) successively in A5.2) beginning with к = 2. This completes
the proof since both continued fractions are regular C-fractions and thus
correspond to uniquely determined power series (Jones and Thron [1, p. 222]).
?
Entry 15 (Second Version)
(i) // the left side o/A5.1) converges to F, say, then the odd part of the continued
fraction on the right side of A5.1) converges to F, and conversely. In

12. Continued Fractions 123
particular, this means that the identity holds in the usual sense if both
continued fractions converge.
(ii) // the left side o/A5.1) converges to F, then the right side o/A5.1) converges
to F, except possibly if h is a limit point of { — B2k/B2k-i}, where, in the
notation A.3) and A.4), An/Bn denotes the nth approximant for the continued
fraction on the left side, and conversely.
Proof. In the notation A.3) and A.4), let the left side of A5.1) have approxi-
mants AjBn and the right side have approximants CJDn. Then Ao = B^ =
О, Л_! = Bo = 1, D_! = 0, C_! = Do = 1, and Co = h. Straightforward calcu-
calculations show that
d = At = ax + h, Di = Bx = 1,
A2 = x(a1 + h), B2 = x + au
C2 = A2 + A1h, D2 = B2 + B1h.
We shall now show, by induction, that for к > 1,
^24-1 = ^2*-l> ^2k-l = B2k-i,
A5.4)
Cik = Auc + A2k-\h, D2k = B2k + Bjk-xh.
For к = 1, each of the last four equalities has been demonstrated. Proceeding
by induction, we find that
and
C2k
= XC;
= xA.
= {x-
-l =
=
Vh)A
C2k-2
A2k-2
-(fl*H
2k-l ^
+ akC2
:*-з — A2k-2
Ь Л)Л21к_3 = .
h n)C2t-2 = X42*-1 •
h h)A2k_2 + /i(y42t_i
1- afcX2)i
-2 = /*2* + *
+ Л
+ (a*
-A
A
2k-3h +
+ h){A
2k-l)
akA
2k-2
2k- 3
+ A2k.3h)
The remaining two equalities in A5.4) may be established in a similar manner.
Both conclusions of the second version of Entry 15 now follow from A5.4).
It also follows that the left side of A5.1) converges generally to F if and only
if the right side converge generally to F. For the concept of general conver-
convergence, see the paper by L. Jacobsen [2]. ?
Entry 16. // neither m nor n is a negative integer, then
f
h (m + k)(n + k)
1 (m + lJ(n + IJ (m + 2J(n + 2J
(m + l)(n +1)+ m + n + 3 + m + n + 5
(от + 3J(n + 3J
+ m + n + 1 +¦¦¦'

124 12. Continued Fractions
Proof. We shall employ the corollary presented in Section 1. Letting ak
(- lf+1/(m + k){n + k) and letting r tend to oo, we find that
& (m + k)(n + k)
(m + I) (" + I) (m + 2Yl{n + 2)"
(m + i)~l(n + lpC" + 3)~*(n + 3)
(m + 2)~l(n + 2) - [m + 3)-1(n + 3)~:
1 (m + \f(n + IJ
(m + l)(n + 1) + (m + 2)(n + 2) - (m + l)(n + 1)
(от + 2J{n + 2J
+ (m + 3)(и + 3) - (m + 2)(и + 2) +•••'
from which the proposed identity readily follows formally.
The continued fraction converges for all m, n such that neither m nor n is
a negative integer (Jacobsen [3, Theorem 2.3]). Since the series also converges
in this domain, the identity is proved. ?
The equality in Entry 17 refers only to the correspondence of the two sides;
neither side need converge.
Entry 17. Write
1 alX a2x a3x »
Т + ~Г + Т + Т + -"ДЛ(~Х)' A7Л)
w/iere Ло = 1. Lei
Pn = ala2--an_1(a1 + a2 + ••• + а„), и > 1.
T/ien
P3 = A3-a1A2,
a2
In general, for n > 1,
^= Z (-1)Ч(пМ„-*> A7.2)

12. Continued Fractions 125
where q>0(ri) = 1 and (pr(ri), r > 1, is defined recursively by
(p,(n + 1) - q>r(n) = a^q,,-^ - 1). A7.3)
First Proof. Let С„ = С„(х) and В„ = В„(х) denote the numerator and
denominator, respectively, of the nth convergent of the continued fraction
A7.1). Then, from A.3) and A.4),
d = C2 = 1, С, = С„_! + an_!xCB_2, 1
В, = 1, B2 = 1 + alX, В„ = В„^ + а„-1ХВ„-2, п > 3.J
By induction, it is easily seen that C2n-i> Qn> and ^2n-i are polynomials in x
of degree n — 1, while B2n is of degree n, where n > 1. Thus, for n ^ 1, set
[n/2]
Д.М = I A(" + l)xk. A7.5)
k=O
We make the convention that fik(n + 1) = 0 if k> [n/2]. From A7.4), it is
obvious that po(n + 1) = 1 for each n > 1. Using A7.5) in the recursion for-
formula for В„ given in A7.4) and equating coefficients of xr, we readily deduce
that
IUn+l)-Pr(n) = an-iPr-i(n-l), r>l. A7.6)
Thus, by A7.3) and A7.6), we see that q>r(n) and Д.(п) satisfy the same recursion
formula. Since furthermore q>0(n) = 1 = f}0(ri), we conclude that q>r(n) = Pr{n),
r > 0. Also note that cpr(n) = 0 if r > [n/2].
Put
where n > 0 and a0 = 1. We shall show, by induction, that (see also Rogers'
paper [2, p. 72, Eq. A)])
Е0В„-Св = (-1Г?0?1---?ях", п>1. A7.8)
Since, by A7.7), Eo = 1/A + xEJ, A7.8) is easy to establish for n = 1. Assume
now that A7.8) is valid for each nonnegative integer up to and including n.
Then, by A7.4),
^oAi+i — Cn+i = Е0В„ — С„ + а„х(Е0Вп_1 — С,^)
= (-1Г1?0?1---?„?и+1х»+1,
and so the induction is complete.

126 12. Continued Fractions
Write
00
Л
?0?i•••?.= L ek(n)xk. A7.9)
Setting x = 0, we find that
eo(n) = e1fl2--a». A7.10)
Next rewrite A7.9) in the form
^ а1Я2'"Я„= > ek(n)x .
1 + x?t 1 + x?2 ' + xEn+i 1=1
Dividing both sides by x and then letting x tend to 0, we deduce that
ei(n)= -a!a2--an(oi + a2 + •¦• + an+1)= -PH+1> A7.11)
for each nonnegative integer n.
In A7.8) replace и by и - 1. Then, by A7.1), A7.5), A7.9), A7.10), and A7.11),
j=0 *=0
Equating coefficients of x", n > 2, yields
which is precisely A7.2). Since the case n = 1 of A7.2) is readily verified, the
proof is complete. ?
Essentially the same proof that we have given above was independently
and almost simultaneously discovered by Goulden and Jackson [2]. They [2]
have also found a beautiful combinatorial proof of A7.3) by enumerating
certain paths. Using a result of E. Frank [1], P. Achuthan and S. Ponnuswamy
[1] have given a very short proof of Entry 17(i).
Before proceeding further, we shall find an exact formula for %(n), defined
by A7.3).
First, it is not difficult to show that
n-2
<jox(n) = ? a,-
and
<Pi{n)= Z_ ataj.
3<j<n-2
We shall show by induction on к that

12. Continued Fractions 127
%(")= .Z._ ahah--aik. A7.12)
l<jk<n-2
We have already indicated that A7.12) is true for к = 1, 2. Proceeding by
induction and employing A7.3), we find that
<Pk(n) ~ <Pk(n - !) = afi-2 Z aHah ''' aJr-i'
q>k(n - 1) - %(n - 2) = an_3 Z ал ah" ajV- >
Adding together all the equalities above, we deduce A7.12). This completes
the proof of the desired exact formula for q>k(ri).
Rogers [2] has expressed cpk{ri) by a determinant.
We are extremely grateful to G. E. Andrews for providing us with the
following elegant, second proof of Entry 17. In fact, this proof was found prior
to the proofs given and mentioned above. The first part of Andrews' argument
was anticipated by De Morgan [1].
Second Proof of Entry 17. We first obtain a recursion formula for the
coefficients Ak, к > 0. In order to do this, we introduce auxiliary coefficients
Ak, k>0, which we now define. Of course, each coefficient Ak can be written
in terms of au a2,.... We define Ak by the same expression for Ak except that
the subscript of each a, appearing in Ak is increased by 1. For example, since
A2 = a\ + a1a2, we define A2 = a\ + a2a3.
Now, by A7.1),
У (-l)kAkxk = 1
K k
1
k=0
where A-t = - \/a1. Multiply both sides of the extremal equation above by
the denominator on the right side and equate coefficients of x" on both sides
to deduce that, for n > 1,
n-l _
Z aiAkAn-k-i =0.

128 12. Continued Fractions
or
Ап = \а1АкАп.к^, и>1, A7.13)
which is the recurrence formula that we sought.
We now show that
An is a homogeneous polynomial of degree n in
the noncommutative variables a1,a2,..., а„,
where the subscripts jx,j2, ¦¦-,}„ of the monomials
comprising An are precisely those sequences of positive
integers starting at 1 for which jk+l —jk < \,jk > 1.
A7.14)
In order to make clearer the assertion above, we record the following
examples:
Ai =au
A2 = a1a2 + axax,
A3 = a1a2a2 + a1a2ai + a1a2a1 + ala1a2 + a1a1a1.
We now prove the assertion A7.14) by inducting on n. By using A7.13), we
easily verify that A7.14) is true for n = 1,2, 3, as indicated above. Assume that
A7.14) is true up to but not including a specific integer n. Let A* denote the
polynomial described by A7.14). We shall show that A* is equal to the
right side of A7.13). Thus, A* = Am which completes the induction. Let us
divide the monomials comprising A* into n classes. The fcth class, 0 < к <
n — 1, consists of all monomials in A* wherein the second appearance of ax
is the (k + 2)nd term in the monomial. (Recall that at begins each monomial.)
Thus, the entries of the /cth class are produced in the following manner. Start
with ax, adjoin a string of к ajs, j > 2, that starts with a2 and follows the
appropriate subscript rules, and lastly adjoin a string of n — к — 1 a/s that
starts with al and follows the prescribed subscript rules. But the entries for
the string of к terms are generated by Ak and the entries for the remaining
n — к — 1 terms are generated by А„-к_^, by induction. Hence, the monomials
in the /cth class are generated by alAkAn-k-1. Summing on k, 0 < к < n — 1,
we find that
n-l _
An = L, aiAkAn~k-i>
k = 0
which, by A7.13), completes the induction.
We now have a combinatorial interpretation A7.14) for An. After finding
combinatorial interpretations for Р„ and (pk{n), we shall use a sieving process
to establish A7.2).
Let us say that a word of the type generated by An, that is, ahah-¦ ¦ ain,
where j] = 1 and7t+1 — jk < 1, with jk > 1, has an "internal drop" if jk+1 —
jk ф 1 for some k, 1 < к < n — 1. Then we see that Pn is the polynomial in al5
a2,..., an composed of all words without internal drops.

12. Continued Fractions 129
From A7.12), observe that %(n) is a homogeneous polynomial of degree к
in the noncommuting variables a1,a2,---, an-2 wherein the subscripts of each
monomial a^a^ ¦ ¦ ¦ ajk satisfy the inequalities ji+l — ;',¦ > 2, 1 < i < /c — 1.
We now begin the sieving procedure. We first examine А„. Recall that an
internal drop occurs when ji+l — jt # 1 and 1 < i < n — 1. Let us call ajk the
"top of the last internal drop" if к is maximal for internal drops; that is, if
ji+i — ji ^ 1, 1 < i < " — 1» then jfc+1 — jk ф 1, 1 < к < n — 1, and i < к. The
top of the last internal drop must be one of the letters al,a2, ¦¦¦, an-i>smce
neither а„_! nor an can be far enough to the left in a word to be at the top of
an internal drop.
In order to eliminate all words from An with internal drops, we take the
words from An_y and insert a,-, 1 <j<n — 2, in the last position where it
forms the top of an internal drop. Thus,
Ля-(а1+а2 + - + an_2)An_x A7.15)
does not possess any internal drops. (Note that we have written A7.15)
commutatively; the correct noncommutative expression would have a,-, 1 <
j < n — 2, inserted as described above.) Unfortunately, there are words in
A7.15) that were not originally in А„. These words arose when the insertion
of an uj produced a subscript increase greater than or equal to 2 from the a,-
immediately to the left of the inserted a-y Of course, we must eliminate these
undesirable words. We do this by taking the words of Л„_2 and inserting pairs
ага} with; — i > 2 so that a, is at the top of the last internal drop. Hence,
An - ъШ.-! + <РгШп-г A7-16)
does not possess internal drops. (Again note that A7.16) is a commutative
representation of what is really a noncommutative polynomial in аь а2,...,
а„.) Unfortunately, we have now introduced some new words that were not
originally under consideration. These new words have triples а^ак with
j — i > 2 and к — j > 2 and with ak at the top of the last internal drop.
We continue the process described above by induction. At each stage we
must introduce a term
to compensate for unwanted terms introduced at the previous stage. For-
Fortunately, <pk(n) = 0 for к > n/2, which is evident from A7.12). Thus, the sieving
process terminates, and we reach the desired formula A7.2). ?
Among others, Muir [1] and Rogers [2] have studied the problem of
deriving a continued fraction expansion from the coefficients of a power series.
Both De Morgan [1] and Rogers [2] have commented on the fact that it is
extremely more difficult to determine the power series coefficients Ak, 0 < к <
oo, from a continued fraction of the form A7.1). Ramanujan's Entry 17 is a
fascinating contribution to this more recondite converse problem.
By a theorem of Euler [1] (Jones and Thron [1, p. 37]) (see also A.2)),

130 12. Continued Fractions
where Bk = Bk(x) is given by A.4) and A7.5). Thus, (- \)"А„ is equal to the
coefficient of x" in
h Bk(x)Bk+l(x)
Obtaining a general formula for А„ in this manner seems hopeless.
However, a very complicated formula for An can be established combina-
torially by counting planted plane trees with respect to their heights in two
different ways. For a nice exposition of this proof, see the book of Goulden
and Jackson [1]. See also a paper of Flajolet [1].
Corollary (i). Write
1 atx a2x
= X Ak{-x)\ A7.17)
1 + b^x + 1 + Ъ2х + 1 + b3x +
where Ao = 1. Define
Р„ = а1а2---ап-1(а1 + bl + а2 + b2 + ¦¦¦ + ап + Ь„), и > 1.
Then, for и > 1,
Рп = X (~ 1)*%(пМл-ь
к = 0
w/iere <ро(и) = 1 and <pr(n), г > 1, is defined recursively by
<РГ(П + 1) — <pr(n) = fen<pr_!(n) + a,,-!^,.-!^ — 1).
As with Entry 17, Goulden and Jackson [2] independently and simultane-
simultaneously discovered the proof that we found and which is recorded below.
Goulden and Jackson [2] have also derived a combinatorial proof. Yet
another proof of Corollary (i) has been found by Achuthan and Ponnuswamy
[1]. McCabe [1] has established an identification of continued fractions of
the type A7.17) with power series of the form Yjk=o^k/xk- Since the proof
below is very similar to the first proof of Entry 17, we give only a brief sketch.
Proof. Let С„ = С„(х) and Bn = Bn(x) denote the numerator and denomina-
denominator, respectively, of the nth convergent of the continued fraction A7.17). Then
C1 = l, C2=l+b2x, С„ = A + Ь„х)С„_1+а„_1хС„_2,1
B1 = \+b1x, B2= 1 +(a1 +bi +b2)x + b1b2x2, )¦ A7.18)
Д. = A +Ь„х)В„-1 +an_!xBn_2,
where и > 3. Observe that С„(х) has degree и — 1 and Bn(x) has degree n,n > 1.
Thus, put
Д,М = ? Pk(n + 1)*\ n>\. A7.19)

12. Continued Fractions 131
By substituting A7.19) into the recursion formula for Bn(x) in A7.18) and
equating coefficients of xr, we deduce that /?r(n) = <pr(n), r > О, и > 2.
The remainder of the proof is exactly parallel to that of the first proof of
Entry 17. ?
Corollary (ii). Let Bn(x) be defined as at the beginning of the proof of Entry
17. Then, for n > 1,
[»/2]
Bn(x) = ? %(n + l)x*.
t = 0
Proof. Corollary (ii) was established in the course of proving Entry 17. In
particular, recall that f}k{n) = q>k{n) and consult A7.5). ?
Example. We have
Proof. Ramanujan evidently intends this example to be an illustration for
Entry 17. In the notation of Entry 17,
_ 1 _3 _ 5 _ \1 _23
1395
°6 ~ ~6256'
Squaring 2^1B. \\ U x), we find, after some laborious computing, that
A ~-1 A -1-1 A - П А -™ А --**
Л1- 2> Л2-32' 3~ 64' 213 ' 214 '
42631
Lastly,
-]Р-ПР-3Р -291 P - Ш and
Л~' Рг~32' Рз~2' П~2^' F5-~214' ad
32337
All these calculations are in agreement with Entry 17, and so A7.22) is indeed
correct. ?
Entry 18. Suppose that x and n are complex numbers such that either x ф
[— 1, 1], or x = +1 and Re и Ф 0, or n is an integer. Then
(x+ l)"-(x- l)"_n и2-12 и2-22 и2-32
(x + 1)" + {x- 1)" ~ x + 3x + 5x + 7x + • • • ' ' ^

132 12. Continued Fractions
First Proof. If we replace x by 1/x in Perron's book [3, p. 153, Eq. (9)], we
obtain a continued fraction representation easily found to be equivalent to
A8.1). By Perron's proof, A8.1) is valid for all complex numbers x outside
[-1,1].
Now suppose that x2 - 1 and n ф 0. If Re и > 0, the left side of A8.1) is
continuous for |x| > 1 and equals +1 at x = +1, respectively. For Re и > 0,
the continued fraction on the right side of A8.1) converges locally uniformly
with respect to x for x > 1 and x < — 1. Thus, by the uniform parabola
theorem (Jacobsen [3]), A8.1) is valid for x = ±1 and Re и > 0. Since both
sides of A8.1) are odd functions of и for x = + 1, A8.1) is valid for x = ±1
and Re n < 0 as well.
Lastly, suppose that и is an integer, and so both sides of A8.1) are rational
functions of x. We already know that A8.1) holds for all x ф [— 1, 1]. Thus,
by analytic continuation, A8.1) holds for all complex x. П
Perron's derivation of Entry 18 arises from Entry 20.
Second Proof. Let
T(^(mx +т-п+ 1))Г(^(отх - m + n + 1))
)
x + m + n+ 1))ГA(тх - m - и + 1))"
Replacing x by mx in Entry 33, with m > 0, we find that, for Re x > 0,
1 - g(m, n, x) _ mn (m2 - I2)(n2 - I2) (m2 - 22)(n2 - 22)
1 + g(m, n, x) mx + 3mx + 5mx + ¦ ¦ ¦
_n A - l/m2)(n2 - I2) A - 22/m2)(n2 - 22)
~x+ 3x + 5x +•••'
A8.2)
Now let m tend to oo in A8.2). By using an asymptotic formula for the
quotient of Г-functions, Lemma 2, Section 24 of Chapter 11, or Stirling's
formula, we find that
lim g(m, n, x) = (x + l)~"(x - 1)".
m->oo
For x exterior to (—oo, 0], by the uniform parabola theorem (Jacobsen
[3, Theorem 2.3]), the continued fraction on the right side of A8.2) converges
uniformly with respect to m in a neighborhood of m = oo. But by Perron's
work [3, p. 153], or by the parabola theorem, the continued fraction in A8.1)
converges for x ф [-1, 1]. Thus, A8.1) holds for Re x > 0 and x ф @, 1]. By
analytic continuation and our argument in the first proof, the domain of
validity can be extended to that indicated. ?
Entry 18 is due to Euler [7] and easily implies a continued fraction expansion
for (x + l)"/(x - 1)" due to Laguerre [1] (Perron [3, p. 153, Eq. A0)]).
If Vn denotes the left side of A8.1), then Ramanujan remarks that Vn +
l/Vn = 2/V2n, a fact that is easily verified.

12. Continued Fractions 133
Corollary 1. Let x be any complex number outside the cuts (—ice, — i] and
[(, ioo). Then
г _x x2 BxJ Cx?
tan x-T + y + -y- + ^ + ...-
For a proof, see Perron's text [3, p. 155]. Early proofs of Corollary 1 were
given by Lambert [2], J. L. Lagrange [1], and Euler [7].
Corollary 2. Let x be any complex number outside the cuts (—oo, —1] and
[1, да). Then
l+x 2x x2 BxJ CxJ
Log
1-х 1—3—5—7—
For a proof, see Perron's treatise [3, p. 154]. Corollary 2 is due to Euler
[7]. For an application of Corollary 2 to product-weighted lead codes, see a
paper of Jackson [1].
Corollary 3. For any complex number x,
x x2 x2 x2
t
Corollary 3 was initially discovered by Lambert [1], [3]. A proof may be
found in Perron's book [3, p. 157].
Corollary 4. For any complex number x,
ex — 1 x x1 x1 x2
ex + 1 2 + 6 + 10 + 14 + • • '
Corollary 4 is due to Euler [5] and a proof may be found in Perron's text
[3,p. 157].
Entry 19. // n and x are arbitrary complex numbers, then
n0Fi(nix) ия-№у/х) и+«+1 +И + 2 + ---'
where Jv denotes the ordinary Bessel function of order v.
First Proof. By a theorem of Euler [5] (Perron [3, p. 281, Satz 6.3]),
c | a a a pF^c/d; a/d2)
c + d+c + 2d+c + 3d+--- C0Fx(cld + 1; aid2)"
where d + 0. Let с = n, a = x, and d = 1 to find that
, xxx
П -\ — = n-

134 12. Continued Fractions
Taking the reciprocal of both sides above and then multiplying both sides by
x, we deduce the desired result. ?
Second Proof. This proof is similar to the proof above, but employs a "finite"
version of A9.1), namely, Entry 24. Simply let r tend to oo in Entry 24. After
multiplying both sides by x/n, we complete the proof, since both sides converge
for all x and n. ?
In fact, Entry 19 is classical; see, for example, the books of Wall [1, p. 349]
or Jones and Thron [1, p. 168].
Entry 20. // x is any complex number outside the interval f — oo, — 1], or if a,
j8,y-aory-0e{O, -1, -2,...}, then
а/?х2^(у-а,/? + 1;у+1; -х)
у 2
afix
у + y+\ + у + 2 + у + 3
(a + 2)(P + 2)x
у+ 4 + ••¦'
B0.1)
This result is very famous and is known as Gauss's continued fraction [1].
A proof may be found in the texts of Jones and Thron [1], Khovanskii [1],
Perron [3], and Wall [1]. The cases when the continued fraction terminates
are discussed by Perron [3, p. 151]. In the case у — /? = a + i, Entry 20 may
be extended to include x = — 1.
It might be mentioned that Gauss's continued fraction may be found in
Carr's book [1, p. 97], which was the most influential book in Ramanujan's
development. Recent work on Gauss's continued fraction may be found in
papers by Belevitch [1] and Ramanathan [1].
Entry 21. We have
— 2Fi(P+l i;y + l;-x)
/?x y(P + l)x Цу-Р)х (у + l)(P + 2)x 2(y + 1 - P)x
у + 1 + y + 2 + y + 3 + y + 4 +•••'
B1.1)
if either хф (-oo, -1], or ft, у or у - 0 e {0, -1, -2,...},
fix (P + l)x 1A + x) (P + 2)x 2A + x)
у + 1 + у + 1 + у +•••'
if either Re x > —\ and not both ft + 1 and у — /J lie
in{0, -1, -2,...}, or fie {0, -1, -2,...} and
у-РФ {0,-1,-2,...},
B1.2)

12. Continued Fractions 135
fix HP + l)x(x + 1) 2(P + 2)x(x + 1)
y + x(j}+l)-7 + l + x(jJ + 3)-y + 2 + x(/* + 5)-"-'
z/ е/г/ier Re x > - \ and not both fi + 1 and у — 0 lie
in{Q, -I, -2,...},or0e{O, -1, -2,...} and
у-/»#{<),-1,-2,...}.
Proof. The expansion B1.1) follows from Entry 20 as follows. First, divide
both sides of B0.1) by a/?. Set p = 0. Then replace a by у - P - 1 and multiply
both sides by p. The conditions on the parameters in Entry 20 are translated
into the new conditions given for B1.1).
An indication of Ramanujan's proof is found in the first notebook (p. 217).
Let
Then
^^(/} + и;у + 1;х) (
7.
B1.4)
l-G
Now in Entry 20, replace f} by fi + 1 and у by у + 1 and then set a = y. This
yields
l-G y+1 2^(/* + 1,1;
0?+l)x l(v-P)x
y+l+y+2+ y+3 + y+4
(y + 2)(j? + 3)x
+ у + 5 +
B1.5)
If we substitute B1.5) into B1.4), we complete the proof of B1.1).
We next prove B1.3). If Re x < 5, a + 1 and у — a are not both non-
positive integers, and f} + 1 and у — P are not both nonpositive integers, then
by a result of Norlund [1] (Perron [3, p. 286, Eq. A0)]),
2F1{a+l,p+l;y+ l;x)
у 2Fl(oc, P; у; х)
1 (« + !)(/? + l)(x - x2)
у - A + a + P)x + у + 1 - C + a + P)x
(a + 2)@ + 2)(x - x2)
B1.6)
Setting a = 0, replacing x by — x, and lastly multiplying both sides by jSx, we
complete the proof of B1.3). The cases when the continued fraction terminates

136 12. Continued Fractions
are established by a familiar argument, since both sides are then rational
functions of x.
Applying Entry 14 and letting и tend to oo, or applying A4.2), we see that
B1.3) is the even part of B1.2). Since B1.2) converges for Re x > -\, the
identity follows.
Under certain conditions B1.1) can be extended to x = — 1 and B1.2) and
B1.3) to x= -i ?
Corollary 1. For every complex number x, we have
X
и
X
и —
X
их
и +
1
+
X
и +
X
2
(и
— и
+ 1)х
+ з
2х
+
2х
п 4-
4_...
Зх
n — x+n+1— x + n + 2 — x + n + 3 — x + • •¦
Proof. To prove the first equality, replace у by и and x by — x/ji in B1.1).
Letting ,S tend to oo, we easily deduce the desired result.
To prove the second equality, employ B1.3) and proceed in precisely the
same manner as above. ?
Corollary 2. // x is any complex number, then
„ ,, , , , 2i 3i 4x 5x
ЛA + 1I+
Proof. Let и = x in the second equality of Corollary 1. ?
Entry 22. Assume that a, /?, and у are complex numbers such that not both /? + 1
and у — /? belong to {0, — 1, — 2,...} and not both — a. and a + у + 1 are in
{0, -1, -2,...}. Suppose that either \x\ < l,or/3e {0, -1, -2, ...},orx = 1
and Re(y - a - ft - 1) > 0. Then
2F1{-a,P;y;-x)
px {0 + l)(a + у + l)x
у - (a + P + l)x + у + 1 - (a + p + 2)x
(/? + 2)(a + у + 2)x
+ у + 2 - (a + J? + 3)x + • • •'
Proof. Since by Entry 19 of Chapter 10,
2^(в, b; c; x) = A - хГв~ь 2^ (с - a, b; с; -^-),

12. Continued Fractions 137
we may write B1.6) in the form
2FXIу - a, p + 1; у + 1;
х- 1
7A - x) (
2FAy- a., p; y;
x- 1,
x (a + !)(/? + l)(x - x2)
- A + a + 0)x + }' + 1 - C + a + P)x
(a + 2)(fl + 2)(x - x2)
provided that Re x < \, not both /? + 1 and у — P belong to {0, — 1, — 2,...},
and not both a + 1 and у — a belong to {0, — 1, — 2,...}. Letting и = x/(l — x),
we find, after simplification, that
u2F1{y-a,p + l\y + 1; -и)
у2рЛу-<х,Р;у, -и)
и (а + !)()} + \)и
-A+« + Р)и + (у + Щи + 1) - C
(а + 2)(Р + 2)и
+ (у + 2){и+ 1) - E + а + у9)и + ¦••'
provided that \u\ < 1. Replacing а by а + у in the foregoing equality, we
readily complete the proof of Entry 22 for |x| < 1.
Lastly, observe that the left side of Entry 22 is analytic in a neighborhood
of x = 1. For x = 1, the continued fraction converges to an analytic function
of a, /?, and у provided that Re(y — а — p — 1) > 0, by the uniform parabola
theorem (Jacobsen [3, Theorem 2.3]). This then completes the proof of Entry
22 for x = 1. ?
Perron [3, p. 306] attributes Entry 22 to Andoyer [1], and, as we have seen,
Entry 22 is equivalent to Norlund's result B1.6). However, R. Askey [2] has
pointed out that Entry 22 is really due to Euler [6], [2]. A somewhat more
detailed discussion of Entries 20-22, along with the associated contiguous
relations, may be found in a paper by K. G. Ramanathan [1].
As with Entry 17, the equality in B3.1) below refers only to the correspon-
correspondence of the two sides, for neither side needs to converge.
Entry 23. Write, for each nonnegative integer n,
a/1 /7 '
bnx + bn+lx + bn+2x + ¦¦¦ k%
where An{0) = 1. Then cncn+l = an,
B3.1)

138 12. Continued Fractions
АпA) + Ап+1A) = ~ = —, B3.2)
cn+i an
AnB) + An+1B) = A2n(\),
AnC) + Ая+1{3) = А„{ЩАЯ{2) - Ая+1{2)}, B3.3)
An{A) + An+lD) = 4,A) D,C) - Ля+1C)} - АяB)Ан+1B),
and, in general, for k>3,
A.(k) + Ая+1(к) = Ая(ЩАя(к - 1) - А„+1(к - 1)} - "f An(j)An+1(k - j).
j=2
B3.4)
Proof. From B3.1),
^ = ся? Ап(к)(-х)к,
bnx + cn+l^An+1(k)(-xf
k = 0
or
an =
? I
k=0 J k=0
= -cHbHfiAK{k-l)(-xf
k = l
+ cncn+1 f X лиGМ„+1(/с-7)(-х)к.
H = 0 j=0
Now equate coefficients of xk, к > 0, on both sides. For /c = 0, we find that
cncn+i = а„, and, for /c > 1, we deduce that
4W + An+1(k) = —An(k - 1) - "f An(j)An+1(k -j). B3.5)
Cn+1 J=l
Letting /c == 1, we immediately deduce B3.2). Using B3.2) in B3.5), we find
that, for к > 2,
An(k) + An+X(k) = {An{\) + Ая+1A)}А„{к - 1) - *? AniJ)An+1{k - j).
Upon simplifying the equality above, we deduce both B3.3) and B3.4). ?
Example. We have
/2x x 2x 3x 4x
/ __
2 + 3+4+
Proof. From Entry 47, for x > 0,
, ехГ(х + 1) Г00 / Л* 2х Зх 4х
1+ —x e '[1+-] dt=l+— — —
xx Jo \ x 2+3+4

12. Continued Fractions 139
Taking the reciprocal of both sides, we find that
,, , x x 2x 3x 4x
L(x) := ^, , „
It therefore remains to show that
L(x)= /—-^- + 0A), B3.6)
v n 3n
as x tends to oo.
Write
L(X) ie
When x is a positive integer, Ramanujan [4], [16, p. 324] derived an asymp-
asymptotic expansion for 6X as x tends to oo. Watson [3] later established the
expansion for general x > 0. See also the corollary to Entry 48 below and
Entry 6 of Chapter 13. Using this asymptotic series D8.4) and Stirling's
formula, we find that
L(x) =
n Ъп
as x tends to oo. Thus, B3.6) is established, and the proof is completed. ?
Entry 24. Let n and x be any complex numbers, and let r be any positive integer.
Let
Then
n x x x x f(n + 1, r — 1, x)
n + n+l+n + 2 + n + 3+--- + n + r f{n,r,x)
Proof. We shall induct on r. For r = 1,
. B4.1)
и + и + 1 n + x/(n +1) x
1 + ¦
n(n + 1)
and so B4.1) is established for r = 1.

140 12. Continued Fractions
Now assume that B4.1) is true when r is replaced by r — 1 for some fixed
integer r, r > 2. Then applying the induction hypothesis with и replaced by
n + 1, we find that
" + n + 1 f(n + 1, r - 1, x)
nf(n + 1, r - 1, x)
nf{n + 1, r - 1, x) + ~^~7f(n + 2,r-2,x)
n + 1
We are thus led to examine, for к > 1,
n(-r + k)k
x f(n + 2,r - 2, x)
B4.2)
(и + \\(-n - r\k\ (n + Щ-п - r)k^(k - 1)!
(-r+ *)*_! f л(-г + 2/c-l
(и + Щ-и - r)t_,(fc - 1)! К-и - г + fc -
( —Г + fcjt.t (w+fc)(-r + fc
(„ + i)t(_n _ r)/k_l(fc _ i). (-n-r + fc-
+ 1
~ (и + 1)*-,(-и - r)tfc! Ш-п - r)kkV
Hence,
лГ(и + 1, r - 1, x) + -4т/(и + 2, r - 2, x) = i/(n, r, x). B4.3)
и + 1
Substituting B4.3) into B4.2), we complete the induction. ?
Entry 24 is a rather remarkable result, for it gives a continued fraction
expansion for the quotient of hypergeometric polynomials,
/1 -r r
2*3 ^—, -~; ~r,n+ 1, -r -n;x
2*31—2 ' ~2' ~Г~ ' "r~n;x
Entry 25. Suppose that either n is an odd integer and x is any complex number
or that n is any complex number and Re x > 0. Then
ГЩх + п+ 1))ГЩх - n + 1))
r(i(x + и + 3))ГШх - и + 3))
_4_ «2-l2 «2-32 n2-52
x — 2x — 2x — 2x — • • •

12. Continued Fractions 141
Entry 25 is originally due to Euler [6, Sec. 67]. Stieltjes [2], [4, pp. 329-394]
derived Entry 25 from Entry 22. Other proofs have been found by Perron
[3, p. 35] and Ramanathan [1]. The hypotheses in these proofs are stronger
than those we have given. Proofs under the stated hypotheses have been
derived by Jacobsen [3] and Masson [3].
First Proof. We offer another proof which is based upon Entry 39, for
Re x > 0, or for either и or <f in {+ 1, ±3, + 5,...}. First, rewrite Entry 39 in
the form
2 lln^! {2-fl з2- з2-/2
or
1
8/P + i(x2 + /2 - n2 - 1)
X2 - 1 + 1 + X2 - 1 + 1 + X2 - 1 + • • •
1 A2-п2)A2-/2) C2 - «2)C2 - /2)
~ x2 - n2 - x2 - /2 - n2 + 9 - x2 - f2 - n2 + 33 - • • •'
by Entry 14. Now take the reciprocal of both sides above and then solve for
P, which again involves taking reciprocals. Hence,
8 (l2-n2)(l2-/2) C2-rc2)C2-/2)
Р
Replacing x by x + t, we find that, either for Re(x + /) > 0, or for n ox
belonging to { + 1, ±3, +5,...},
Щ(х + п+ 1))ГЩх - и + 1)) r(j(x + 2/ + П
(^(х + и + 3))r(i(x - и + 3)) r(i(x + 2/ + и + 3))r(i(x + 2/ - и + 3))
i(x2 + 2x/ - n2 + 1) - x2 + 2x/ - n2 + 9
C2 - n2)C2 - /2)
- x2 + 2x/ - n2 + 33 - ¦ • •
8 (I2 - n2)(l//2 - 1)
iBx + (x2 - n2 + 1)//) - 2x + (x2 - n2 + 9I f
C2 - n2)C2//2 - 1)
- 2x + (x2 - n2 + 33)// '

142 12. Continued Fractions
Now let ? tend to oo. By using the reflection formula for the gamma function
and Stirling's formula, we deduce that
ГЩх + 2/ + П+ 1))ГЩх + 2/ - n + 1))
Л™ ГШх + It + n + 3))r(i(x + 2^ - n + 3)) '
and so Entry 25 readily follows, since the continued fraction above converges
uniformly in a neighborhood of t = oo under the stated hypotheses. ?
We next offer another proof of Entry 25 that is due to D. Masson [3]. In
[1]> [2], and [3], Masson employs second-order linear recurrence relations
and a theorem of Pincherle [1] to represent a general class of continued
fractions by quotients of hypergeometric functions. He also determines the
rate of convergence of the continued fractions and establishes connections
with several types of orthogonal polynomials. However, we shall not discuss
the latter topics here.
Consider the recurrence relation
Xn+1 - bnXn - anXn.x = 0, B5.2)
where an = — (an2 + bn + c) and bn = z — dn; here, a, b, c, and d are constants.
A solution A^s) is said to be subdominant if for any other linearly independent
solution JfJ") of B5.2),
lim X^X™ = 0.
Pincherle's theorem [1] then states that K(an/bn) converges if and only if there
exist linearly independent solutions X{ns) and X^] of B5.2), as described above.
Moreover,
K(an/bn)=-X[°yxp. B5.3)
We now quote Masson's primary theorem [2], [3] for us.
Theorem 1. Let a, d2 — Aa + 0. Then B5.2) has linearly independent solutions
where
fi = (d2- Aa)vl, - n/2 < arg ц < тс/2,
and a and f} are defined by
a(n + <x)(n + /0 = an2 + bn + с

12. Continued Fractions
143
Moreover, there exists a subdominant solution if and only if
Re
Re
a + b\ d_ z_
la~)ji+ji
and it is given by
(Х„+, if Re(d/n) <Oorif Re{d/n) = 0 and Re(y+ - y~) > 0,
{X;, if Re{d/n) >0orif Re{d/n) = 0 and Re(y+ - y~) < 0.
We shall apply Theorem 1 to prove the following theorem from which
Entry 25 follows as a corollary.
Theorem 2. // ± Im z > 0 and и е С, then
1
CF:=-
z —
= 2
\
B5.4)
Proof. For brevity, we set 6 = 6± and у — y±.
Comparing the left sides of B5.3) and B5.4), we see that a = 1, b = — 1,
с = A — n2)/4, and d = 0. In the notation of Theorem 1, we then find that
ц = 2i, S = i у = ±2/Bi), a = (и - l)/2, and y? = -(n + l)/2. If ±Im z > 0,
then by Pincherle's theorem and Theorem 1, there exists a subdominant
solution of B5.2) such that
-A2-п2)/4 C2-n2)/4
z — z — ¦ •¦
Since Zj = zX0 — afiX-i, we may rewrite B5.5) in the form
X$
where, by Theorem 1,
= ±-
B5.5)
B5.6)
B5.7)
It remains to evaluate this quotient of hypergeometric functions.
Recall that (Erdelyi [1, p. 104, Eq. E1)], Bailey [4, p. 11, Eq. C)])
2F1 (a, 1 -a;c;^) =
It follows immediately that
B5.8)

144 12. Continued Fractions
B5.9)
In order to evaluate 2Л(а, ft', У; г)» we must use contiguous relations to
obtain functions evaluable by B5.8). Using a common abbreviated notation
in Erdelyi's compendium [1, p. 103], we solve C1) there for F(a - 1) and then
replace F by an expression for F obtained from C2). Accordingly, after
simplification, we find that
F(a - 1)
(b{c -2a-(b- a)z}F(b + 1) + a(a + b - c)F(a + 1)).
(b - a)(c - a)
We apply this formula with F(a - 1) = 2-Fi(a> P', У'Л\ After considerable
simplification, we deduce that
M + 1, j? + 1; y, i) + —, \ г- 2Fi(« + 2, P; y;
by B5.8). Putting this and B5.9) into B5.7), employing B5.7) in conjunction
with B5.6), and lastly taking reciprocals of both sides, we conclude that
Taking the reciprocal once again, we deduce B5.4). ?
Second Proof of Entry 25. It is now easy to deduce B5.1). We remark at
the outset that each of the two equalities in B5.4) yields B5.1).
Let x = iz, where Im z < 0. Then Re x > 0. From B5.4), we easily deduce
that
1 n2 - I2 n2 - 32
2.x 2x — 2x —
. J(iC + n + х))ГШЗ - n
= -i —ix-Ai-
T(i(l + n + х))Г({A - n + x))
After taking the reciprocal of each side, we find that

12. Continued Fractions 145
»2-l2 n2-32 _ ГЩЗ + и + х))ГЩЗ - и + x))
2x - 2x -
Taking the reciprocal of both sides once again, we complete the proof. ?
Masson's proof is very interesting because it brings the hypergeometric
functions out of the closet. Thus, there is a connection with the previous entries
that was not heretofore noticed. Although Ramanujan probably did not derive
in this way the many continued fractions for gamma functions that appear in
this chapter, we have gained some insight into why these continued fractions
exist.
Corollary 1. // Re x > 0, then
Г2Щх + 1)) _ 4 21 H ^1
Proof. Set n = 0 in Entry 25. ?
Corollary 1 was first proved by Bauer [2] in 1872 and was communicated
by Ramanujan [16, p. xxvii] in his first letter to Hardy. Corollary 1 was also
recently submitted as a problem by W. B. Jordan [1].
If we put x = 1 in Corollary 1, we obtain Lord Brouncker's continued
fraction for я,
4 ^2 22 52
For a very interesting historical account of Brouncker's continued fraction,
see Dutka's paper [2].
Corollary 2. // Re x > 0, then
ГЩх + 3))ГЩх + 1)) _ 8 1-3 57 9-11
ГЩ* + 7))Г(?(х + 5)) х + 2х + 2х + 2х +•••'
Proof. Replace x by x/2 and и by \ in Entry 25. ?
Entry 26. Suppose that n is an odd integer and x is any complex number or that
n is an arbitrary complex number and Re x > 0. Then
Г2Щх + n + 1))Г2Щх - и + 1))
Г2(|(х + n + 3))Г2(±(х - и + 3))
8 I2-и2 I2 32-и2 З2
8 I 12-п2 3j= 32-п2
~ Ых2 -пг~^\) + I + "х2- 1 + Т + х2- 1 + • • • ¦

146 12. Continued Fractions
Proof. To obtain the first equality, set ? = 0 in Entry 39. The second equality
follows from Entry 39 by letting и = 0 and replacing t by n.
Alternatively, the two continued fractions can formally be shown to be
equal by an application of Entry 15. Let h = — n2 and ak = Bk — lJ, к S: 1,
and also replace x by x2 — 1 in Entry 15. The desired equality easily follows,
since both continued fractions terminate if n is an odd integer, and since both
continued fractions converge for Re x > 0, otherwise. ?
Corollary. If Rex > 0, then
Г4Шх+1)) 8 I2 I2 32 32
Г4(|(х + 3)) \(x2 -1)+ 1 + x2 - 1 + 1 + x2 - 1 + • • ¦'
Proof. Set и = 0 in Entry 26. ?
The next theorem is found in Ramanujan's [16, p. xxix] second letter to
Hardy. The first proof in print was provided by Preece [1]. Entry 27 can also
be found in Perron's book [3, p. 37, Eq. C1)]. A very instructive proof of Entry
27 has been derived by Ramanathan [1].
Entry 27. Suppose that x, у > 0. Then
A + y? + n C + yJ + n E + yJ + n
X +
2x + 2x + 2x + • • •
A + xJ + и C + xJ + и E + xJ + и
~y 2y + 2y + 2y +¦¦¦¦
For an improved version of Entry 27, see Jacobsen's paper [3].
Entry 28. Let Re x > 0 and |arg n\ < n/2 — S, for some positive number д. Then
n2 + I2 n2 + 32
Proof. Apply Entry 25 with n replaced by in to find that, for Re x > 0,
T(i(x + in + 1))Г(?(х - in + 1)) _ 4 n2 + I2 n2 + 32
or
in + 3))Г(?(х - in + 3)) x + 2x + 2x
r(j(x + in + 3))ГЩх - in + 3)) _ n2+ I2 n2 + 32
F(i(x + in + 1))Г(?(х - in + 1)) ~ X 2x + 2x +

12. Continued Fractions 147
Next, apply Entry 25 with x and и interchanged to obtain, for Re и > 0,
(я - x + 3)) n x2-!2 x2-32
r(i(n + x + 1))ГШй - x + 1)) ~ П In - In -¦¦¦'
Now, for |arg n\ < я/2 — 3,
гЩх + in + 3))rq-(x -m + з))гщ« + x + i))rq(n - x + 1»
я™ ГШх + in + 1))Г(±(х - in + 1))Щ(п + x + 3))ГЙ(и - x + 3))
= 1,
where we have applied Stirling's formula for the quotient of two gamma
functions (Lemma 2, Section 24, Chapter 11). Thus, we have shown that
n2 + I2 n2 + 32
B8-2)
However, for Re и > 0,
x2-l2 x2-32
—- /1{г~ хг2П_ъг"" = L B8-3)
because the numerator and denominator above are both of the form и +
ОA/и) as и tends to oo. Combining B8.2) and B8.3), we deduce B8.1). ?
In his first notebook (p. 160), Ramanujan states a more precise version of
Entry 28,
и2+12 и2 + 32
x H
2x + 2x
x2 - 12 x2 - 32 1 - 2e~nnS2 sin(Wx/2) + e~m'
In + In + ¦¦¦
Ramanujan probably intends the right side to be an approximation to the left
side for и large. However, the right side is 1 + О(е~жп/2) as n tends to oo. A
close analysis of our proof of Entry 28 shows that the left side of B8.1) is of
the form 1 + O(l/n) as и tends to oo and that the estimate O(l/n) cannot
be improved. Thus, Ramanujan's claim does not appear to have a valid
interpretation.
Entry 29. Let n be an odd integer and x complex, or let n be complex and
Re x > 0. Then

148 12. Continued Fractions
,
x-n + 2k- 1
-I l2~ ?I 32-n2 4^ 52-n2
x + x +x+ x +x+ x +•••
We provide two proofs under more restrictive hypotheses than what we
have given. Jacobsen [3] has proved Entry 29 with the stated conditions.
First Proof. Our first proof merely consists of a reformulation of a result
found in Perron's book [3, p. 33, Eq. A2)],
4 l2^"! 2* 32- 42 52-n2
X + X +X+ X + X + X + •••
'x + n + 3\ /x_n + 3\ , /x + n+l\ ,/x-n+l
4
B9.1)
where x > 0 and n2 < 1. Now employ @.1) and simplify to complete the proof.
?
In fact, Entry 29 was first proved in print in 1953 under these stronger
hypotheses by Perron [2] who derived it from Entry 34 below.
Second Proof. Since
we find that, for Re x > — 1,
TV72dt = Z ]_ ,b' , • B9-2)
1 + r k0 x + 2k + 1
JO 'Ti lc = O X + 2fc +
Then for Re(x + n) > — 1,
H(x + n) + H(x-n)= I ' v~ ' — dt
Jo I + t
Jo
coshы
where we have made the change of variable t = e u. But for x > 0 and n2 < 1,
Rogers [3] has shown that
c.xucosh(nu) ^ = 1 !2-n2 2^ 32-n2 4^
о cosh u x+ x +x+ x + x + • • • '
Employing B9.2) and B9.4) in B9.3), we arrive at the desired formula. ?

12. Continued Fractions 149
Corollary. // Re x > 0, then
oo C_11* + l 1 I2 ?2 T,2 42
у
2k — 1 x+x+x+x+x+---
Proof. Set и = 0 in Entry 29. ?
Entry 30. Suppose either that n is an integer or that Re x > 0. Then
1 1
C0.1)
- n + 2k + 1 x + n + 2fc + 1
_n I2(l2-n2) 22B2-n2) 32C2-n2)
x + 3x + 5x + 7x
First Proof. Letting
R = ГШ* + m + n + 1))ГШх - m - n + 1))
and
T = Г(±(х + их - и + 1))ГШх - т + п + 1)),
we first write Entry 33 in the form
R-T _mn (m2 - 12)(и2 - I2) (m2 - 22)(и2 - 22)
R + T~ x + 3x + 5x +¦¦¦'
where x, m, and n are complex numbers such that Re x > 0, or either m or n
is an integer. Thus,
'tl: ¦'|1'-') ^1^> . C0.2)
+
т-*0^^т^ Х-Г JX -г JX
On the other hand, a direct calculation with the use of L'Hospital's rule shows
that
,. I R-T 1 , fx + n+ 1\ 1,/jt-n+r
„fflR + Г 2r\ 2 / 2r\ 2
1 1
= ttt,
{x - и + 2fc + 1 x + n + 2k +
Combining C0.2) and C0.3), we finish the proof. ?
Second Proof. This proof requires that n2 < 1 and x > 0. Proceeding in
somewhat the same way as in the second proof of Entry 29, we find that, for
Re(x ± ri) > -1,
Jo 1 - r t=o (.x - и + 2k + 1 x + n + 2k +
On the other hand, letting t = e~" and using a theorem of Stieltjes [1],

150 12. Continued Fractions
[4, pp. 378-391], which was also proved by Rogers [3], we find that, for x > 0
and n2 < 1,
sinh и
n 12A2 — и2) 22B2-n2)
x + 3x + 5x +
Combining C0.4) and C0.5), we complete the second proof. ?
Corollary. If Re x > 0, then
» 1 _ 1 I4 24 34
&, (x + 2k + IJ ~ x +3x + 5x +7x +¦¦¦'
Proof. Divide both sides of C0.1) by n and let n tend to 0. ?
If we set x = 1 in the corollary above, we deduce that
1 n2 1 I4 24 34
-ГП\ — = -
2^y ' 12 1+3+5+7 +•¦•¦
For a simple proof of this expansion, see a note by Madhava [1].
In fact, the corollary above is due to Stieltjes [1, Eq. B2)], [4, p. 387].
Entry 31. Suppose that n is an even integer or that Re x > 0 and n is any
complex number. Then
f [ (-1)" (-1)*
k=o \x - n + 2k + 1 x + n + 2k + 1
n 22 - n2 22 42 - n2
x2- 1 + 1 +x2 - 1 + 1 +x2- 1 +•••
First Proof. From Entry 36, if Re x > 0 or if n is an even integer,
-P n 22-n2 22 42 - n2 42
C1.1)
lim - —
<f->c
C1.2)
x2 - 1 + 1 + x2 - 1 + 1 + x2 - 1 +
On the other hand, a direct calculation with the use of L'Hospital's rule gives
.. 1 1 - P 1 f . fx + n + 1\ /x - n + 3N
-Щ
4 J \ 4
х-и + 1\ , fx + n

12. Continued Fractions 151
-H- '
x + n + 4fc - 3 x - n + 4k - 1
1 1
x - n + 4k - 3 + x + n + 4k - 1J ' ^' '
Equalities C1.2) and C1.3) taken together yield C1.1). ?
Second Proof. This proof requires more severe restrictions on x and n. As
in the second proofs of Entries 29 and 30, we easily find that, for Re(x + n) >
_ j
f f (D (D 1 Г
(xn + 2lt+l x + n + 2k + 1J J
+ n + 2k + 1J Jo cosh u
But Stieltjes [3], [4, pp. 402-566] and later Rogers [3] have shown that, for
x > Oand n2 < 1,
00 _xusinh(nu)
e " du
, cosh u
n 22-n2 22 42-n2 42
x2- 1 + 1 +x2 - 1 + 1 +x2 - 1 +¦¦¦¦
The foregoing two equalities imply C1.1). ?
Corollary. // Re x > 0, then
» (_!)* i 22 22 42 42
k% (x + 2k + IJ x2 - 1 + 1 + x2 - 1 + 1 + x2 - 1 + • • •'
Proof. Divide both sides of C1.1) by и and then let n tend to 0. ?
If we put x = 2 in the foregoing corollary, we obtain the following elegant
continued fraction for Catalan's constant G:
oo ( —l)k 1 22 22 42 42
k=0Bk + IJ 3 + 1 + 3 + 1 + 3 + • • • '
Of course, similar continued fraction expansions for G can be obtained by
setting x = 2n, where n is any positive integer, in the corollary above. This
same infinite set of continued fractions for G was independently found by H.
Cohen (personal communication) who obtained them from a different formula.
Entry 32(i). If Re x > 0, then
»• {-If _ 1 1-2 2-3 3-4
ъ^л y -I- ?t y 4- y 4- y 4- y -4- ¦ ¦ •

152 12. Continued Fractions
Proof. Let
p_p( _ ГЩх + n + 3))r(j(x - n + 3))
[X' ' Г(|(х + п+ 1))Щ(х - n + 1))'
Then by Entry 25, for Re x > 0 and и # 1,
12-п2 32-n2 52-n2
4P = x + -— -— -—
2x + 2x + 2x + • • •
or
4P - x _ 1 + n 32 - n2 52 - n2
fi 'У v* _1_ О у _1_ 'У v* _1_ • • .
— ft Z.-K \^ ^.Л, i^ ^.Л> i^
C2.2)
Note that P(x, 1) = x/4. We now let n tend to 1 in C2.2) and apply
L'Hospital's rule on the left side. We then find that
,/x \ /xYl 2 2-4 4-6 6-
- ф T + 1 - ф
4 / r\4 7 V4/J 2x+ 2x + 2x + 2x +•••
Simplifying each side above, we arrive at C2.1). ?
Entry 32 (ii). // Re x > 0, then
t=i (X + fcJ X + X + X + X + X + X + • • • '
Proof. Let
Then from B9.1), we find that, for Re x > 0 and n ф 1,
4/p _ x i + „ 22 32 - n2 42 52 - n2
1 — n x +x+ x +x+ x +••¦
C2.3)
Observe that P(x, 1) = 4/x. Letting n tend to 1 in C2.3) and employing
L'Hospital's rule, we find that
4_P
on
P2
{x,n)
(x, 1)
2
CO
22
f 1
t(x + 4/c)
2-4
2
42
(x +
4-
2
4fe
6
-2J
1
+ 4k
-4J
X+X+X +X+X +
Replacing x by 2x, we deduce that

12. Continued Fractions 153
2 » (-If _ 2 22 2-4 42 4-6
1 + X h (x+W = 2x+2x+lx~ + 2x+^x~ + ---'
Simplifying the right side, we complete the proof. ?
If we set x = 1 in Entry 32(ii), we deduce that
1+1+ 1 +1+ 1 + 1 + ¦•¦
Putting x = 5 in Entry 32(ii) yields another continued fraction for G,
1 I2 1-2 22 2-3 32
2G=l+y -j- — -j- — x ___.
Entry 32(iii). // Re x > -\, then
CC,x+!):=? -4-3
1 I3 1:
' 2x(x + 1) + 1 + 6x(x + 1) + 1 + 10x(x + 1) + • • •
1 l6
: 2x2 + 2x + 1 - 3Bx2 + 2x + 3)
26 36
5Bx2 + 2x + 7) - 7Bx2 + 2x + 13) - • • ¦
C2.4)
Proof. In Entry 35, replace x by 2x + 1. Then у = 4x(x + 1) + 2m — m2, and
we need to require that Re x > -\. Also let f = n = m. Noting that t = 0
and using the second continued fraction of Entry 35, we find that
Г(|Bх + 2 + Зш))Г3(|Bх + 2-т))
1 -P _ ГЩ2х + 2 - Зш))Г3(^Bх + 2 + m))
1 + P ~ ^~"r(jBx + 2 + Зш))Г3(|Bх + 2 - m))
ЩBх + 2 - Зш))Г3ABх + 2 + m))
2nx3 2A - m)(\2 - m2) 2A + m){\2 - m2)
у - 2mi + 1 + Ъу
2B_- m)B2 - m2) 2B + m)B2 - m2)
C2.5)
+ 1 + 5y +•••¦
Now divide both sides of C2.5) by m3 and let m tend to 0. On the right side,
we arrive at
__2 2j_P 2-13 2-23 2-23
4x(x + 1) + ~\~ + 12x(x + 1) + 1 + 20x(x + I) + • • •'
Simplifying above, we obtain the former continued fraction of C2.4).

154 12. Continued Fractions
Next, write the aforementioned continued fraction in the equivalent form
1 P 13/3 23/3 23/5 375
2x(xTT) + T + 2x(x + 1) + T~ + 2x(x + 1) + T~ + • • • '
Applying Entry 14 to this continued fraction, we deduce the equality between
the continued fractions of C2.4).
For brevity, set z = x + 1. For Re z > \, it remains to examine, by C2.5),
.. 1 1-P
lim —=- ——
m-*0
m3 1 + P
x |r(z) + ™r(z) + ^-Г"(
2m3
F
ЗГ"(г)Г(г) HzK
T3(z)
\d4T'(z)
The proof is now complete. ?
Ramanujan's second continued fraction in Entry 32(iii) is slightly in error
(p. 149).
We might compare Entry 32(iii) with another continued fraction for ?C, x),
x + x + x + x + x + ¦¦¦
where, for к > 1,
k\k + 1) . k(k + IJ
P* = ^T" and ft = ^4fcTT-
The last result was discovered by Stieltjes [1], [4, pp. 378-391].

12. Continued Fractions 155
Setting x = 1 in Entry 32(iii), we deduce the following beautiful continued
fraction for CC): / ' , • : ! .
^- - ^ - —
2-2 + 1 +6-2+ 1 +10-2 + •••¦
This continued fraction also follows from work of Apery [1] and was of crucial
importance in his famous proof that ?C) is irrational.
Entry 33. Let x, nx, and n be complex. If either morn is an integer or ifRex> 0,
then
- m - n + 1)) - Г(|(х + m- n+ 1))Г(|(х - m + n + 1))
+ m + n+ l»r(i(x - m - n + 1)) + T(i(x + m-n + 1))Г(^(х - m + и + 1))
mn (nx2 - \2){n2 - I2) (m2 - 22)(и2 - 22)
x + 3x + 5x
(m2 - 32){n2 - 32)
+ 7x + • • •
Proof. Set
_ Г^х + <? + n + l) + m)T(^(x - t - n + 1) + nx)
(Ш) ~ ГШх - t + n + 1) + т)ГШх + i - n + 1) + m)
and
_ Г(^(х + t - n + 1))Г(^(х - ? + n + 1))
T = T(i(x + ( + n + 1))Г(|(х - / - n + 1))'
Suppose that nx is a positive integer in Entry 35. Replacing x by x + nx in Entry
35, we find that
1 - R{m)T 2tmn 4(/2 - I2)(m2 - I2)(n2 - I2)
1 + R{m)T
x2 + 2nxx - ?2 -
+ 5(x2 + 2mx -
in
[i2 - 22)(n2 -
- n
'*'
-2
2 + l
22)(n
2-n2
f l)/2
2)A-
+
2 _
+
m 4
-22
3(x2
22)
13) -
-3(x
Im2)
+ 2nxx - t2
- \2){n2 - 1
+ (x2-/2-
-п2л
2Ш-
-n2 +
-5)
1/m2)
5)/2m)
+ 5(x + (x2 - B - n2 + 13)/2nx) + ¦ • •' v '
Now let nx tend to oo in C3.1). By Stirling's formula, R(m) tends to 1 as nx tends
to a). The continued fraction converges uniformly with respect to ra in a
neighborhood of nx = со if Re x > 0. Hence,

156 12. Continued Fractions
1 -Г _ fn_ Qf2 - I2)(n2 - I2) Qf2 - 22)(n2 - 22)
1 + Г ~ х + 3x + 5x +•••'
Replacing <f by m above, we complete the proof. ?
In fact, Entry 33 was first proved in print by Norlund [1] under more
restrictive hypotheses.
The continued fraction in Entry 33 is a special case of a more general
continued fraction for a quotient of two integrals involving hypergeometric
functions that was discovered by Stieltjes [1], [4, p. 389, Eq. B9)].
Entry 34. Suppose that n is an odd integer or ? is an even integer, or assume
that Re x > 0 with t and n arbitrary complex numbers. Define
_ ГЩх + / + n + 1))Г(|(х + /-п+ 1))Щх - ( + n + 3))r(j(x - ( - n + 3))
~ f (i(x - ( + n + 1))Г(|(х - ( - n + 1))Щх + ( + n + 3))ГЙ(х + ( - n + 3))'
Then
\-P г I2 - и2 22 - <f2 32 - n2 42 - t2
I + P x + x + x + x + x +¦¦•
Entry 34 was stated by Ramanujan [16, p. 350] in his first letter to Hardy.
The first published proof was provided by Preece [2]. Another proof has been
devised by Perron [1], [3, p. 34, Eq. A5)]. These two proofs require stronger
hypotheses. Jacobsen [3] has shown how to establish Entry 34 under the given
assumptions on the parameters.
Corollary. Suppose that Re(x/y) # 0. Put
[x+
Then
F(a, /?) + F(fi, a) = 2F{i(a + 0), ±(a + 0)}.
The corollary was communicated by Ramanujan [16, p. 353] in his second
letter to Hardy. Again, the first published proof was given by Preece [2], and
indeed this result is a corollary of Entry 34.
Entry 35. Let x, t, m, and n denote complex numbers and put у = x2 — A — mJ
and t = (и2 - /2)A - 2m). Define
_ Г(|(х + ( + m + n + 1))ГЙ(х + ( - m-n + \))Щ(х -t + m-n+ \))Щ(х -t-m + n + 1))
i -{ -m -Т+1))Г(|оГ- { + m + n + l))r(i(x + t - m + n + 1))Г(^(х + t + m - n
Then if either ?, m, or n is an integer or if Re x > 0,

12. Continued Fractions 157
4Qf2 - \2){m2 - I2)(n2 - I2)
4(<f2 - 22)(m2 - 22)(n2 - 22)
+ 5(х2 - /2 - т2 - п2 + 13) +•••
limn 2A - m)(l2 - и2) 2A + m)(l2 - t2)
y + t- 2{2т + 1 + Ъу + t
2B - т)B2 - и2) 2B + ш)B2 - f2)
+ 1 + Ъу + t +¦¦¦'
C5.1)
Proof. The first equality was shown by Watson [8] to be a corollary of Entry
40. If either t, m, or n is an integer, Watson's limiting process is trivially
justified. If <f, m, and n are nonintegral, then the limiting process is more
difficult to justify. We refer the reader to Jacobsen's paper [3], where this
justification is carefully presented.
To prove the second equality, we employ the following generalization of
Entry 14 but special case of A4.2). If the former continued fraction converges,
then
Ol «2 ^ «4 ^2k-l a2k
x1 + 1 +x3+ 1 +--- + x2k-!+ 1 +•••
a2 - x3 + a3 + a4 - x5 + a5 + a6 -
Thus, with ax = limn, a2 = 2A - m)(l - n2), ... and x1 = у + t - lt2m,
*з = 3y + t,..., we find that
Ifmn 2A - m)(\ - n2) 2A + m)(l - t2) 2B - m)B2 - n2)
y + t- lt2m + 1 + 3y + t + 1
2B + m)B2 - f2) l(k - m)(k2 - n2)
+ 5y + t +•••+ 1 +••¦
Itmn 4A - m2)(l - /2)A - n2)
~ x2 - i2 - m2 - n2 + 1 - 3(x2 - <f2 - m2 - n2 + 5)
4B2 - m2)B2 - /2)B2 - n2)
- 5{x2 - t2 - m2 - n2 + 13)
4((fc - IJ - m2){{k - IJ - I2)((k - IJ - n2)
- Bk- l)(x2 - <f2 - m2 - n2 +lk2 -2k+ 1) +••
where we have used the easily proved identity
C5.2)

158 12. Continued Fractions
B/ + l)y + t + 2{j + m)(j2 - t2) + 2{j + 1 - m)((j + IJ - n2)
= B/ + l)(x2 -{2-m2-n2 + 2j2 + 2/ + 1).
This establishes the second equality in C5.1). D
Entry 36. Suppose either that n or ( is an even integer or that Re x > 0 and n
and Л are arbitrary complex numbers. Let
p = r(l(x + e +«+ з))гЩх -е-п + здгЩх + e - n
~ Г
+ t + n + 1))ГЙ(х - t - n + 1))Г(|(х + t-n + 3))r(i(x -t
Then
\-P in 22 -n2 22 - t2 42 - n2 42 -
TT7; x2 - l -T2 + i + x2- l + I + x2- l + • • • '
Proof. In the second equality of Entry 35, let m = j and replace x, n, and t
by x/2, n/2, and (f/2, respectively. After simplification, the proposed identity
follows. ?
Entry 37. Suppose that either f or n is an integer or that Re x > 0. Then
2| V 2 / \ 2
, (x + S + n + l\ fx-f-n+l
V 2 / \ 2
2A2-п2) 2A2-
x2 - 1 + n2 - t2 + 1 + 3(x2 - 1) + n2 - t1
4B2 - n2) 4B2 - Q
C7.1)
Proof. Taking the second equality in C5.1), divide both sides by m and then
let m tend to 0. Applying L'Hospital's rule on the left side, we readily deduce
the desired formula with no difficulty. ?
Entry 38. Assume that either n is an integer or that Re x > 0. Then
Д 1 « 1
ttb (x-n + 2k + IJ ~~ ktb (x + n + 2k + IJ
n 2(l2-n2) 2-12
x2 - 1 + n2 + 1 + 3(x2 - 1) + n2
4B2-n2) 4-22
+ 1 +5(x2- 1) + и2 +•••
n 4A2 - n2)!4 4B2 - n2J4

12. Continued Fractions 159
Proof. To prove the first equality in C8.1), divide both sides of C7.1) by 2?
and let {tend to 0. Applying L'Hospital's rule on the left side, we easily achieve
the desired equality.
The second equality in C8.1) is also easily established. First, divide both
sides of the first equality in C5.1) by m and then let m tend to 0. Of course,
this gives a second continued fraction for the left side of C7.1). Now divide
both sides by t and let t tend to 0. ?
Entry 39. Let f and n denote arbitrary complex numbers. Suppose that x is
complex with Re x > 0 or that either n or ( is an odd integer. Then
p _ _
Г(±(х + ( + п + 3))Г(|(х - ( + п + 3))ГЙ(х + t-n + 3))r(i(x - ( - и + 3))
8 12-п2 I2-/2 Ъ2-п2 З2-/2
(х2 - i2 + п2 - 1)/2 + 1 +х2-1+ 1 +х2-1+---'
C9.1)
Proof. We shall prove Entry 39 for — oo < i2, n2 < 1 and x > 1. An argu-
argument of Jacobsen [3] can then be used to extend the domains of convergence
for <f, n, and x to those indicated.
To prove Entry 39, we employ the following theorem found in Perron's
text [3, p. 27, Satz 1.13]. Suppose that all the elements are positive in both
continued fractions below. Assume also that each continued fraction con-
converges. Then
h +"! ^ ^i
0 Ь1+Ь2+Ьъ + •••
fot + rt + b2 + r2 — ro(p2/(p1 + b3 + r3 — гг (р3/(р2 + ¦¦¦'
C9.2)
where
<Pk = °k ~ h-dW + rk), fe>l. C9.3)
(The parameters гк, к > 0, have no restrictions other than those imposed
above.)
Let
F(x) = F(x, /, и) = Z +
2 ' 1 + x2 - 1
32 - n2 32 - t2
+ 1 + x2 - 1 +
C9.4)
In the notation above, a2k = Bk — IJ - t2, a2k^ = Bk — IJ - n2, b2k =
x2 — 1, and fo2t_! = 1, where к > 1. Write
r2k = dik + ct and r2k^ = d2k + c2, k>l. C9.5)

160 12. Continued Fractions
Our first goal is to determine c1; c2, dx, and d2 so that cpk is constant for
fe> 1.
From C9.3), C9.5), and the aforementioned formula for a2k, it follows that
d,d2 = 4. C9.6)
Thus, from C9.3), we find that
ц>2к = B* - IJ - /2 - (d2? + c2)(x2 -l+dtk + c,)
= -{4 + d2(x2 - 1 + Cl) + cjdjfe + 1 - /2 - c2(x2 - 1 + Cl) C9.7)
and
?„_! = Bfc - IJ - n2 - {dt(k - 1) + Cl} {1 + d2k + c2)
= -{4 + d1(l+c2)-d2(d1-c1)}fc
+ 1 -n2 + (dt -cj(l + c2), C9.8)
where fe > 1. By our prescriptions, we require that
<*2(x2 - 1 + Cl) + с2^! = -4 = ^A + c2) - d2(dx - Cl). C9.9)
Using C9.6) and simplifying the extremal equality above, we find that
d\ - 4d, + 4A - x2) = 0.
We shall choose the positive root dt = 2x + 2. Thus, by C9.6), d2 = 2/(x + 1).
Since we wish <pk to be constant, by C9.7) and C9.8), we need to stipulate
that
1 - i2 - c2(x2 - 1 + Cl) = 1 - n2 + (d, - Cl)(l + c2).
Simplifying, we find that
c, - c2(x + IJ = /2 - n2 + 2(x + 1). C9.10)
On the other hand, from C9.9),
ct +c2(x+ 1J= -(x+ IJ. C9.11)
Adding C9.10) and C9.11), we deduce that
c, = \{B - n2 - x2 + 1),
and so
/2 - n2 - x2 + 1
C2 = ~l 2(^TT? •
Hence, we have determined the parameters cu c2, du and d2 so that q>k is
constant, namely, from C9.5),
r2k = 2(x + l)fe + W2 - n2 - x2 + 1), к > 1, C9.12)
and

12. Continued Fractions 161
/2 _ и2 — y
Let us set q>k = a. By C9.8) and our determinations above,
ix = l -n2 + (d1 -cJO +c2)
_ 4A - n2)(x + IJ - 4(x + l)(<f2 - n2 - x2 + 1) + {{2 - n2 - x2 + IJ
_ 4(x + IJ
_ -4n2(x + IJ + {t2 - n2 - x2 + 1 - 2{x + I)}2
_ 4(X + IJ
\{2 - n2 - x2 + 1 - 2A + n)(x + 1)} {/2 - и2 - x2 + 1 - 2A - n)(x + 1)}
4(x + IJ
The numerator above is a polynomial in x of degree 4. It is easily checked that
the four roots of this polynomial are x + 1 = ± ( + n, where all four possible
combinations of signs are taken. Hence,
_ (x + 1 + t + n)(x + 1 - ( - n)(x + 1 + ? - n)(x + 1 - { + ri)
a= 4(x + IJ '
C9.14)
Recalling the definition C9.4), applying C9.2), and employing C9.12) and
C9.13), we have shown that
a l2-n2
F(x) =
n2 - x2 + \ + x2 - I + 2x + 2
x + 1 2(x + IJ
l2-/2 32-n2 32-/2
+ 1 + 2/(x + 1) + x2 - 1 + 2x + 2 + 1 + 2/(x + 1) + •••
a(x + IJ l2-n2 I2-/2
{{x +Tf - S2 + n2 - l}/2 + i + (лГ+ 1)(х +Y)
32 _ n2 з2 _ /2
+ 1 + (x + l)(x + 3] +
a(x + IJ
For brevity, set, for any function /,
C915)
= f(x + k+/? + n)f(x + k-f- n)f(x + к + t - n)f(x + к - t + n).
Hence, from C9.14) and C9.15),

162 12. Continued Fractions
and so
F(x)F(x + 2)
F(x + 2)F(x + 4)
By iteration of this formula, we find that, for each positive integer m,
F(x + 4m) Lo[[±
= J_ т-г mf\ (i(x + l ± f ± n)
ffl k — 0 \4\-^ i -^ ~ ^ i 'v "^ l^)lll-llt
Hence,
F(x)m2 T(\(x + 3 + / + nil 1
1 1± FYl/v- _|_ J ^_ ? _(_ и\) р
From the definition of F(x) in C9.4), we easily see that
2 i
lim = -. C9.17)
F(x + 4m) 8 v ;
Combining C9.16) and C9.17), we deduce C9.1). ?
H. Cohen has communicated to us a similar proof of Entry 39. His proof is
based on Apery's method for accelerating the convergence of a continued
fraction. For a complete description of this method, see Cohen's seminar notes
[1]. Accounts are also given in papers by Apery [1] and Batut and Olivier [1].
The equality C9.2) is called the Bauer-Muir transformation. Jacobsen [5]
has shown that the conditions for its validity can be considerably relaxed.
We might note an interesting consequence of Entry 39. From Malmsten's
integral representation for Log T(z) (Whittaker and Watson [1, p. 249]), we
find that
Г00 /V e-(x±f±n + l)t/A _ у e~ ^{
where ?± indicates a sum of four terms with each possible combination of
signs taken. Simplifying, we find that
f *> (e-'xtA cosh(/r/4) cosh(nt/4) _,\ dt
LogP=2J0 {—^m e h
= 2 e 1 2
Jo V cosht Jt Jo
0 v cosh г /f Vх/
by Frullani's theorem (Edwards [2, pp. 337-342] or Part I [9, p. 313, Eq.

12. Continued Fractions 163
B.15)]). Exponentiating C9.18) and combining the result with C9.1), we
deduce that
x2/2 l2-n2 I2-/2 32-n2 32-/2
(x2 - t1 + n2 - l)/2 + 1 + x2- 1 + I + xT^\ + • • ¦'
C9.19)
where 0 < \?\, \n\ < 1 and x > 1.
The expansion C9.19) appears to be new. It generalizes a result of Rogers
[3] and is similar to results of both Rogers [3] and Stieltjes [1], [4,
pp. 378-391].
Entry 40. Let
P = U r(i(a ± P ± У ± s ± ? + !))'
t/ге product contains eight gamma functions and where the argument
of each gamma function contains an even number of minus signs. Let
e = П Ш« ± P ± у +s ±? +!))-
where the product contains eight gamma functions and where the argument
of each gamma function contains an odd number of minus signs. Suppose that
at least one of the parameters /?, у, д, е is equal to a nonzero integer. Then
P + Q
1 \l{tx -f p •
+ 3{2(a4 + /
4- / + <54 -
64(a2 - 1
f4 + / + <!
64(a2 -
f 64 +
i W -
>4 + ?4
22)(/?2
1) - (a2 + ,
- I2)(y2 -
+ 1) - (a2
- 22)(y2 -
P2 + y2 + i
\2){52 - I2
+ P2 + y2 -
22)E2 - 2
i2 + e2 - 1)
)(e2 - I2)
f 52 + ?2 -
2)(e2 - 22)
2 _
5J
22}
-62}
+ 5{2(a* + Рл + у* + 5A + ел + \)~ (a2 + p2 + y2 + d2 + s2 - 13J - 142} + • ¦ ¦'
D0.1)
Entry 40 is certainly one of Ramanujan's crowning achievements in the
theory of continued fractions. Watson [8] has given the only published proof
of Entry 40.
In an address before the London Mathematical Society in 1931, Watson
[7] discussed Entry 40 but incorrectly wrote 9 and 10 instead of 13 and 14,
respectively, in the last recorded denominator above. In a footnote of [8],
Watson remarked: "Through an error in copying which occurred when I
previously published an enunciation of the theorem...." However, Watson
did copy the result faithfully; Ramanujan had made the same error (p. 152).

164 12. Continued Fractions
Throughout the notebooks, Ramanujan normally did not completely state
identities involving sequences, but he did usually give enough terms to deter-
determine the sequence. In particular, if a sequence is linear, Ramanujan often gave
only two terms, while if a sequence is quadratic, he would give three. In the
first notebook, he only stated two terms of the sequences 2n2 + 2n + 1 and
2n2 + 2n + 2; that is, 1,5 and 2, 6, respectively, that occur on the right side
of D0.1). This was probably carelessness on his part for he most likely knew
the quadratic patterns of the sequences. When he wrote his second notebook,
a revised enlargement of the first, he decided to add one more term. However,
he evidently did not rederive his identity and erroneously assumed that the
two sequences are linear. Ironically, Watson's statement of Entry 40 in [8]
also contains a misprint. Watson [8] also obtained a ^-analogue of Entry 40.
It is natural to ask if the hypotheses on /?, y, 5, and ? can be relaxed. Jacobsen
[3] has answered this by proving the following theorem.
Theorem. The continued fraction on the right side of D0.1) converges to a
meromorphic function F(a., /?, y, S, e) in (€ъ. Furthermore, F ф (P — Q)/{P + Q).
The identity of F is not known.
Entry 41. Let x and у be complex numbers such that either \x + 11 > 1 or у is
a nonnegative integer. Then
(/» + l)x + 1 - у - (P + 2)x + 3 - у
2B - y)(x + 1)
1A -y)(x+ 1)
D1.1)
Proof. From Erdelyi's treatise [1, p. 108, formula B)],
_Г(-Р- 1)ГG + i) l_rp + 2. _ 1/x)
с
^1 х-2РЛ-Р,-Р-г,-Р;-Ух)
Г(-р)Г(у)х
Г(р + 1)ГG 4
D1.2)
Now apply B1.3) with /?, у, and x replaced by — у, /? + 1, and l/x, respectively,
to deduce that

12. Continued Fractions 165
y/x 1A - y)(l + l/x)/x 2B - y)(l
/?+l+(l-y)/x- /J + 2 + C-y)/x - /J + 3 + E-y)/x -•¦•
D1.3)
Translating the conditions under which B1.3) is valid, we find that D1.3)
holds if Re 1/x > — j with not both 1 — у and P + у + 1 belonging
to {0,-1,-2,...}, or if у is a nonnegative integer and P+y+l$
@, -1, -2,...}. Combining D1.2) and D1.3), we obtain D1.1) under the
conditions given in the previous sentence. Now Re 1/x > — \ if and only if
|x + 1| > 1. Lastly, Jacobsen [3] has employed the uniform parabola theorem
to remove the extraneous conditions on /? and у given above. ?
Entry 42. // n is a nonnegative integer, or if x ф (—go, 0], then
ехГ{п + 1) n 1-й 1 2-й 2 3-n 3
x" x+ 1 +x+ 1 +x+ 1 +*+•••
ехГ(п +1) n 1A - n) 2B - и) 3C - n)
x" x + 1 — n — x + 3 — n — x + 5 — n — x + 7 — n — •••
D2.1)
Proof. In Entry 41, replace x by x/p and у by n. If 11 + x/p\ > 1, or if и is a
nonnegative integer, we find that
1)(х/рТ
1A - и)A + x/p)
{p + l)x/p +l-n-{p + 2)x/p + 3-n
2B - n)(l + x/P)
- (P + Ъ)х/Р + 5 - n - ¦ ¦ ¦'
Since the continued fraction above converges uniformly with respect to /? in
a neighborhood of /? = go, we may let /? tend to со to deduce the second
equality in D2.1).
To obtain the first equality in D2.1), apply Entry 14. ?
Entry 42 was first discovered by Legendre [1]. See also Nielsen's book
[1, p. 217] for a proof.
Corollary. // either x is exterior to(—go, 0] or if n is a positive integer, then
» (-xf _ T(n) e~x 1-й 1 2-й 2
k% k\(n + k) ~ ~x" ~x + 1 +x+ 1 + x + • • •'

166 12. Continued Fractions
Proof. Multiplying both sides of D2.1) by e~x/n and comparing the resulting
equality with that above, we see that we must show that
e"x X -^— = I , ¦ D2.2)
Applying Entry 21 of Chapter 10 with x replaced by — x, n replaced by n + 1,
and m = n, we deduce D2.2). Q
Entry 43. If x is any complex number outside (-co, 0], then
,„ 1 12 3 4 5
k% 1' 3 ¦ ¦ • Bk + 1) V 2x x + 1 + x + 1 + x + 1 + • • •
tV „„ 1 1-2 3-4 5-6
7 _i_t _i-^_i__i_Q j_ 1 
D3.1)
Proof. Putting n = j in Entry 42, we find that
S Bx)k [Ке*-1И lH I У1 1 ^
kSH-3---Bfc + 1)" V4^ ^ ^ +~Г + х+~Г + х+ l +••¦'
Replacing x by x/2, we obtain an equivalent form of the first continued fraction
of D3.1).
The second continued fraction in D3.1) follows in the same way from the
second continued fraction of D2.1). Alternatively, apply Entry 14 to the first
continued fraction in D3.1). ?
Corollary 1. For Re x > 0,
F(x) := e dt =
J
2 2x + x + 2x + x + 2x H
Proof. By D2.2), for n > 0,
e"''" df = t fe^C = С У ^Р-. D3.2)
Г
Jo
Let и = з and replace t by t2 and x by x2. Applying Entry 43, we then find
that, for x2 exterior to (-co, 0],
F(x) = xe'xl Yj 7зТ~ = xe~*z J]
= xe
' 4x2 2x2 + 1 + 2x2 + 1 + 2x2 + 1 + ¦ ¦
which is equivalent to the proposed formula. ?

12. Continued Fractions 167
Corollary 2. Let x be real. Then as x tends to go,
T dt 2 B
where F is defined in Corollary 1 and у denotes Euler's constant.
Proof. Integrating by parts, we find that
C
= F{x) Log x - e~'2 Log t dt
Jo
e~' dt- e~' dt) Log x
e~'2 Log tdt - \ e '2 Log t dt
Logx- e~' Log tdt + o(l), D3.4)
2 jo
as x tends to go.
From the integral definition of Г(х), for x > 0,
f00
Г(х) = 4 e-'V* Log t dt.
Jo
In particular,
„-t2
'(i) = 4 e"'2 Log t dt = - v^(y + 2 Log 2), D3.5)
which was established by Ramanujan (p. 92) in Chapter 8. (See our book
[9, p. 184, Cor. 3(i)].) Employing D3.5) in D3.4), we deduce D3.3) at once.
?
Entry 44. For x > 0, define
dt.
x + t
Then for x > 0,
Г
J
dt = ? ^—yn = 7 + Log x + е-'ф), D4.1)
0 t fc=l "!"
w/iere у denotes Eulefs constant.
Proof. At the outset, we remark that essentially the same calculations are
made in slightly more detail in our edited version of Chapter 4 [9, p. 103].
The first equality in D4.1) is readily established by writing the integrand as
a Maclaurin series and inverting the order of summation and integration.

168 12. Continued Fractions
Next, making a simple change of variable in the definition of cp and using
a well-known integral representation for у (Olver [1, p. 40]), we find that
f" e~' Г1 1 — e~' f00 e~' fx dt
e~x<p(x) + у + Log x = —dt + \ dt - —dt + —.
Jx t Jo l Jl l Jl l
Upon simplification, we complete the proof of the second equality in D4.1).
?
Entry 44(i). Let x be real. Then as x approaches со,
Entry 44(i) was established by Euler, and a rigorous discussion of it can be
found in Hardy's book [5, pp. 26, 27]. Ramanujan also stated this result in
Chapter 4 (p. 44); see our book [9, pp. 101-102].
For Entry 44(ii), we quote Ramanujan (p. 153).
Entry 44(ii). q>(x) lies between 1/x and l/(x + 1) and very nearly equals
/
Proof. Letting n tend to 0 in the corollary of Section 42, we find that, for
x>0,
x
= 7 + Log x + e~xf(x), D4.2)
where
x + 1 +x + 1 +x + •¦¦
Comparing D4.1) and D4.2), we deduce that f(x) = (p(x).
Now from D4.3), it is immediate that cp{x) < 1/x. Next, if
2 2 3 3
F = x + - - - -
1 + x + 1 +x + •••
we can write D4.3) as
Thus, Ramanujan's upper and lower bounds for cp(x) are established.
Squaring the asymptotic series from Entry 44(i), we find that, as x tends
to со,

12. Continued Fractions 169
1 _2_ _^_16
x2 x3 x4 x5
On the other hand, also from Entry 44(i), as x tends to oo,
<p{x + 1) 1 1__ 2
x x(xTl) ~ x(x + IJ + x(x + IK +
_ЛГ|-1+Л-Л^-Л('1--+Л
x V x x
1 2 5 10 /1
= -2 3+^: 5 + 0 — .
2 3 4 5
Thus, the initial three terms of the asymptotic expansions for <p2(x) and
cp(x + l)/x agree. Hence, Ramanujan's approximation for <p(x) is reasonable.
?
Entry 44(iii). For x > 0,
1112 2 3 3
«>(x) = —- — - — - —
x + 1 +x + 1 +x + 1 +x + •••
1 I2 22 32
x + 1 —x + 3 — x + 5 — x + 7—¦¦¦
Proof. The former continued fraction was established in the course of prov-
proving Entry 44(ii) (see D4.3)). To obtain the latter continued fraction, apply
Entry 14. ?
In fact, Entry 44(iii) is valid for all complex x outside (—oo, 0) (Jacobsen
[3]).
The second continued fraction above was first derived by Tschebyscheff
[ID-
Entry 44(iv). Let x be any complex number exterior to (—oo, 0], and let nbe a
natural number. Then
1 n + 1 2(n + 2) 3(n + 3)
Proof. Integrating by parts n times, we find that

170 12. Continued Fractions
k = 0 X Jo \x + 4
D4.4)
k=C
where we have used the equality (Perron [3, p. 219])
1 f°° e~'tbl 1 '*oc "~'*a~1
dt =
r(b)J0 A+tr r(a)J0 A+0" "•'
However, for x ? (-oo, 0] (Perron [3, p. 219, Eq. A2)], Khovanskii [1, p. 148,
Eq. A1.17)]),
e-'t" , 1 n + 1 2(n + 2) 3(n + 3)
— «t = г ^ r
D4.5)
Substituting D4.5) into D4.4), we deduce the proposed identity. ?
Corollary 1. Let
Then if x > 0,
00 H,xk
-jr = ex(Logx + y) +
k = l K-
Corollary 1 is also given by Ramanujan in Chapter 4 (p. 44). See the author's
book [9, p. 103] for a proof.
Our formulation of Corollary 2 corrects that given by Ramanujan (p. 153).
Corollary 2. For \h\ < 1 and n > 0, define f(h, n) by
rn(l-h) j _ g-l
dt = y + Log n + e~"<p(n) - e~"f(h, n). D4.6)
Jo (
Then
Proof. First, if /i = 0, we see from Entry 44 that /@, n) = 0. For brevity, set
g(h) = f(h, n). Clearly, we shall be finished if we can show that
0<*+i>(O) = fc! (en -И), к > 0. D4.7)
V j=o У- J
First, difl'erentiating D4.6), we find that

12. Continued Fractions 171
e-g'(h) = h{ . D4.8)
Setting h = 0 in D4.8), we deduce D4.7) in the case к = 0. For к > 0, we apply
Leibniz's rule to D4.8) to find that
where ёк = 1 and 5i, = 0, 0 <j < к - 1. Thus,
Equality D4.7) now follows upon replacing j by к — j above. ?
Ramanujan concludes Section 44 by recording the values <p(l) = 0.5963474
and <p($) = 0.9229106. From D4.1),
/ 00 f_
and
Using calculated values for 7, e, yje, and Log 2 (Abramowitz and Stegun
[1, pp. 2, 3]) and 11 and 9 terms, respectively, from the two sums above, we
can readily verify that Ramanujan's calculations are correct.
Entries 45(i), (ii). Consider the continued fraction
1 x x 2x 2x 3x 3x (n — l)x nx
T + T + T + T + T + T + T + ---+ i +T'
Then in the notation of A.4), for n > 1,
В2„М:=В2„=1Цр D5.1)
n-l Г —nl2 / k\
В2я.1(х):=В2я.1 = I L^iji _*b D5.2)
k=o к! \ nj
Proof. We shall induct on n. For n = 1, both D5.1) and D5.2) are easily seen
to be correct.

172 12. Continued Fractions
We shall thus assume that both D5.1) and D5.2) are true up to a specific
positive integer n. By A.4),
B2«+i(x) = B2n(x) + nxB2n_i(x)
X~< \ ^vfc -^ \~" V ft)k — lv^ К ~r i)X
= k% ~~kl + h (k - 1)! '
But, for 1 < к < n,
it]
{(n-
fe! V "+1
Hence, we have established D5.2) with n replaced by n + 1.
By A.4) and the proof just completed above,
В2я+2(х) = B2n+l(x) + (n + 1)xB2h(x)
к
-xk.
But, for 1 < к < n + 1,
k\
n + l n + lj k\
Hence, D5.1) is established with n replaced by n + 1. ?
We have slightly rearranged the ordering of the formulas in Section 46.
Entry 46(i). For\x\<\,set
j(=o к!
Define (р„{х) as the constant term in the Laurent expansion of xT(l — p)/p",
0 < |p| < 1, where n is a nonnegative integer. Then, if x ф О,
(^- D6-2)
Furthermore, define ф„(х), n>0,by
Then, for n > 1,

12. Continued Fractions 173
D6.4)
Proof. First, for \p\ < 1, by D6.1),
хрГA - p) J ? Pk Log" x
p" r=o Г! \fc=o
Equality D6.2) is now immediate.
Using D6.2) in D6.3) and differentiating both sides with respect to x, we
find that, for n > 1,
The proof of D6.4) is now complete. ?
Entry46(ii). Forn> 1,
" in — IV
= Z —^
k=i (n - /c)!
vv/iere Лк is defined by D6.1), Sx = y, and Sk = ?{k), k>2, where С denotes the
Riemann zeta-function.
Proof. Entry 46(ii) is a reformulation of a well-known result that can be
found in Luke's book [1, p. 27]. Namely, if
Г(х + 1) = ? bkxk, \x\<l,
k = 0
then, for n > 1,
nbn=?(-l)kSkbn.k. D6.6)
Translating the recursion formula D6.6) in terms of the coefficients Ak, we
readily obtain D6.5). ?
We state Entry 46(iii) as recorded by Ramanujan. Afterward, we discuss
the accuracy of his numerical calculations.

174 12. Continued Fractions
Entry 46(iii).
Furthermore,
In the notation
if we write
Г(х + 1) =
ь,
b-
bt
1 +
D6.6),
, = -0.5772156649,
i = 0.9890560173,
, = -0.9074790803,
v = 0.9817280965.
btx + b2x2 + b3x3 + fr4
x4
1 + вхх'
D6.7)
i -i- axx
then
0O = 1.00027,
0t = 51/52,
в2 = 77/82,
06 = 5/68,
07 = -1/38
"nearly."
The coefficient b^ is equal to — 7, and the numerical value that is given is
correct. The given values for b2, b3, and b4 do not seem to be correct. We have
employed D6.6) along with values of Sk given in Abramowitz and Stegun's
tables [1, p. 811] and have found that
b2 = 0.9890559953,
b3 = -0.9074790762,
fe4 = 0.9817280865.
Evidently, we are to interpret 9X to be that unique number yielding an
equality in D6.7). The values given by Ramanujan are rational approxima-
approximations. The value for в0 is enigmatic, because, for x = 0, в0 is not well defined.
In the table below, we give the calculated values of the right side of D6.7) using
Ramanujan's determinations and also our determinations of b2, b3, and bA.
x вх T(x + 1) Ramanujan's Value Our Value
1 51/52 1 0.999990949 0.999990967
2 77/82 2 1.999702292 1.999702625
6 5/68 720 719.9611865 719.9612493
7 -1/38 5040 2623.541808 2623.542013

12. Continued Fractions 175
Thus, the values for ви в2, and 96 give good approximations, but the value
for 97 certainly does not.
We are very grateful to Henri Cohen for motivating the proof of Entry
46(iv) below. In particular, he informed us of formula D6.20). As in Entry 17,
the equality below refers only to the correspondence between the two sides.
The left side is a power series, and the continued fraction on the right side is
the (unique) C-fraction corresponding to the power series.
Entry 46(iv). If n is a nonnegative integer, then
\ D6.8)
+ 2 + 6x + 10 + •
Proof. From D6.4), it is clear that ф„(х) can be expressed as a power series
in 1/x. Putting
k=0 X
we then write D6.4) in the form
- aM - (A: - 1)ак-Лп) _ Z ак-Лп - 1) ,
Zj vk + L к ~ L vk ' к*0-*)
k = O X k = 2 X k=l X
where n > 1. It follows immediately that ao(n) = 0 if n > 1, a^n) = 0 if n > 2,
and
ak(n) + (k- l)aft_1(n) = ak.x{n - 1), D6.10)
for к > 2 and n > I. Now assume that, up to some fixed integer к — 1,
flt-i(n) = 0 if n > k. Thus, ak^(n - 1) = 0 if n > fc + 1. It follows from D6.10)
and our inductive assumption that ak(n) = 0ifn>fc+l. Hence, we shall
rewrite D6.9) in the form
» Ш « (n + k-l)bk-M _ « ^(n - 1)
kkx"+k + kk x»+k h x-+k '
Hence, for n > 1,
&о(и) = bo(n - 1) D6.11)
and, for k, n > 1,
bk(n) + (n + k- l)bk-M = bk(n - 1). D6.12)
From the definition D6.3) of фп{х), it is easy to see that фо(х) = 1- Hence,
by D6.11) and induction, we find that
bo{n)=l, n>0. D6.13)
Next, in D6.12), let к = 1 and replace n by j. Since bo(j) = l,j> 0, we find
that

176
12. Continued Fractions
bid) + j = b^j — 1), j > 1. D6.14)
Summing both sides of D6.14) for 1 < ; < n and recalling that b^O) = ax@) =
0, we deduce that
or
bi(n) = -hn(n + 1).
Put к = 2 and n =j in D6.12) to obtain the equality
b2U) + U+ i)bi(y) = b2u-i).
D6.15)
D6.16)
Sum both sides of D6.16) onj, 1 <j < n. Using the fact that b2@) = 0 as well
as D6.15), we find that
+ l)(n + 2)Cn + 5). D6.17)
Lastly, we set к = 3 and n = j in D6.12) and find that
h(j) + (j + 2)b2(j) = b3(j - 1). D6.18)
Summing both sides of D6.18) for 1 < j < n and employing D6.17), we find
that
Z4 j=i
Ю + 1H" + 2JC; + 5)
+ l)(n + 2J(n + 3J,
D6.19)
after a lengthy calculation. (Formulasforsumming^1<J<n7'[, 1 < к < 5,may
be found in Gradshteyn and Ryzhik's tables [1, pp. 1, 2].)
In conclusion, from D6.13), D6.15), D6.17), and D6.19), we have demon-
demonstrated that, for x sufficiently large,
1
n(n + 1) n(n + \){n + 2)Cn + 5)
n{n + l)(n + 2J(n + 3J
48x3
Now, by D6.8), we wish to prove that
D6.20)
= x +
= x
n 5n + 10 41n + 58
2+ 6x + Ш +••¦
n 5n + 10 41n + 58
2x 12x
T+ P
60x
1

12. Continued Fractions 177
In fact, it will be slightly more convenient to show that the reciprocals of the
expressions above ae equal. Hence, we shall prove that
n 5n+ 10 41и + 58
In order to establish D6.21), we shall first compute the power series for
{х"ф„(х)}1/(л+1) in powers of 1/x. By D6.20) and the binomial theorem, we find
that, for x sufficiently large,
1 / n(n + 1) n(n + l)(n + 2)Cn + 5)
- H : г I z l~
n + 1V 2x ' 24x2
n{n + l)(n + 2J(n + 3J
48x3
n{n + 1) n(n + l)(n + 2)Cn + 5)
2(n + IJ V 2x 24x2
nBn+l)
We now compute the coefficients ct(n), сг(п), and c3(n) of 1/x, 1/x2, and 1/x3,
respectively. Clearly, c:(n) = — n/2. Second,
n + 5) и3 пA1и + 10)
24^T 24 ¦
Third,
n(n + 2J(n + 3J n\n + 2){3n + 5) n4(
+
Сз(")=
48+4848
n(9n2 + 20n + 12)
~ 16
Hence,
We now employ Entry 17 to compute the continued fraction representation
D6.21). In the notation of Entry 17, by D6.22),
n n(lln + 10) J n(9n2 + 20n + 12)
X Aand А
Аз16
First,
a, = A, = H-. D6.23)

178 12. Continued Fractions
Second,
P2 = a^a, + a2) = Л2 =
Using D6.23) and solving for a2, we readily find that
Lastly,
n(9n2 + 20n + 12) n2
a2 + a3) = A3-a1A2 =
16 24
Solving for a3 and employing D6.23) and D6.24), we find, after a mild calcula-
calculation, that
41n + 58
fl3 = -60"- D6-25)
Employing D6.23)-D6.25) in Entry 17, we complete the proof of D6.21). ?
Example. For x > 0, let
1 e
F(x)= I ——dt.
Jo f
T/ien
Proof. First, from Entry 44,
if2(x) = iy2 + i Log2 x + у Log x + o(l), D6.26)
as x tends to oo.
Next, integrating by parts twice and using Entry 44, we find that, as x tends
to со.,
C 1 e
= F(x) Log x - Log t dt
Jo l
= (y + Log x) Log x + o(l) - ^A - e~x) Log2 x + H e~' Log2 t Л.
Jo
D6.27)
Combining D6.26) and D6.27), we deduce that

12. Continued Fractions 179
-(^dt- |F2(x) = -\y2 + U e~' Log2 tdt + o(
Jo
= ~b2 + К"A) + 0A), D6.28)
as x tends to oo.
By Entry 26 of Chapter 7 (see the author's book [9, p. 176]),
Hence, after two differentiations,
T"/ I 1 \ 00
T(x + 1) k = 2
and so
Г"A) = ф2{1) + CB) = у2 + n2/6.
Substituting the value for Г"A) found above into D6.28) and letting x tend to
oo, we complete the proof. ?
Entry 47. // n is any complex number outside of (— oo, 0], then
GO
eTx(l + x/n)" dx
:i + ? + lv, + 2^ + 3(^)+ D?1)
, n- 1 l(n-2) 2(n-3) 3(n-4)
D7.3)
4+ 6
2n 3n 4n 5n
„- 2 + 3 + 4 + 5 +••••
Proof. In B1.2), let x = y/п and fi = —n. Thus, under certain restrictions on
у and n arising from B1.2),
- n, 1; у + 1; -y/n)
B - n)y/n 2A + y/n)
У + 1 + 7 + 1 +
Now, for Re(y/n) > 0 (Bailey [4, p. 4]),
¦ D7.4)
- n, 1; у + 1; -y/n) = у Г A - t)""^! + ty/пГ1 dt
- t)""^
o
у
= A - и/уУ'1^ + u/n)"'1 du. D7.5)
Jo

180 12. Continued Fractions
Thus, letting у tend to со in D7.4) and D7.5), we find that, for n exterior to
Integrating by parts once, adding 1 to both sides, and writing the right side
above in an equivalent form, we see that
n 1-й 1 1-n 2
e""(l + ujrif du=\+- —— - —— -
о n+ 1 +n+ 1 + и + •••
n n - 1 2(n - 2) 3(n - 3)
~ +I + ~3~+ 5 + 7 +•••'
by Entry 14. This completes the proof of D7.1).
Second, let x = y/n and jff = 1 - n in B1.2). Then, for Re(y/n) > 0,
n — 1
—— 2^1 B - и, l; ? + l; -r/n)
= (n - 1)у/и B - и)у/н 1A+y/n) C - n)y/n 2A + y/n)
у + 1 + у + 1 + у +•••'
Now proceed as above and let у tend to oo to find that, if n is outside (—oo, 0],
n- 1
n
e~'(\ + t/n)"'2 dt
о
(n - l)/n B - n)/n 1/n C - n)/n 2/n
n-1 2-й 1 3-n 2 4-n
n + 1 + n+ 1 +n+ 1 +
n - 1 n - 2 2(n - 3) 3(n - 4)
2 + 4 + 6^ + 8 +•••'
D7.6)
by Entry 14.
Assuming that n is any complex number outside (—oo, 0] and integrating
by parts twice, we find that
e'\\ + t/ri)"-2 dt= -2+ \ e~'(l + t/nf dt.
n Jo Jo
Substituting the formula above into D7.6), we establish D7.2).
Third, setting x = t — n, we find that
|*0O gfl 1*00
e~x(l + x/rif dx = — e-'t" dt
JO П Jn

12. Continued Fractions 181
_ е"Г(п + 1) » nk+l
n" *=o(n+l)*+i
епГ(п + 1)
= hi — iFi(l; n + 1; n),
n
where in the penultimate line we employed D3.2). Applying Corollary 2 in
Section 21, we complete the proof of D7.3). ?
In essence, Entry 47 is due to Nielsen [1], [2]. Equality D7.1) may be
derived from [2, p. 46, Eq. F)]. Equality D7.2) can be deduced from [2, p. 47,
Eq. A1)]. Lastly, equality D7.3) can be proved by using [1, p. 219, Eq. (8)].
Note that, by Corollary 2 in Section 21, the continued fraction in D7.3)
actually converges for all complex n.
Entry 48. As n tends to oo,
16 8992
The asymptotic expansion given above first appeared in Ramanujan's
solution to an ultimately famous problem proposed by Ramanujan [4],
[16, pp. 323, 324] in the Journal of the Indian Mathematical Society. In
addition to Ramanujan's (formal) solution, later proofs were given by Watson
[3] and Szego [1]. In fact, the last displayed term on the right side of D8.1)
has not been recorded by any of the aforementioned authors. Further coeffi-
coefficients have been calculated by Bowman et al. [1] and Marsaglia [1].
The corollary below is similar to the aforementioned problem posed by
Ramanujan [4], [16, pp. 323, 324]. A version of this corollary was also
communicated by Ramanujan [16, p. xxvi] in his first letter to Hardy.
Corollary. Define 9 = 9nby
n-i nk n" e"
Then
A 4- 1 Sn
D8.3)
Proof. As Ramanujan [4], [16, p. 324] easily demonstrated,
„ e"T(n -
+ 1 - e~'(\ + x/nf dx, D8.4)
and so D8.1) may be reformulated as

182 12. Continued Fractions
as n tends to oo. On the other hand,
4+15n 1 4 32
as n tends to oo. Thus, в* is a fairly good approximation to в. ?
In 1983, a problem similar to the corollary above was published, and a
lengthy discussion, with three solutions, was given in a later issue of the
Mathematical Gazette [2]. In particular, suppose that each of the n inde-
independent random variables Xk, 1 < к <п, has a Poisson distribution with
parameter 1. Then Sn := ?"=1 Xk has a Poisson distribution with parameter
n. Thus,
After applying the central limit theorem, we conclude that
lim P(Sn <n) = \.
n~*GO
For further connections of the aforementioned corollary to probability, see
the papers by Bowman et al. [1] and Lawden [1].
The integral of Entry 48, as well as a generalization, arises in a solution of
the famous "birthday surprise" problem. See the delightful paper by Blaum
et al. [1] where earlier work of Klamkin and Newman [1] is corrected and
greatly extended.
A result analogous to D8.5) has been obtained by Copson [1] for e~". More
precisely, if <р„ is defined by
6 k k\ + п!
then
1 1 1
as n tends to со.
Generalizations of Ramanujan's and Copson's theorems have been estab-
established by Buckholtz [1] and Paris [1]. The commentary in Szego's Collected
Papers [2, pp. 151, 152] provides a good summary of the literature on
generalizations and related problems. Another proof of Ramanujan's result
D8.5) as well as some related results may be found in Knuth's book [1,
pp. 112-117]. Carlitz [1] has examined a class of functions arising in the work
of Ramanujan, Copson, and Buckholtz. Jogdeo and Samuels [1] considered
a binomial analogue of D8.2).

12. Continued Fractions 183
Ramanujan concludes Section 48 with the following table.
n
0
i
2
1
3
2
2
00
0.50000
0.37750
0.35914
0.35146
0.34726
0.33333
в:
0.50000
0.37705
0.35849
0.35099
0.34694
0.33333
Of course, when n = 0, it is trivial that 90 = 9% = \. From D8.5) and D8.6),
it is clear that 6X = в* = ?. The proposed values for 0,, 0*, 02, 0J, 0*/2, and
0|2 are easily corroborated by using the definitions of 9n and #* given in D8.2)
and D8.3). It remains to examine the values of 91/2 and 93j2.
In order to calculate 01/2 and 93/2, we shall employ D8.4) and the continued
fraction D7.3). Hence,
,n=1_^±J) + ^ 34 ^ ~ , «>0. D8.7)
2n" 2 + 3 + 4 + 5+---
In the notation of A.3) and A.4), when n = \,
Ak = (k+ \)Ak^ + |(k + 1)Л_2, к > 1,
and
Bk = (k+ 1)Вк_! + i(fc + l)Bk_2, fc > 1.
By successive calculations, we eventually find that
— = 0.4106925, — = 0.4106857, — = 0.4106862.
B5 B6 B7
Thus,
2/2 3/2 4/2 5/2
Since
= 0.410686.
2 + 3+4 + 5 +¦••
/^= 1.033182838,
2 V 2
we conclude from D8.7) that Ramanujan's proposed value for 9l/2 is correct.
If n = |, again, from A.3) and A.4),
Ak = {k + 1K_! + |(fc + \)Ak-2, k>l,
and
Bk = (k + Wu-i + Ш + l)Bfc_2, /c>l.
Iterated calculations yield
~ = 0.972952, — = 0.972930, — = 0.972933.
°i B8 B9

184 12. Continued Fractions
Proceeding as above, we find that 93/2 = 0.35145, which differs slightly from
the value given by Ramanujan.
Ramanujan [4], [16, p. 324] conjectured, probably partially on the basis
of his calculations above, that 9n always lies between \ and \. This conjecture
was proved by both Watson [3] and Szego [1].
Entry 49. For each integer n>2, define 9 = 9nby
» nk Л ft! (n-1)! Л
r + Log n + ? — = e" ? Zi+r + L-ir
k=l K\K \k=0 П П
where у denotes Euler's constant. Then, as n tends to oo,
We are very grateful to F. W. J. Olver for providing us the following
solution based on material from his book [1].
Proof. First, observe that, for n > 0,
By combining D9.1) with a familiar formula for у (Olver [1, p. 40]), we readily
find that
7 + Logn+ ?— = PV\ -dt=:Ei(n),
where n > 0. Olver has calculated an asymptotic series for Ei(n), and in the
notation of his text [1, p. 529, Eq. D.06)], в = CH_i(n). By [1, p. 529, formula
D.07)],
D9.2)
k=o yn - i)
as n tends to oo, where the first three values for yk{\) are given by (see
[1, P- 530])
7o(D=4 ^ = Ш' and У^=~Ш
Putting these values in D9.2), we deduce that
2 4 76
9 = - +
3 135(n - 1) 2835(n - IJ
_2 4
~3 +
from which the proposed asymptotic expansion follows. ?
For much of the theory of Ei{n), see Nielsen's book [2].

CHAPTER 13
Integrals and Asymptotic Expansions
In assessing the content of Ramanujan's first letter to him, Hardy [9, p. 9]
judged that "on the whole, the integral formulae seemed the least impressive."
Later he added that Ramanujan's definite integral formulae "are still inter-
interesting and will repay a careful analysis" [9, p. 186]. Indeed, a dismissal
of Ramanujan's contributions to integration would have been decidedly pre-
premature. First, we might recall that this first letter contained several remarkable
formulas on series and continued fractions. In evaluating infinite series and
deriving series identities, Ramanujan had no peers, except for possibly Euler
and Jacobi. Ramanujan's work on continued fraction expansions of analytic
functions ranks as one of his most brilliant achievements. Thus, if Ramanujan's
contributions to integrals dim slightly in comparison, it is only because the
glitter of diamonds surpasses that of rubies. Indeed, there are many elegant
and important integrals that bear Ramanujan's name. (See, for example,
Entry 22.)
Chapter 13 is largely devoted to integrals. In this chapter, we find some of
Ramanujan's more prominent integral evaluations. In particular, many of
the integrals from [8], [16, pp. 53-58] are found here. But much more
importantly, Chapter 13 contains some absolutely remarkable results not
heretofore observed. Entry 6 gives an asymptotic expansion of a certain inte-
integral and provides a generalization of a famous question posed by Ramanujan
[4], [16, pp. 323, 324] in the Journal of the Indian Mathematical Society.
The latter problem and related asymptotic expansions may be found at the
end of Chapter 12. Entry 7 is a highlight of Chapter 13 and a truly remarkable
formula. Ramanujan offers here an asymptotic expansion of a certain integral
as two parameters tend to oo. From both theoretical and computational
standpoints, Entry 7 was very difficult for us to prove. As a by-product of
Entry 7, we obtain an asymptotic expansion for the hypergeometric function

186 13. Integrals and Asymptotic Expansions
1, m; m — n; I as m, n, and m — n tend to oo. Such an expansion
does not appear to have been previously given in the literature. Another
elegant asymptotic formula for an integral appears in Entry 8. This expansion
is related to the confluent hypergeometric functions Ф(а, с; z) and ^(a, c; z)
(Lebedev [1, pp. 260, 263]). We have proved a generalization of Entry 8 in
Section 10 (see A0.22)). Entry 5 is a very unusual integral formula that has
its roots in a favorite theorem of Ramanujan, an interpolation formula in
the theory of integral transforms. Special cases of Entry 5 are formulas for
K-Bessel and confluent hypergeometric functions.
In addition to theorems on integrals, Chapter 13 contains material on
infinite series. Undoubtedly, the most impressive results on series appear in
Section 10. Entry 10 offers an extraordinarily beautiful asymptotic expansion
for series that are remindful of hypergeometric series. We know of nothing
like it in the literature. Corollary (i) is also a very interesting result which, in
a special case, is related to Entry 8 and therefore to confluent hypergeometric
functions.
It should be remarked that none of Ramanujan's integral evaluations or
asymptotic expansions is accompanied by conditions of validity. Particularly
in Entries 5, 7, and 10, the determination of these conditions was not an easy
task.
For an enlightening discussion of several of Ramanujan's asymptotic ex-
expansions and for some further generalizations, see Evans' paper [1].
As might be expected, several of Ramanujan's integral evaluations are
classical. It would be very difficult to determine the original discoverers of
these results, and so we usually content ourselves with just pointing out their
appearances in the tables of Gradshteyn and Ryzhik [1].
Occasionally, we shall write expressions such as
By this we mean that
fix) = g(x)h(F(x)),
where F(x) has the asymptotic expansion
F(x)~a0 + — + Ц +
as x tends to oo.
Entry 1. Letn>0 and put N = [n + 1]. Then
TV" ? Ak(-x)kdx = (-l)N ["х-"*"-1 ? AN+k(-xfdx,
Jo k=0 Jo k=O
when the right side is meaningful.

13. Integrals and Asymptotic Expansions 187
Ramanujan does not intend Entry 1 to be a theorem, but instead he is
defining the integral on the left side by the expression on the right side. To
illustrate Entry 1, Ramanujan gives the example
e~x
]o x
which is to be interpreted as
e~x2 - 1 + x2
A.1)
0
This result is easy to establish either directly or by using the general formula
(Whittaker and Watson [1, p. 243])
T(z)=\ t*-i e-« - ? 5-± Д A.2)
Jo \ *=o K\ J
due to Cauchy and Saalschiitz, where the integer n is chosen so that
— n — 1 < Re(z) < — n. Hence, employing A.2), we find that
6 X ~* ' " • * ' -'"¦ -'-\ + t)dt
which establishes A.1).
Corollary. If a,n > 0 and b is real, then
f°° -„ .-icos(bx) T(n) cos(« tan-1 (b/a))
Jo 6 X sin(fex) X {a2 + b2)"/2 sin(n tan-1 (b/a)Y
These two formulas are well known (Gradshteyn and Ryzhik [1, p. 490]).
Ramanujan furthermore remarks that the integrals above "for negative values
of n are known." Indeed, Ramanujan's definition in Entry 1 assigns a meaning
to these integrals for negative values of n. In fact, these same formulas still
hold if n < 0, provided that n is not a negative integer. To that end, using
Ramanujan's definition from Entry 1 and A.2) and defining the nonnegative
integer m by — m — 1 < n < — m, we find that
f°° 1 f °°
e "x" cos(bx) dx = - x" 1(ex( a+bl) + ex{ " l))dx
Jo 2 Jo
f»
= Ща - bi) " + (a + bi) "} x"~1e~x dx
Jo
= Ща - ЫУп + (a + Ы)-"} \ x" ( e~x - ? i~^-) dx
Jo V *=o ki )
= (a2 + b2ynl2 cos(n tan-1(b/a))r(n).
A similar argument holds for sin(bx) in place of cos(fex).

188 13. Integrals and Asymptotic Expansions
Entry 2. Let cp have m + 1 continuous derivatives. Then
(i) <р{х)е-ю dx = -e~nx ? ^У + -^ Ф<т+1»(х)е-« dx,
J n n J
(ii) I cp(x) cos(nx) dx
^2(-l)V2*>(x)
) ? () ?
/ |jm/2 + l Г
H ^pt— <p<m+1)(x) sin(nx) dx, if m is even,
fc=O
+ ^— <p{m+i)(x) cos(nx) dx, if m is odd,
(iii) I cp(x) sin(nx) dx
. /"Z2-1 (-1)V:
() I 3ST2() Z I
k=0 И 4=0 П
/ iyn/2 |*
H ^j— (p(m+1)(x) cos(nx) dx, if m is even,
I 124 + 2 C0S("X) I ЗШ
t=o 1 k=o и
/_ jym+l)/2 [*
+ ^j— <p(m+1)(x) sin(nx) dx, if m is odd.
All the equalities above may be established by successively integrating by
parts.
Entry 3. Let n,x>0 and define 9 and r by 9 = tan'^n/x) and r = (n2 + x2I'2.
Suppose that m is any positive integer. Then as x tends to oo,
e~'2 cosBnt) dt
bcosBBx + Bfc + l)fl) _2m_
Proof. Upon successively integrating by parts, we find that, for x sufficiently
large,

13. Integrals and Asymptotic Expansions 189
<T'2 cosBnt) dt
е'"г Г00 , 1 , Г00
е~ (Г00 е~' . Г00 е~г
4 \.)(x-inJx/t J(x+inJ
1 f00 e"' , 1 Гж <Г' ,
f
4 \x-in x + in 2 J(x_inJ f3/2 2 J(jc+(nJ13/2
i /„-x2 + 2inx „-x1-2inx „-x2 + 2inx g-x2-2inx
4\ х - in x + in 2(x - inK 2{x + inK
(x-inJ L J(x+inJ L
+i$ „-2inx-ie „2inx + 3i6
2r3 2r3
2r5
22r5 ' 22r
z U(x-inJ ' J(x+inJ l
It is now clear that, after m integrations by parts, we may easily deduce the
desired formula. ?
Entry 4. Suppose that (p is entire, n is real, and that the integrals and series
below converge. Then
e~x2{e2"x(p(x) + e~2nx(p( — x)} dx = e~x2{<p(n + x) + cp(n — x)} dx
Jo Jo
00
el
Proof. Letting / denote the integral at the far left side, we find that
e'x2+2nx(p(x) dx = e \ e'^2cp(x) dx
= e e'x2(p(n + x)dx= e-x2{(p(n + x) + <p{n - x)} dx,
J-oo JO
and so the first equality of Entry 4 is established.
Expanding <p(n + x) and <p{n — x) in power series, simplifying, and inverting
the order of summation and integration by a theorem in Titchmarsh's book

190 13. Integrals and Asymptotic Expansions
[1, p. 47], we find that
е~'г{ф + x) + ф - x)} dx = 2 e~*2 ? ^-^- dx
J B/c)!
from which the second equality of Entry 4 easily follows. ?
As an example, if we put (p(x) = ex in Entry 4, we find that
Гж f00 /пеп2+п+1/*
e~*2 cosh(Bn + l)x) dx = e»2+»-*2 COsh x dx = ^ .
Jo Jo 2
In order to state Entry 5, we first need to enunciate a theorem due to
Hardy [9, p. 186, formula (A)]. See also Part I [9, p. 299]. Let s = a + it with
a and t both real. Let Я(E) = {s: a > —d}, where 0 < 5 < 1. Suppose that
\j/{s) is analytic on Я(E) and that there exist constants C, P, and A with A < n
such that
\4>{s)\ < CePa+AM, E.1)
for all s e Я(E). For x > 0 and 0 < с < д, define
ico л
V ^-s)x~* ds. E.2)
ITU Jc_to SinGTs)
If 0 < x < e"p, an application of the residue theorem yields (Hardy [9, p. 189])
»P(x)= t
k = 0
Finally, if 0 < a < 5 (Hardy [9, pp. 189, 190]),
y?(x)xs'1 dx = -r^M-s). E.3)
Entry 5. Let \l/(s) satisfy the hypotheses of Hardy's theorem given above for
some д > j. Put i//(s) = ^s+i/Hs + 1). and so, in the notation above,
Suppose that for a = 28 > 1, xa/2lP(x2) e L2@, oo). Then
k=0

13. Integrals and Asymptotic Expansions
191
Ramanujan (p. 156) states Entry 5 in the form
1 \K
e "~ } -
0
k\
2 ?
{-2fAk
fc!
Although 4*(x) has been defined for x > 0 by E.2), there is no guarantee that
its power series converges for all x.
Proof. First, for 0 < a < 23,
E.4)
by E.3).
Second, for a < 0,
f
Jo
~u du = \T(-
E.5)
We now apply Parseval's theorem for Mellin transforms (Titchmarsh
[2, p. 95]). Using E.4) and E.5), we find that, for a > 1,
In order to evaluate the integral on the right side above, we examine
—— Us = -^
2 / 2m г
E.7)
where CM N is a positively oriented rectangle with vertices a ± iM and
— N ± iM, where M, N > 0 and N = |(mod 1). By hypothesis, the only
singularities of the integrand for a < a are at s = 1 — k, where к is a non-
negative integer. Thus, by E.7) and the residue theorem,
By E.1), for <7< a,
r'i
I\I,N = \Jn
S- 1
E.8)
0<k<N + l
ns n(s — 1)
csc у csc—-—Ax_s
<Cn2
ns n(s — 1)
csc — csc
3-s
From the upper bound above and from Stirling's formula, we easily see, by

192 13. Integrals and Asymptotic Expansions
first letting M tend to oo and then letting JV tend to oo, that
Hm IM,N = -L Г" 1г(^_1+1г(Ц1) is
2 J 4 \2) \ 2 )
2тп Ja_ice 4 \2)
2
E.9)
by E.8). Substituting E.9) into E.6), we complete the proof. ?
As a first illustration of Entry 5, we note that (Gradshteyn and Ryzhik
[l.p.307])
J. .-—¦*-*-•-•
For a second example, take Ч'(.х) = A + x)~", where fi> j. Then
An application of Entry 5 then yields
fc!
where in the last line ^(a, c; z) denotes the confluent hypergeometric function
mentioned in the introduction to this chapter.
The theorem of Hardy that we quoted above is a rigorous reformulation
of one of Ramanujan's favorite theorems. It is Entry 11 of Chapter 4 and
also appears as Theorem I in his quarterly reports. See our first volume
[9, pp. 105, 298] on Ramanujan's notebooks, where many applications of
Ramanujan's theorem are also found. Hardy's book [9, Chapter 12] also
contains several applications. According to J. Edwards [2, p. 213], a special
case of Ramanujan's theorem, or the case s = \ of E.3), was established by
J. W. L. Glaisher.
An alternative approach to Entry 5 is now sketched. Suppose that we
expand exp(—1/x2) in a power series, invert the order of summation and
integration, and apply the aforementioned favorite theorem of Ramanujan.
Accordingly, we find that
1 со |_|V'
2) dx Y 'r
Y
1 oo ( _ 1
du
_ ° 22M2J
2 Д B/)! '

13. Integrals and Asymptotic Expansions 193
Thus, we obtain the "wrong" answer; the odd indexed terms do not appear!
Now, in fact, Ramanujan used this same type of argument in many similar
instances; see our account of the quarterly reports in [9]. Despite the non-
nonrigorous nature of the procedure, Ramanujan possessed extraordinary intuition
in determining when the process leads to the correct formula and when it
leads to an incorrect formula.
The case h = 0 of the asymptotic expansion in Entry 6 below is essentially
a famous problem that Ramanujan [4], [16, pp. 323, 324] submitted to the
Journal of the Indian Mathematical Society. See also Entry 48 of Chapter 12
for the case h = 0. Watson [3] has made a more detailed study of this
asymptotic expansion, and we shall use some of his analysis in our proof of
the generalization below.
It should be remarked that the first integral below is equal to иЧ*A, п +
2-h;n) (Lebedev [1, p. 268, formula (9.11.6)]), where *?(a, c; z) denotes the
confluent hypergeometric function.
Entry 6. Let n > 0 and suppose that m is a positive integer. Then
k=o
nm Jo V "
_e"T(n-h + l) A1 A2
2n n n
as n tends to oo. Here,
. 2 4 h\\ - h)
A H A
H> AiW5'
8 2AA - A) A(l - h2)B - 3h2)
+
z 2835 135 45
Proof. The first equality in Entry 6 follows by successively integrating by
parts m times.
We now establish the asymptotic expansion. Putting x = (U — V)n and
x = un, respectively, in the two integrals below, we find that
-x(л x\"~h , eT(n-h + 1)
e x 1 + - dx —.
о \ «/ 2n"
= n I en^-U)Un-h dU--\ en{1-u)un-h du
li 2 Jo

194 13. Integrals and Asymptotic Expansions
2jx 2JO
dt, F.1)
where we have made the changes of variables el VU = e ' and e1 "и = е ',
respectively, in the two foregoing integrals.
From Watson's paper [3], for t sufficiently small,
It follows that
dU 1 2 BtI/2 At BtK'2 At2
+ + + +
Л BtI/2 3 12 135 864 2835
and
| 2t ( (ItK!2 | j
+ - -^ ip" + y + -
A(A + 1)(A + 2)(A + 3)(A + 4)
120
The expansion for u(t) in ascending powers of yjt is the same as that for
U(t), except that the coefficients of odd powers of y/t are of opposite signs.
Omitting all the algebraic calculations, we find that
...JV 1
dt BtI/2 \3 7 ' ' V 135 3 3
4 2h AAh{h
2835 + 135 + 135
7fc(A + 1)(A + 2) A(A + l)(fc + 2)(A + 3)
A(A + 1)(A + 2)(A + 3)(A + A)\
30 ]t

13. Integrals and Asymptotic Expansions 195
where cl5 c2, ... are certain constants, depending on h but not on t. The
expansion for u~h du/dt is the same as that above, except that the coefficients
of odd powers of ^ft are of opposite signs. By the same justification as in
Watson's proof [3], we thus obtain the following asymptotic expansion as
n tends to oo:
2 Jo V dt dt
8A(A+1) 2h(h + l)(ft + 2)
/8 Ah %%h{h + 1) \Ah(h + \)(h + 2)
+ \2835 + П5 + 135 9
2h(h + 1)(/t + 2)(h + 3)
15
2 1/ 4 /т2 Л3
--/i + - h
3 n\ 135 3 3
1 / 8 Ah _ 2h2 Л3 _ h5
I35i ~ I35 + J ~ 15
By F.1), this completes the proof. ?
Entry 7. Let m> n + \. If m and n tend to oo while m — n remains bounded,
then
/:=/(„, „):=(«-»-!) Гп^Т^- t ^ + 0A). G.1)
Put
n(m — n)
If m, n, and m — n tend to oo, implying that also R tends to oo, then we have
the asymptotic expansion
T(m-n+ 1)
1Г-" } + Al+A2 + A3 + AA + -t G.3)
where Ak, 1 < к < oo, is a rational function of m and n such that
Ak = 0(mRl-k), G.4)
as m, n, and m — n tend to oo. Moreover,

196 13. Integrals and Asymptotic Expansions
. 2(m + n)
G.5)
4(m + w)(m - 2n)(m - j
_
2 "
Цт3 + n3)(m - 2n)(r,
A-, = —
n (m — n)
and
_ 16(m3 + n3)(m - 2n)(m - \n)(m2 - mn + n2)
A* — ОСПС...З..З/ \3 • У'-°>
Proof. Replacing x by nx in G.1), we find that
/ = (m - n - \)n\ (I + x)"(l + nx/m)"m dx. G.9)
Using a standard integral representation for the hypergeometric function
2^@, b; c; z) (Luke [I, p. 57, Eq. B)]), we deduce that
, m; m - n;m П). G.10)
m I
We first suppose that as m and n tend to oo, m — n < В for some constant
B. ByG.lO),
I = nf
k=o (m - n)k mk '
and so
f (m ~ ")* f (m - n)"(
mf f
n t=0 (m - fi)t k=0 (^ - п)ктк
n
To prove G.1), we shall show that the left side of G.11) is bounded as m and n
tend to oo, by proving that
(m - nUmX - m*)
for all m and к sufficiently large.
Clearly,
(m — n)k k\
Next, by the mean value theorem,
' ( '
G.13)
m" x \ m

13. Integrals and Asymptotic Expansions 197
Thus, by G.13), G.14), and Stirling's formula, for к and m sufficiently large,
1к<1\к2 <2 ' lffc~m'
and
Tk<-k2[—\ <2'\ if m<k.
k\ \m/
Hence, G.12) is established, and therefore the proof of G.1) is complete.
Second, we suppose that m, n, and m — n tend to oo. For brevity, set
тт+1Г(п+ 1)Г(т-п+ 1)
ij — ——
2nT(m + Щт - пТ~я
Employing a basic integral representation for the beta function (Gradshteyn
and Ryzhik [1, p. 948, formula 3]), we see that
_ mm(m -n- \)T(m -n- 1)Г(и + 1)
S ~ 2n{)m-n
mm(m-n- 1) f00 x" dx
2nn{m - n)m'n l Jo A + x)m
mm(m -n-\)n f00 t"dt
2(m-nT Jo
m — n
Combining G.9) and G.15), we obtain the representation
G.16)
The former integrand in G.16) is decreasing on @, oo), while the latter integrand
is increasing on (— 1, 0). In order to see this, define
^, G.17)
and observe that
()
m \m

198
13. Integrals and Asymptotic Expansions
where R is defined by G.2). Now replace both integrands in G.16) by e~' to
obtain
G.19)
By the inverse function theorem, for t > 0 and t sufficiently small,
00 CO
u(t)= ? aktki2 and U(t)= ? {-\faktkl1, G.20)
where the coefficients ak, 1 < к < oo, are functions of m and n with
«! = -R/2. G.21)
Recalling G.17), we observe that, for \u\ < 1,
t =/(«):= -Loge(u) = mLog(l +-%)-nLog(l + u) = Ru2 У скик,
\ m ) k=0
G.22)
G.23)
where
Note that
c0 = 1, Cl = -
Thus, for |z| <5,
> 1
(fc + 2)m*(m - n) '
2(m + ri)
k>0.
-, and \ck
k>0.
G.24)
G.25)
We next proceed to show how the coefficients ak in G.20) are related to the
coefficients ck defined in G.22) and G.23).
For t > 0 and t sufficiently small, let
g(t) = t a^k-
k = l
From G.22), t2 = f(u(t2)) = f(g(t)), and so t = Jf(g(t)). Applying g to the last
equality, we find that
G.26)
« = gWfiu)),
for u < 0 and и sufficiently close to zero. Let R(F) denote the residue of
a function F(z) at a pole z = 0. Then by the Lagrange inversion formula, C6.8)
of Chapter ll,forfc> 1,
ak =
G-27)

13. Integrals and Asymptotic Expansions 199
By G.26) and G.27), for к > 1,
ak = -^^W*72) = ^(/(z)-^). G.28)
Now by G.22) and G.25), for 0 < \z\ < |,
|/(z)|>R|z|2/3>0. G.29)
Hence, by G.28), G.29), and the residue theorem,
f(zrk'2dz, k>\. G.30)
Finally, by G.29) and G.30), for к > I,
1 In „„ 1 /27Y/2
*' 2тг^Т м P l-/( " Vc\~Rl ¦ ( ^
jL.ilК J Z =1/3 ¦-"*¦ \ 1\ /
It follows that the expansions for u{t) and ?/(f) given in G.20) are valid for
0 < t < R/30.
By G.20), G.21), and G.31), there exists a positive number S <j$ such that
uER)<-5 and UER)>5, G.32)
since |u(<5R)| and | UER)\ both exceed
k=2 k=2
= Jo -= > д.
Now return to G.19) and write
I~S = H + J, G.33)
where
and
dR
Fix a positive integer K. By G.20),
where
Я = (m - n - l)n e~r X k'W* dt = H,+ H2, G.34)
Jo *=i
Hl=(m-n- l)n Y ka2k e~4k'x dt G.35)
*=i Jo

200 13. Integrals and Asymptotic Expansions
and
H2 = (m-n-l)n e~' ? ka^t"-1 dt. G.36)
Jo k=K+l
By G.31),
H2<(m-n)n\ ^ I (=Л ^
«R -t
o
<2mR[—J A-27.5)-1 е-'г*Л
Х G-37)
as m, n, and m — n tend to oo. Thus, by G.34)—G.37),
к
H = Hi+ 0(mR~K) = (m - n - \)n Y alkk\ + 0(mR-K), G.38)
as m, n, and m — n tend to oo.
Define, for k > 1,
Л, = n(m - и)а2к/с! - na2k.2{k - 1)!, G.39)
where a0 = 0. Then, by G.2) and G.31), for k < K,
Ak = 0(mR ¦ R~k) +
as m, n, and m — n approach oo. Thus, G.4) holds.
By G.31), G.38), and G.39),
Я = ? Л, + O^R"*), G.40)
as m, n, and m — n tend to oo. In order to prove G.3), it suffices, by G.33) and
G.40), to prove that
J = O(me-R«\
for some fixed positive constant g. Since (m — n — \)n = 0(mR), it suffices to
show that, for some constant g > 0,
fee
e-'U'it) dt, e-'n'(t) dt = O{e-Re\
SR J3R
as m, n, and m — n tend to oo. Changing variables, using G.32), and recalling
the remark made after G.16), we see that it suffices to show that, for some g > 0,
(n \'m f'd In \~m
(U + 1)" [-U+ 1 dU, (m+1)" l-u+1) du = O(e~Re).
\m ) J-i V" /
G.41)

13. Integrals and Asymptotic Expansions 201
Now let ц= +д. By G.29),
I/Ml > R<52/3- G.42)
By G.17) and the aforementioned remark prior to G.17), 0 < Q(/i) < 1. Thus,
by G.22), f(n) > 0. Thus, by G.42), Дц) > Rg, with g = 52/3; that is,
e'fM < e'Re
Hence,
Q(u) du < e~Rg,
I-i
since the integrand is increasing on (— 1, — 6). Similarly,
Q(U)dU <3e~Rg.
Thus, by the last two inequalities, to complete the proof of G.41), it suffices
to prove that
Q(U) < U-Rl2, G.43)
when U > 1, for then
Q(U) dU < U-R/2 dU = — = 0(e-R'2).
J3 J3 R-2
By G.18), for U> 1,
R 2RU
dU ЬУ *У " 2U , (n
(U + 1) -I/+ 1
\m
R
Thus, UR/2Q(U) is decreasing for U > 1. Moreover, with w = m/n,
since A + l/w)w is increasing for w > 1. This completes the proof of G.43)
and consequently of G.3) as well.
In order to calculate At, A2, A3, and A4, by G.39), we need to determine
a2, a4, a6, and a8. To do this, we employ G.28). From G.28) and the value
of Cj given in G.24), it is easy to see that
2(m + ri)
°2 3n(m - n)'
However, the calculations of a4,a6, and a8 rapidly increase in difficulty. After

202 13. Integrals and Asymptotic Expansions
very many hours of excruciatingly laborious calculation, we found that
2(m + n)(m2 — 25mn + n2)
 = 135mn2(m - nf '
4(m + n)(m4 — \4m3n + 267m2n2 — 14mn3 + n4)
U6= 8505m2n3(m - nK '
and
2(m + n)(m6 - 3m5n - 12m4n2 + 389m3n3 - 12m2n4 - 3mn5 + n6)
aK =
25515m3n4(m - rif
The values G.5)-G.8) now follow from G.39) and the evaluations given above.
?
Customarily, Ramanujan provides no hypotheses for Entry 7. Only the
expansion G.3) is given, and G.1) is not found in the notebooks. Although
Ramanujan was very familiar with the Lagrange inversion formula, it is very
doubtful that our proof is substantially like that found by Ramanujan. In
particular, our calculations of A3 and A4 were so involved that Ramanujan
must have had a proof wherein the coefficients Ak arise more naturally with
less computation.
By combining G.3) and G.10) with Stirling's formula, we obtain an asym-
asymptotic expansion for 2^i(l> m; m — n; (m — n)/m), as m, n and m — n tend to
oo. The asymptotic behavior of this 2^i function for general m > n > 0 with
m tending to oo is discussed in the paper by Evans [1, Theorems 15-17].
A vast literature on asymptotic expansions of hypergeometric functions exists,
but this asymptotic expansion appears to be new.
Entry 8. As n tends to oo,
f00 f пхГ(п + 1) / х\ . е"Г(и+1) 6n nt ,.-
Jo \T{n + x + 1) V n/ ) » 12n + 1
Before proving Entry 8, we indicate its connection with the confluent hyper-
hypergeometric functions Ч*(а, с; z) and Ф{а, с; z). As mentioned prior to Entry 6,
пЧ>A, n + 2; n) = e'x(l +-j dx.
Also from Lebedev's text [1, p. 263, Eq. (9.10.3)] and the definition of Ф
[1, p. 260, Eq. (9.9.1)],
иЧ»A, n + 2; n) = пГ(~П~1)ФA, n + 2; n) + Г(" +п 1)Ф(-п, -и; и)
Г(-п) nn
n ? nk e"T(n + 1)
"f z
n + 1 *=o (n + 2)»

13. Integrals and Asymptotic Expansions 203
k (n + i)k + й-
пкГ{п + 1) е"Г(п
+
k^o l\n + к + 1) n"
Thus, Entry 8 may be rewritten in the form
— -dx — ? —: г = —z 1- O(n~3/2). (8.1)
Proof. From Stirling's formula, as n tends to oo,
nxT(n + 1) _ e*n"+*+1/2 f 1 _ 1 /1 \1
T(n + x + 1) ~ (n + x)n+x+1'2 \ + 12n ~ 12(n + x) + [n2)]'
uniformly for 0 < x < oo. Thus,
nxT(n+\) , Г00 е-„"+-+1/2 f i i /i
о Г(п + х+1) Jo (n + x)"+x+1/2( 12и 12(n + x)
= {i+T^ + o(iHn ra + tr^L e.,,.!> л
_±r-
12 Jo
= Л + h, (8.2)
say.
As ? increases from 0 to oo, e'/(l + tI+< decreases monotonically from 1 to
0. To apply Watson's lemma (Copson [3, p. 49], Olver [1, p. 113]), set
v = (l + t) Log(l + t)-t
_ f (-tf
For v sufficiently small and nonnegative, let
t = f ckvkl2. (8.4)
Now substitute (8.4) into (8.3) and solve for clt ..., c4. After a lengthy
calculation, we find that
t = BvI'2 + l-v- ^v*2 +^-v2 + --., (8.5)
5 Zb Ijj
and so
dt 1 1 Bt;I/2 4
for v sufficiently small and v > 0.

204 13. Integrals and Asymptotic Expansions
Again, from (8.5), for v sufficiently small and nonnegative,
(y 6+ 12 1080
Thus, by Watson's lemma,
103
103
12 ' "\nj/l\j2n 6n ' 24n3'2 ~1080^
1 1 /2л 118
as n tends to со.
Next, from (8.5) and (8.6),
t 3B^2
1 7
for i; > 0 and v sufficiently small. Hence, by Watson's lemma,
-A{Br)+ } +
Hence, from (8.6) and (8.7), for v > 0 and v sufficiently small,
\-чгА±- 1 _I B^I/2 ЮЗ
( +t) dv~Bvy2 6+
as n tends to со.

13. Integrals and Asymptotic Expansions 205
Putting (8.8) and (8.9) in (8.2), we conclude that
nxT(n +1) fnn 1 1 J2n 13
V ' dx-= / + /
о Г(и + х + 1) V 2 6 т 24 V п 1080и т V/2
е"Г(п + 1) 1 13
6 1080»
as n tends to oo.
Next, from Entry 6, as n tends to oo,
Combining (8.10) and (8.11), we deduce that
» <- «*г(И +1) . .__ . ..,..J .,.. е-г(И + 1) , i_j_ + 0/i
о [Г(п + х+1)^ и-г-/»^-- „„ -2 24n-"V«3/V'
as n tends to oo. Since, as n tends to oo,
6n 1 1 /1 \
~\2~n^\~2~2An+ \y)'
we conclude the proof of Ramanujan's approximation. ?
For a generalization of Entry 8, see A0.22).
Entry 9. If
lo i -t- л
and if \m\ > \n\, where m and n are real, then
e~m2*2 g-
2 cosBmnx) rfx = —-— {<p(m + n) + <p(m — n)}. (9.1)
о 1 + x 2
Proof. First, note that (9.1) is trivial for n = 0. Assume next that 0 < n < m.
Then
1
1 f / 1 1 \
- cosBmnx) dx = -I I I e~m2*2 cosBmnx) dx
4jV^ + i xij
r cosBmnx) dx I I
о 1+x2 4j_00V^ + i x-i
Ц
x-i
¦4Ч/1 " W (9-2)
say.

206 13. Integrals and Asymptotic Expansions
Let p = n/m, so that 0 < p < 1. By integrating e~imz~inJ/(z + i) around
a rectangle with vertices + N and + N + ip, applying Cauchy's theorem, and
letting N tend to oo, we find that
e-m2x2 m+n Лоо e-<m+nJH2
U = I 77, 7 dx = du
Д + il+P m J-xm + n m + n.
u + 1
m m
л» e-(«+.)V Г» (и - ,>-<»+•>'«'
= — du = : г du
= -2i<p(m + n). (9.3)
Proceeding in the same fashion as above and setting x = (m — n)u/m, we
find that
i)e~<-m~nJu2
rf" 2'P(m n). (9.4)
l+u2
Subsituting (9.3) and (9.4) into (9.2), we easily deduce (9.1) for 0 < n< m.
Observe that both sides of (9.1) are even functions of n. Hence, (9.1) holds
for —m<n<m. Since the left side of (9.1) is an even function of m and
since (p(r) is an even function of r, we see that (9.1) is valid for \n\ < \m\. By
continuity, (9.1) holds for \m\ = \n\ as well. This completes the proof. ?
We now find the analogue of (9.1) when \n\ > \m\. Suppose that 0 < m < n.
As before, It is given by (9.3). But, letting R(i) denote the residue of
e'(mz~inJ/(z - 0 at the simple pole z = i, we find that
x + i(p — 1)
o-(n-mJu2
-CO « +
(U - i)
du + 2nie(m-H)[
- du -
, l+u2
= -2i<p(n - m) + 27iie(m-"J. (9.5)
Hence, substituting (9.3) and (9.5) into (9.2), we easily find that
2 cosBmnx) dx = je~{(p(m + ri) — q>(m - и)} + \nem2~2mn.
о 1 + x
(9.6)
By the same arguments are before, (9.6) is valid in general for \n\ > \m\ > 0.
Another proof of Entry 9 can be given by combining a result of Binet
(Burkhardt [1, p. 1154]) with some formulas in Nielsen's book [2, pp. 18,19,

13. Integrals and Asymptotic Expansions 207
Eqs. E), A3)]. Entry 9 can also be derived by an appropriate application of
Parseval's theorem.
Entry 10. Let a, /?, y, and S be fixed real numbers with y>6>0. Assume that
for some fixed d > 0, (p(x) is analytic and nonzero in the disk \x\ < d; (p(x) and
(p'{x) are positive for x > —d; and there exists a constant M > 0 such that
x<p'(x) > M(p{x) A0.1)
for all x > d. Let h > 0. Then as h tends to 0,
— o)(p @) 3 у — о I (p @) j у — 5
Two functions (p(x) that satisfy Entry 10 are ex and A + x)n, n > 0 (see
Corollary (i) below). Observe that if <p satisfies Entry 10, so do e and <pc,
for any с > 0. Also, if q>x and <p2 obey the hypotheses of Entry 10, then <pj q>2
does as well. Entry 10 is truly a remarkable theorem, and there does not appear
to be anything like it in the literature. The form of this asymptotic formula is
reminiscent of the asympto tic formulas that arise in the method of stationary
phase and in other asymptotic estimates of integrals.
Proof. Let L(x) = Log q>(x) and w = [JT3/5]. Write
s = sx + s2,
where
S = У П v№±Wl nd
1 &}Jiq>{hp + hjy)
We first examine S:. Choose h so small that
\ha + hjd\, \hfi + hjy\ < d,
for each;, 1 < j <, w. Since, for |x| < d,
L(x) = L@) + L'@)x + ±L"@)x2 + 0(x3),
we find that
si = 1 exP(t iW* + hjS) - L(hp + hjy)}j
= I expf ? {L'@)M« - P +j(d - y)) + $L"@)h2(E2 - y2)j2
k<w \j = l
¦A3O(j3)}

208 13. Integrals and Asymptotic Expansions
= X exp(L'@)A((« - № + US - У)(к2 + к))
k<w
+ %L"@)h2{\E2 - У2)к3 + 0{k2)) + h3O(kA))
= ? exp(-Ahk2 + Bhk + Ch2k3 + O(h2k2) + O(h3k4)),
k<w
where
A = i(y - 5)L'(p) > 0, В = ?1'@)Bа - 2fi + 8 - y),
A0.3)
С = gL"@)(<52 - y2).
Since ex = 1 + x + O(x2), whenever x = 0A), we deduce that
S1 = X e~Ahkl{\ + {Bhk + Ch2k3 + 0{h2k2) + 0{h3k4)) + 0{h4k6)}
fc< w
= To + BhT, + Ch2T3 + 0{h2T2 + h3T4 + hAT6), A0.4)
where
Tr = ? e~Ahk2kr, r > 0.
k=0
Furthermore, define
К
Then
¦-Г
Jo
and, for r > 2,
r —
Kr " 2Ah ''-
Recall now the Euler-Maclaurin summation formula (Olver [1, p. 285]).
Let a and b denote nonnegative integers with b > a. Suppose that /Bm)(t) is
absolutely integrable over [a, b], where m is a fixed positive integer. Then
t № = I"At) dt + i{/(a) + f(b)}
t
k=a J
- /B*~1)(«)} + Rm, (Ю.5)
where
Here B,- denotes the jth Bernoulli number and ВДх) denotes thejth Bernoulli
polynomial, 0 <j < oo. The Euler-Maclaurin summation formula was the

13. Integrals and Asymptotic Expansions 209
focus of much of Ramanujan's work. In particular, see Chapters 6-8 of Part I
[9] and Chapter 15 in this book.
Applying the Euler-Maclaurin formula A0.5) with f(t) = exp( —/Ш2)(г,
a = 0, b — w, and m = 1, we easily find that, as h tends to 0,
To = Vo + i + 0{y/h)
and, for r > 1,
Tr=Vr + O(h~rl2).
Thus, by A0.4),
Comparing the right sides of A0.2) and A0.7) with the help of A0.3), we find
that they agree. Thus, it remains to show that S2 = О(^/Й), as h tends to 0.
Let N + 1 denote the smallest integer;',;' > 1, for which x + jd < ft + jy.
Then
S2 = I expf ? {ЦЫ + hjS) - L(hp + hjy)})
k>w \j=l J
exp( Y {ЦЫ + hjS) - L(hp + hjy)}
j=N+l
E exp(- t+i L'idjMp - a +j(y - 5))\
where we have applied the mean value theorem, and so
/i(a + jd) < dj < h(P + jy). A0.8)
Since L'(x) is continuous for |x| < d and L'(x) > 0 for x > —d, there exists
a constant Q > 0 such that L'(x) > Q whenever \x\ < d. The terms with
h(a + jd) <d< h(fi + jy) make a total contribution that is less than 1 to each
summand on k. Hence,
S2 « X exp ( - Y Qh(P - a + j(y - S))
- ? L'iOjMP - a + j(y - S))
j=N+l
- <*+j(y-&))),
where g - [_{dh~l - /?)/}>] and/= [{dh'1 - a)/5] with the understanding that
/= oo if ,5 = 0.

210 13. Integrals and Asymptotic Expansions
Now by A0.1), xL'(x) > M > 0, for x > d. Thus,
слч Mh(P - a + j(y - 5))
Mh(P - a + j(y - 8))
- h(p+jy)
for some constant R > 0, where we have used A0.8). Thus,
(min(k,g) к
-Qh X @_ a+;(?-*))-Я X
j=N + l j=f +
= Pt + P2 + P3,
where
P2= X
and
+ 0 - N2 - N)} - R(k - /)).
It is not difficult to see that there exist positive constants Qlf Q2, and Q3
such that, as h tends to 0,
and
Thus,
k>w
P2 « fexp{-Q2hg2) « /е-Оз/* =
P3
as h tends to 0. This completes the proof. ?
Corollary (i). Let n > 0. Then as x tends to oo,
¦ + — + O(x-1/2).
k% [T(x + к + 1)J V In Ъп
We first offer a short proof for the case when и = 1 and x is a positive
integer. Using the corollary and D8.5) in Section 48 of Chapter 12 and

13. Integrals and Asymptotic Expansions 211
Stirling's formula, we find that
" x*T(x + 1) x! " xx+k x! " x*
x\ f x «-1 *'
vx)e Z- 1,1
x i k=o Hi
exx\ 1 /1
2x* 3 Vx
as x tends to oo. The result now follows in the case that n = 1 and x is a positive
integer.
Second, we remark that a more precise version of Corollary (i) in the case
n = 1 has essentially already been proved in this chapter. By combining (8.1)
and (8.10), we deduce that
» xT(x + 1) _ jnx 1 1 12л 4 /1
?i> Г(х + к + 1) = \JY + 3 + 24 V "x" + 135x + \x~^
as x tends to oo.
We next give two proofs of Corollary (i), in general. The first uses Entry 10;
the second is ab initio.
First Proof. In Entry 10, let <p{t) = A + tf, a = ft = 6 = 0, у = 1, and
x = l/h. Brief calculations of the expressions on the right side of A0.2)
complete the proof. ?
Second Proof. For и > 0, set
A0.9)
)
By Stirling's formula, as x tends to oo,
M
+-
Hence, with t = u/x,
12x ' 288x2
J
- A0.11)

212 13. Integrals and Asymptotic Expansions
To apply Watson's lemma (Olver [1, p. 113]), we set v = A + t) Log(l + t)
— t and proceed as in the proof of Entry 8. Using (8.5) and (8.6), we find
that for v > 0 and sufficiently small,
dt
Hence, from A0.11) and A0.12), as x tends to oo,
f °° dt
f(u) du = (x + 0A)) e-"x"(l + t)""/2— dv
о Jo dv
= (x + 0A)) I e~nxv\{2v)~il2 + (\-'l) + -..}dv
0{x'n (ШЗ)
For each pair of nonnegative integers k, r, let Akr(z) denote a function
with an asymptotic expansion
as z tends to со, where the coefficients at,i> 0, may depend on к and r, and
where, for each positive integer;', A0.14) becomes an asymptotic expansion
of Ak^r(z) after j-fold term by term differentiation with respect to z. Using
A0.10) and induction on r, it can be shown that, for each positive integer r,
/(r)(u) has the form
/w(m) = f(u) ? AKr(x + u)(x + M)-[("+D/2] Log'"" A + -Y A0.15)
k=0 \ Xj
as x tends to oo. In particular,
I IVJ — uyx j, ^iu. ioj
as x tends to oo, and
/(r)(")-0, A0.17)
as и tends to a).
Applying the Euler-Maclaurin formula A0.5) with f(u) defined by A0.9),
a = 0, and b = oo, we find that, in view of A0.17),
oo (*» 1 m-l P
k=o Jo 2 jt=i
where
Rm=\ Blm ^2"([ [f^/B"»(t)A. A0.19)
j о 1т)

13. Integrals and Asymptotic Expansions 213
By A0.13) and A0.18) with m = 1, it remains to show that R, = O(x/2), as
x tends to oo. We shall show more generally that, for each integer m > 1,
Rm = 0{xll2-m), A0.20)
as x tends to oo. Observe that A0.16) and A0.18)-( 10.20) imply the interesting
infinite asymptotic expansion
/(«) du ~ - - X ^/'2'^) (Ю21)
k=-0 Jo ^
as x tends to со.
By A0.15) and A0.19), as x tends to oc,
Rm « l/Bm>(«)l du « f(u) X (* + u)-[((c+1)/21 Log2 1 + - )du
Jo Jo t=o \ x/
2m
Г / u\
x((t+1>/21 /(и) Log2"* [l+ )du.
J x
o
u
Set t = m/x and apply A0.10) to deduce that
Rm « 'f x1-»^1! fЖ A + Г"/2 jM g' ,Г Log2-"^! + t) dt.
*=o Jo U1 + f) J
Setting v = A + t) Log(l + t) - t> we then obtain
2ra
-*A + t) dv.
*=o Jo "y
By A0.12), (8.5), and Watson's lemma,
2m
Rm« I x
k=o
2m f
V xl-[(*+l)/2] g-nx^ym-H + D/2 ^
« У x(k+1|/2~[(l+1>/21~m = 0(x1/2~m),
k = 0
as x tends to oo. This completes the proof of A0.20). ?
The second proof above is substantially due to F. W. J. Olver (personal
communication), who established A0.20) in the case m = 1. By an extension
of his ideas, we have proved A0.20) for all m in order to obtain the asymptotic
formula A0.21). As an application of A0.21), we demonstrate that
where n >0 and x tends to oo. Observe that A0.22) generalizes Entry 8,

214 13. Integrals and Asymptotic Expansions
since in the case л = 1, A0.22) implies (8.1). To verify A0.22), logarithmically
differentiate with respect to и in A0.10) to obtain
/(и) 2(и + х) °\ xJ { 12(х + и)
1 1"Ч 1 1
288(х + иJ J { П(х + иJ 144(х + иK
Thus,
/'@) = -?- + О(х~2)
as х tends to oo. Since B2/2! = Д, A0.22) therefore follows from A0.16) and
A0.21).
Corollary (ii). // n is a positive integer, as x tends to oo,
It is tempting to conclude that the sum in the exponent on the right side
is equal to — Log( 1 — l/(nx)). However, then we would have an exact formula
rather than an asymptotic formula, and it is clear that this exact formula could
not possibly be true for n > 1.
For n = 1, A0.23) is trivial. For n = 2, the left side of A0.23) is equal to
/0Bx), where /0 is the Bessel function of imaginary argument of order 0. In
this case, the first three terms
16x 512x2
agree with the asymptotic expansion for I0Bx) found in Watson's treatise [9,
p. 203, Eq. B)]. The case n = 5 was communicated by Ramanujan in his first
letter to Hardy [16, p. xxvi] and was proved by Watson [2].
Proof. Ramanujan's result follows easily from a general result proved by
Barnes [1, p. 115]. Accordingly, Barnes showed that (see also Watson's paper
n2 - 1 (n2 -
+ 24nx + 1152nV +'
as x tends to oo. Expanding the exponential on the right side of A0.23), we
find that Ramanujan's result is in agreement with that of Barnes (for the first
three terms). ?
For another approach to Corollary (ii), when n is any positive number, see
the text by Olver [1, pp. 307-309].

13. Integrals and Asymptotic Expansions 215
Entry ll(i). As x tends to oo,
ex^k
Proof. We shall apply a general asymptotic formula
^ _ ^M + x „(x) + x „,w + x» >
t=o к! zoo
•, A1.1)
as x tends to oo, that is found in Chapter 3, Entry 10 of the second notebook.
The function q>(x) = ехГ(х + l)/xx is easily seen to satisfy the hypotheses of
a rigorous formulation of this theorem (Part I [9, pp. 57, 58]). Thus, by A1.1)
and Stirling's formula,
139
L [ ь I ~ ^2n \ x + Uxw + 288x3'2 51840x5/2
x r--A 1
2^ ^
5
+"
32x7/2
15 35
15 \ x2 f^-( 105
24л/?7Г|
x3 r~( 945
as x tends to oo.
On the other hand,
1 1 /1 1 \ 1
24x
_
48? \36 + 5760Ух3 + + 2-242x2
1 1
_
24-48.? + 6-243x3 +
1 23 11237
210-34-5x3

216 13. Integrals and Asymptotic Expansions
Comparing the two asymptotic expansions found above with that in Entry
1 l(i), we complete the proof. D
Ramanujan's asymptotic formula A1.1) is very useful and powerful. In
addition to Part I, see the paper by R. J. Evans [1] for several applications.
Corollary 14 of his paper provides a solution to a previously unsolved problem
of Appledorn [1].
Entry 11 (ii). .4s n tends to oo,
_ Cx x"'1 dx /1 1
jo Z (ФУ V
11 = 0
Proof. By A1.2),
"(Ю л" dx
= e" I —
0 IJex/kf
e ' ' 23 11237
о
e" Гж ,,, f 1 25 11957
It seems convenient to express each of the gamma functions above in terms
of Г(п + ?) and then use the asymptotic series (Olver [1, p. 295]),
as n tends to со. Hence, as n tends to oo,
25
24(n-_) 27-32(n-f)(n-f)
11957

13. Integrals and Asymptotic Expansions 217
1 1 1003
"
If 1 1 1 If,1
{1 j|1 +
25 25 11957
Collecting together the coefficients of l/n\ 1 < к < 4, we complete the proof.
?
Entry ll(iii).
( —1)*
:= L°8
? Reg * LogB*j
Entry 11 (Hi) was, in fact, submitted as a problem by Ramanujan to the
Journal of the Indian Mathematical Society [12], [16, p. 333].
Proof. We shall show by induction on n that, for n > 0,
A 1 S Log2 2 » (-l)'Log2
й k(k + 1) »й fc LogB"&) LogB"+1/c) ^ A: LogB"/c) * " '
By definition of S, A1.3) is valid for n = 0. Now,
- Log22 + «(-!)* Log 2
LogB"fc) LogB"+1k)
- {(-1)" + П Log2 2 + (- l)k Log 2 LogB"&)
Log2 2» (-l)*Log2
+
LogB"+1k) LogB"+2A;) + ?2 к
1 « Log2 2 » (-l)*Log2
(n + l)(n + 2) *t'2 fc LogB"+1fc) LogB"+2fc) к^2 к
which completes the induction. Letting n tend to 00 in A1.3), we easily
conclude that
00 1
s; - V - 1 n
Ramanujan begins Section 12 by briefly describing Entry 10 of Chapter 3.
He concludes this section by giving an example that is an elaboration of
Example 2, Section 10 of Chapter 3. Pollak and Shepp [1] have proposed an
equivalent asymptotic expansion, but with less terms.

218 13. Integrals and Asymptotic Expansions
Example. As x tends to со,
™Log(k+l)xk ( 1 1 1 19
I' (L°gX + + + +
I*!
Proof. As in the proof of Entry ll(i), we apply Entry 10 of Chapter 3.
However, in addition to the seven terms displayed in A1.1), nine more terms
are needed. Thus, as x tends to oo,
_x ,?, Log(& + l)x*
T
X
4(x + IL
25x2
6(x + IN
8x2
' (x + IO
189x5
1
X
X
2x2
' (x + IM
15x3
1
(x + IO
245x3
4(x + I)8
1 1
2x2 3x3
X
5x3
2(x + lN
8(x + I)8
140x4
' (x + I)9
1 1
4x4 ' 5x5
5(x
6(x
3x
: +
X
+
X
+
2
¦0*
iM
iN
J_/ 2 Ъ^_А_ 5_ \ J_/,3
2x \ xx2 x3 x4 / Зх2 \ x
6 10 \ 3 / 4 10 20
x2 x3 / 4x2\ x x2 x3
— (\-- — \ "Li _5 15
4y \ Y Y^ / Y^ \ Y Y^
5 /. 6 21 \ 1 / 5
— (\ \ l 8 245 140
189
2x5
Collecting the coefficients of x~\ 1 < к < 5, we arrive at Ramanujan's asymp-
asymptotic expansion. ?

13. Integrals and Asymptotic Expansions 219
Entry 13. Let a, p, y, and S be any complex numbers. Then
dx
о {x2 + a2)(x2 + p2)(x2 + y2)(x2 + 82)
n (« + p + у + SK - (a3 + p3 + y3 + S3)
y)(y + a)(a + d)(p + S)(y + 6)'
A3.1)
Corollary. // a, P, y, and 6 are the roots of the polynomial x* — рхъ + qx2 —
rx + s, then
dx л 1
о (x2 + a2){x2 + P2)(x2 + y2)(x2 + S2) 2s _ ps
-. A3.2)
q-r/p
Ramanujan's formula A3.2) is the same as an evaluation in Gradshteyn and
Ryzhik's tables [1, p. 218, formula E)]. Since p = a + P + y + S, q = ap +
ay + aS + Py + PS + yd, r == aPy + aP5 + ay5 + Руд, and s = a/fy<5, formula
A3.2) can be rewritten in the form A3.1), after a tedious calculation.
Entry 14. // x is arbitrary and а ф 0, — 1, — 2,..., then
Г2(а)
A proof of Entry 14 was published by Ramanujan [8], [16, p. 53].
R. Askey has pointed out the following observation. Letting b = ix and
с = — ix and using a value for 5^ found in Wilson's paper [1, Eq. B.4)], we
find that the sum in Entry 14 equals
а Г 2a, a + 1, a + ix, a — ix
a2 + x2 4 3 \_a, a + I + ix, a + 1 - ix '
2a, a + I, a + b,a + c,a + d
а Г 2a, a + 1, a + b, a + с, а + d
(a + b){a + c) dlmx 5 4 [a, a + 1 - b, a + 1 - c, a + 1 - d '
(a + b)(a + c)
lim Па + 1 - b)T(a + 1 - с)Г(о + 1 - d)T{l - a - b - с - d)
lim
X dimx TBa + 1)ГA -b- с)ГA -b- d)T(l - с - d)
Entry 14 now follows after computing the limit above.
Example. For n,a>0,
' cos(nx) dx n
„ a2 + x2 ~ 2a

220 13. Integrals and Asymptotic Expansions
This formula is well known and was established by Ramanujan in his
quarterly reports (Part I [9, p. 322]) via the Fourier cosine inversion formula.
Entry 15. For a > 0 and n real,
Г 00
|Г(а + ix)\2 cosBnx) dx = %JnT{a)T{a + i) seen2" n.
Jo
Entry 15 was proved by Ramanujan in [8], [16, pp. 53, 54].
We note the following generalization of Entry 15. If a > 0 and |Re y\ < \,
then
Лес
|Г(а + ix)\2e2yx dx = ^/лГ(а)Г(а + i) sec2" y.
J-00
For this and substantial ramifications, see Wilson's paper [1].
Entry 16. For a and n both real, and n integral in (it;),
Г00 sinh(ax) 1 sin а
(l) ———- cos(nx) dx = - — , а < n,
Jo sinh(^x) l ; 2 cosh n + cos a
C™ cosh(ax) . 1 sinh n
(n) ^-r?—: sin(nx) dx = - —-—¦ , a < n,
Jo sinhGcx) 2 cosh n + cos a
1 1 1
- 1 4п
n>0,
—гт dx = (— 1)"Е2„, п>0,
с/2)
where Bk and Ек, 0 < к < оо, denote the kth Bernoulli and Euler numbers,
respectively.
In each case below, [1] refers to the tables of Gradshteyn and Ryzhik.
Both (i) and (ii) can be found in [1, p. 504]. Ramanujan has stated (iii) in
[8], [16, p. 56] but does not give a proof. Formula (iii), however, is easily
derived from [ 1, p. 481, formula 3.911, No. 2]. Both integrals in (iv) are classical
[1, pp. 1076, 349].
Entry 17. Let q>(z) be analytic for a < Re(z) < n, where a is a nonnegative
integer. Suppose that
lim \(p{x ± iy)\e~2ny = 0,
uniformly for a < x <n. Then

13. Integrals and Asymptotic Expansions 221
t<P(k)= [\(u)du
k=a Ja
_ . Г °° Ф
Jo
+ iu) - cp(n - iu) - cp(a + iu) + q>(a - iu) ^
е2ки 1
Entry 17 is the famous Abel-Plana summation formula (Henrici [1, p. 274],
Whittaker and Watson [1, p. 145]). For the history of this formula and some
of its applications, see Lindelof s book [1, Chapter 3]. Ramanujan's formula-
formulation of Entry 17 is not as precise as that given above, because all those
expressions that are independent of n are not explicitly given.
Corollary. For each positive integer n,
tan'^x/n)
Log n! == n Log n - n + \ LogB7tn) + 2 | —^ — dx.
This corollary is easily established by setting q>(x) = Log x in Entry 17.
Details may be found in Lindelof s text [1, pp. 69,70]. Whittaker and Watson
[1, pp. 250, 251] give another proof and attribute the result to Binet in 1839.
Entry 18(i). Let t > 0, and fix a positive integer n. Set x = tn, and put q>(z) =
f(t + tz) — f(tz) for a given function f. Suppose that (p(z) satisfies the hypo-
hypotheses of Entry 17 with a = 0. Then
f{x) + \ф) = i{/@) + f{t)} + ф) du
Г
Jo
. cpin + iu) — <»(n — iu) — cpiiu) + cpi — iu)
- '¦ !o г^\ —du-
Proof. Apply Entry 17 to <p(z) with а = 0. Now observe that the left side of
A7.1) is equal to
л
k=0
After some rearrangement, we deduce the desired result. ?
Our formulation of Entry 18(i) is rather different from that of Ramanujan
since he does not record those parts of the formula that do not depend on x.
Furthermore, there are two misprints in his statement (p. 159). In order to
prove Entry 18(ii), which is likewise not properly stated by Ramanujan, we
need to establish a lemma that is similar in character to Entry 17.
Lemma. Let n = 2m be an even positive integer. Suppose that <p(z) is analytic
on 0 < Re(z) < и and that

222 13. Integrals and Asymptotic Expansions
lim \<p(x± iy)\e~"y'2 =0,
y-*CC
uniformly for 0 < x < n. Then, provided that the integrals below exist,
<p(n - iu)
Proof. Let CN denote the positively oriented rectangle with vertices ± iN and
n ± iN. By the residue theorem,
2ni ]Ck cosGtz/2) h
If we let N tend to oo and invoke our hypotheses, we find that
2i Jn_i00 cosGtz/2) 2i J-ioo
Letting z = n + iu and recalling that n = 2m, we find that
_
(n + щ) + <p(n - iu)
The remaining integral in A8.1) can be transformed in a similar fashion. The
desired result now follows. ?
Entry 18(ii). Let t > 0, and fix an even positive integer n = 2m. Set x = tn and
define cp(z) = f(tz + t) + f(tz — t) for a given function f. Suppose that q>(z)
satisfies the hypotheses of the previous lemma. Then
;:
provided that the integrals above exist.
Proof. Apply the previous lemma and observe that
m
2 ? (-l)VB/c - 1) = 2(-l)m/(x) - 2/@).
The desired equality now follows. ?

13. Integrals and Asymptotic Expansions
223
-f
Jo
ф(п) = cp(x) cos(nx) dx,
Entry 19. Let n>0.If
then
(i)
if
then
(ii)
Entry 19 follows easily from the Fourier integral theorem (Titchmarsh [1,
pp. 432-435], [2, pp. 16, 17]) and is valid when <p is continuous and of
bounded variation on [0, hi]. Entry 19 is also given in Ramanujan's quarterly
reports (Part I [9, p. 333]).
Corollary. If a > 0 and n is real, then
ф(х)
ф{х)
cos(mx)
ф(п) =
sin(mx)
dx= <
Г V(x)
[o
dx= -
n
2
n
4
.°>
sin(nx) dx,
''n
2
n
4
0,
m
}ti
m
m
m
m
<
>
<
=
>
К
h,
h;
К
К
h.
This result was proved by Ramanujan [8], [16, p. 54] by means of the
Fourier inversion formula and Entry 15.
We note the following generalization of the previous corollary. If a >
I Re у |, then
Г
4
ГBо)
To see this, observe that
sech2" x e2yx dx =
+
2a
e2yx dx
dt

224 13. Integrals and Asymptotic Expansions
= 22a t2y+2a-l{l +t2r2adt
Jo
'¦^(l + u)~2adu
у)Г(а - у)
= 22
ГBа)
where we have employed a familiar integral representation for the beta-
function.
Entry 20. Ifn>0 and 0 < a < n, then
30 sinh(ax) dx _ " (-1)*+1 sin(fea)
0 sinhGcx) 1 + n2x2 k=i 1 + nk
This result is classical (Gradshteyn and Ryzhik [1, p. 352]) and is easily
established by contour integration.
Entry 21. Let p, q, and n be real. Suppose that (Pj(p, x) and F(nx) are continuous
for a.j<x< fij, where j = 1, 2. Define ф^{р, п) and \j/2{p, ri) by
<Pi(p, x)F(nx) dx = фх{р, ri) and \ (p2(p, x)F(nx) dx = ф2(р, ri).
Then
С Pi
<PiiP, х)ф2(й, nx) dx = I q>2(q, х)ф^р, пх) dx.
Entry 21 is easily established by inverting the order of integration.
The following corollary, which is Parseval's theorem for cosine transforms,
is formally a special case of Entry 21. However, since the intervals of integra-
integration are not finite, different hypotheses, which we have taken from Titchmarsh's
book [2, p. 54], must be assumed. Both Entry 21 and the corollary below were
proved formally by Ramanujan in [8], [16, pp. 55, 56].
Corollary. Let p, q, I, and n be real. Suppose that (p(p,x)eL@, oo) in the
variable x and that limx_0+ q>(p, x) exists. Define
ф(р,п)= <р(р, x) cos(nx) dx,
Jo
which we assume is integrable over any finite interval in 0 < n < oo. Also
suppose that ф(р, ri) tends to 0 as n tends to oo. Then
-- <p(p, x)cp(q, Ix) dx = ф{ц, х)ф(р, Ix) dx.
<- Jo Jo

13. Integrals and Asymptotic Expansions 225
The corollary above and the example below were communicated in
Ramanujan's first letter to Hardy [16, p. 350]. Earlier, Ramanujan [6] had
submitted this example as a problem to the Journal of the Indian Mathematical
Society. Ramanujan also established this example in [8], [16, p. 55].
Example. // a/? == я/4, then
e~*2 dx r- f°° e*2 dx
0 cosh(ax) Jo cosh(/?x)
Ramanujan's next statement is enigmatic. He says that the example above
can be derived from the formula
"which is obtained from the theorem"
f (-l)t+V(/c)= ? {-Ifcpi-k). B1.2)
k=l 11=0
Equality B1.1) is really just a very special case of the Poisson summation
formula (see Corollary (i) in Section 31) when the functions appearing in the
formula are self-reciprocal Fourier transforms of a special type. Formula B1.2)
was stated by Ramanujan in Chapter 4, Section 9, Example 2 and, as to be
expected, is valid only under severe restrictions (Part I, p. 97).
Entry 22. Ifa,b> 0, then
о, Г * и* + щг«, + m
Jo
if 0 < a < b + i, then
(ii)
Г(а + be)
T(b + 1 + ix)
dx =
- a + j)
2T(b+ 1)Г(Ь + |)Г(Ь - a + 1)"
These two beautiful formulas were derived by Ramanujan in [8], [16,
pp. 57, 54] and are perhaps his most famous integral evaluations. It should
be mentioned, however, that Barnes [2, pp. 154,155] established an extension
of (i) at roughly the same time that Ramanujan discovered Entry 22. R. Roy
[1] has employed Mellin transforms to give a proof of (ii). For ramifications
of these results, see papers of Wilson [1] and Askey and Wilson [1].
Entry 22(i) can be generalized in the following way. If we apply Parseval's
theorem (Titchmarsh [2, p. 5]), the corollary in Section 19, and Legendre's
duplication formula, we find that

226 13. Integrals and Asymptotic Expansions
2 |Г(а + ix)T(b + ix)\2 cos(xy) dx
J
o
sech2" ^ sech2» Ц^ du,
where у > 0. Glasser [1] has shown how to evaluate integrals like that on the
right side above.
Entry 23. Let a > 0, m < 1, and m + n > 0. Then
ncsc(nm) ™ (-n)t
0 Г(х + а + л + 1) Г(п + 1) к% к\(а + к)т '
We have not been able to find this result in the literature. Ramanujan has
also obtained this integral formula in his quarterly reports, and a complete
proof may be found in Part I, pp. 303, 304.
Entry 24(i) offers the triviality
И 00 00
Zj Ak= 2_, An-k — 2_, A^-k->
k=0 fe=0 fc=l
which is followed by a corollary in which Ak above is replaced by Ak/T(k + 1).
The intent of Entry 24(ii),
N N
lim Yj <p(x + ^) = lim ]T <p(y + k),
N->oo k = -N JV->oo k = -N
is indeed unclear. What can be said?
Ramanujan, in a corollary, claims that
h\
1) F(-fcn + \)
where и < 1 and x and й are arbitrary. Although these equalities are true for
h = 0 and л = 1, they certainly are false in general, because the far left side is
a nonconstant function of л and the expressions to the right are not. Moreover,
the series diverge if л is not an integer.
In Entry 24(iii), Ramanujan offers the equality
Instances when an integral is equal to the corresponding sum are rare. For
examples of this phenomenon, see papers by Boas and Pollard [1], Krishnan
[1], and Forrester [1]. See also Entries 5(i), (ii) and Entries 16(i), (ii) in
Chapter 14.

13. Integrals and Asymptotic Expansions 227
In Corollary (i), Ramanujan claims that
dx = e",
which follows formally from B4.1) by setting q>(x) = ax. However, if a = 0,
Ramanujan's claim is clearly false, and if a is real and nonzero, the integral
diverges.
In Corollary (ii), Ramanujan asserts that
- = A + a)", B4.2)
_„ Г(х + 1)Г(и - x + 1) v
which follows formally from B4.1) by letting q>(x) = axT(n + 1)/Г(п — x + 1).
Again, B4.2) is false for a = 0, while the integral diverges for real а ф 0, +1.
If a = e", | a | < n, with a real, then B4.2) is valid and, in fact, was proved by
Ramanujan in his paper [14], [16, pp. 216-229, Eq. A.2)].
Entry 25(i) is the special case b = 0 of Entry 25(ii).
In Entry 25(ii), Ramanujan writes
ab+x аь~х
-.= eacos"cos(asinn - nb)
and
Гсс / ab+x аЬ-х \
Jo \Г(Ь + х+ 1) {b-x + 1)/
= eacosnsin(a sin n - nb),
where presumably a is real.
It is easy to see, by Stirling's formula, that both of these integrals diverge
if а ф 0. But let us discern how Ramanujan reasoned. By simple changes of
variable and B4.1),
einx dx
du
nkoin{k-b)
= exp(ae'" — inb).
Equating real and imaginary parts, we complete Ramanujan's formal
derivation.
The content of Entries 23-25 perhaps served as the seed for Ramanujan's
beautiful paper [14] on integrals involving the gamma function.

228 13. Integrals and Asymptotic Expansions
Example (i). The maximum value of ax/T(x + \)is equal to
¦+" О I —г
( |)[ ^ 1152a
when a is large.
Proof. Differentiating ах/Г(х + 1) with respect to x, we find that ах/Г(х + 1)
achieves its maximum when
ф(х + 1) - Log a = 0, B5.1)
where, as usual, ф(х) = Г'(х)/Г(х). Now in Entry 15 of Chapter 8 (p. 95), (Part
I, p. 194), Ramanujan derived an asymptotic expansion for the root x of B5.1)
in descending powers of a as a tends to oo; namely,
Letting
we find that, for a large,
x a-l/2+e+O(a->) a-l/2+s Г / 1 )
l+0-г ¦ B5.2)
Г(х + 1) Г(« + i + e + O(aA)) T{a + \ + e)
From Lemma 2, Section 24 of Chapter 11,
Ца + i) V 2a' 24a2
/+|!+ , + /i
as a tends to oo. Using the expansion above in B5.2), we deduce the desired
approximation. П
Our version of Example (i) is different from that of Ramanujan, who writes
that the maximum value of ах/Г(х + 1) is
aa-l/2
a
^—гттехР
"very nearly." This agrees with our statement, except for the appearance of
the expression 323.2a, which is apparently incorrect.
In Example (ii), Ramanujan states a version of the Euler-Maclaurin sum-

13. Integrals and Asymptotic Expansions 229
mation formula A0.5) and remarks that it "is very useful in evaluating definite
integrals."
Entry 26(i). Let n > 0 and suppose that m is a nonnegative integer. Then
f °° cosBnx) dx nnme~2n ™ (m + k)\
Jo (ГГхгГ" = 2ml k%j4nf{m - k)\k\'
This result is classical (Gradshteyn and Ryzhik [1, p. 413]) and can be
established by contour integration.
Entry 26(ii). Let p > 0 and suppose that m and n are nonnegative integers with
m < n. Then
» x2m cos(px) (-IT™'' - „_r
n^2^+T dx = 7,+iw, I ArP > B6.1)
A + X ) Z П\ r=0
where, for r > 0,
Г)! -lnO;,MLt(_r)|t(_w|)|t(_H)|fc
, (П
r 2'г!(и-г)! ib (-n-rJk/c! "
Proof. First, for n = 0, the proposed formula is readily established, for exam-
example, by the calculus of residues. Thus, in the sequel, we assume that n > 0.
We shall induct on m. For m = 0, formula B6.1) is seen to be valid by Entry
26(i). Now it is easy to see that
I{m + 1, n) = I(m, и - 1) - I{m, n), B6.2)
where m > 0, n > 1. Inducting on m, we shall employ B6.2) to show that B6.1)
is true with m replaced by m + 1. To that end,
и + 1 -гJкк\
r% 2rr\(n-r)\ k% {-n-rJkk\
_ (~l)mne-p " p"~r Шп - 2 + r)\
~ ~21^rnT~ re0 (n - r)! | 2r~1(r~ 1)!
i (^n-r + 2J,/c! TrV
min(r. m) Д^*' — г\.( ууЛ, ( — t

230 13. Integrals and Asymptotic Expansions
(-1Г + 17Г<Г" " „_г (И+Г)!
-)п+1и| L Р
2n+lnl r% (n-r)!2rr!
*p™) 4k(-r)k{-m)k{-n)k 4nr
(-n-rJkk\ {n + r)(n + r
>4k-\-r + l)t_!( -m)^ -п + 1)к_
(-])т+1пе-р • п_г (и + г)!
(n-r)!2rr!
_ mi"<^"+i> 4k(-r)k(-m)k(-n)k
kk (-» - rJk(k - l)\
Since
k\ (Jt-1)! fc!
the desired formula, B6.1) with m replaced by m + 1, follows. ?
Entry 27. // n is an even positive integer, then
Чр. coshB7ix sin(B/c — \)п/п)) — cosB7ix cl^vv_.. -,~i--,,
t = l 27C X
Proof. We have
^ coshB7ix sin(B/ — 1)я/и)) — cosB7ix cos(B/ — 1Oс/и))
sinh(mxe''IiB-/-1)/")sinh(-mxe'tIBj~1)/")
2 -2niBj-l)ln\ / 2 2.iBrl
)(¦—?-)¦ B7-2)
Comparing B7.1) with B7.2) and replacing x/k by x, we find that it remains
to show that
It is easily checked that the 2и roots on the left side are precisely the same as
the 2n roots on the right side, and so the proof is complete. ?

13. Integrals and Asymptotic Expansions 231
Corollary (i). // n is arbitrary, then
l + (-Yl = r> + !) sinh(WH>/3)
V
Г(ЗП
Corollary (ii). If n is arbitrary, then
^ + к ) ]
n + к ) ] ГCп + 2)тг
The latter two formulas were proven by Ramanujan in [9], [16, p. 51].
Entry 28. // m and n are positive integers and x is arbitrary, then
oo y"'1 mn — 1
mnY = У excosBltt/")cos(xsinB7cfc/n)).
k% {nk)\ t=0
Proof. Letting к =jn + r, 0 < j < m — 1,0 < r < n — 1, below, we find that
m m
k=0 k=0
= "l "Z
r=0 j=0
n-i с
r=0 j=0 J-
CO v- j' П — 1
1=0 j\ r=0
The last inner sum is equal to 0 unless n\j in which case it is equal to n. The
proof is now complete. ?
Entries 29(i), (ii). Suppose that p > 0, / is a nonnegative integer, and n is a
positive integer with n> I. Then
) . . .
^ cos(px) dx
e + T e coJ p sin
In n & \ n V n
1 f »^
- p sin^Y ifnis odd,
_ (
In v In
if n is even.
The integrals above may be evaluated by the calculus of residues, although
the initial form of the answer obtained might be different from that stated by

232 13. Integrals and Asymptotic Expansions
Ramanujan. See also Gradshteyn and Ryzhik's tables [1, p. 414, formula 3.738,
No. 2], where again the evaluation is given in a different formulation and a
bracket {is misplaced. Since a similar calculation is performed in the proofs
of Entries 33 (i), (ii), we suppress the details.
Entry 30(i). If n is a nonnegative integer, then
f
Jo
2n\
A proof of Entry 30(i) may be found in Fichtenholz's text [1, p. 656]. These
integrals actually are special cases of Entries 16(i), (ii) in Chapter 14. For
further references to Entries 30(i), (ii), see a problem of Wang [1].
Entry 30(ii). // p > 2 and n - p + 1 > 0, then
(p - \)(p - 2)<p(n, p) = n(n - \)(p(n - 2, p - 2) - n2q>(n, p - 2),
where
f00 sin" x
<p(n,p)= ~-—dx. C0.1)
Jo x
Proof. Integrating by parts twice, we find that
(n — 1) sin" x cos2 x — sin" x
f00
J.
which is easily seen to be equivalent to the proposed recursion formula. ?
Corollary (i). // n is a nonnegative integer, then
where q> is defined by C0.1).
Proof. By Entries 30(ii) and (i), respectively,
q>Bn + 3, 3) = ?Bи + 3)Bи + 2)фBп + 1, 1) - \{2n + 3JфBп + 3,
which, upon simplification, yields the desired result. ?
Corollary (ii). If n is a nonnegative integer and q> is defined by C0.1), then

13. Integrals and Asymptotic Expansions 233
Proof. The proof is like that of Corollary (i); simply apply Entries 30(ii) and (i)
and then simplify. ?
Example (i). IfO < p < n + 1 and q> is given by C0.1), then
l Г00 Г00
q)(n, p) = sin" x e~txtp 1 dt dx.
T(P)JO Jo
Proof. From the definition of the gamma function,
1 1
1 dt, x,p>0.
*p Г(р) Jo
The desired result now follows from C0.1). ?
Examples (ii), (iii). If a > 0 and n is a nonnegative integer, then
lo (a + 1 )(a +3 )¦•
and
Bn)!
sin2" x dx =
a{a2 + 22)(a2 + 42)---(a2 + Bn)
These formulas are classical (Gradshteyn and Ryzhik [1, p. 478]) and fol-
follow readily by induction.
In the sequel, a prime (') on a summation sign, Y^a<k<bf(k\ indicates that if
a and/or b is an integer, then only ^f(a) and/or ^f(b), respectively, is counted.
Entry 31 (i). Let h, a, /? > 0 with a/? = 2л:. Let q> be a continuous function of
bounded variation on [0, h]. Define
ф(г) = I (p(x) COS(rx) dx.
Then, if n is real.
a X' P(a/c) cos(anfc) = ф(п) + ? {ф(Рк + п) + ф(Рк - и)}. C1.1)
0<k<h/x к = \
Proof. We shall employ the Poisson summation formula in the form
I' f(k) = \ fix) dx + 2^ Г f{x) cosi2nkx) dx, C1.2)
a<k<b Ja k=l Ja
where / is a continuous function of bounded variation on [a, b~]. In C1.2),
let a = 0, b = h/a, and fix) = q>iax) cos(anx). Thus, putting и = ax, we find
that

234 13. Integrals and Asymptotic Expansions
Y' (p(<xk) cos(anfc)
1 2 °° f*
/ф)
(p(u) cos(nu) cos(Pku) du
=1 Jo
1 l » f»
= -ф(п) + - Y \ <p(u){cos(/?/c + n)u + cos(/?/c - n)u) du.
Upon using the definition of ф, we complete the proof of C1.1). ?
Entry 31 (H). Let q> and ф be defined as in Entry 31(i). Let h > 0, and assume
that n is an integer. Then
— <p(x) dx = n Y! (-^)к(Р{кп) cos(kmz)
О Sln x 0<k-?h/n
-2Уф(п + 2к+1). C1.3)
k=0
Proof. We shall induct on n. First, in C1.1), put a = n, j? = 2, and n = 1 to
obtain
But this equality is precisely C1.3) in the case n = 0.
Second, let a = n, j5 = 2, and n = 0 in C1.1) to find that
00
k=l
This equality is easily seen to be equivalent to C1.3) in the case n = 1.
Now assume that C1.3) holds up to a fixed integer и. Then, by induction,
"h sin(n + 2)x , ч , f1 sin(nx) , ч , ^ Г" / , ч ,
: <p(x) dx = —: (p(x) dx ± 2 cos(n + l)x (p(x) dx
0 Sln -^ Jo sm ^ Jo
: Y' (—l)k<p(kn) cos(knn)
0<k<h/n
-2 У \^(n + 2k + 1) + 2^(n + 1)
= n Y' (~1) (p(kn) cos(k(n + 2)n)
0<k<h/n
- 2 f Ф(п ± 2 + 2k + 1),
k=0
which is C1.3) with n replaced by n + 2. П

13. Integrals and Asymptotic Expansions 235
Corollary (i). Let a, fi > 0 with a/? = 2n. Let q> be a continuous function of
bounded variation on @, oo). Suppose that q> is integrable over @, oo). Put
ф(г) = (р(х) cos(rx) dx.
Jo
Then
CO GO
a ?' (p(afc) = i/KO) + 2 ^
Proof. InC1.1),letn = 0 and let h tend to oo.(Tojustify this, see Titchmarsh's
book [2, pp. 61,62].) ?
Corollary (ii). Under the assumptions of Entry 31 (ii),
lim [ sn^) ф} dx_n ^ (-lfq>(kn) cos{knn) j = 0.
Proof. The desired result is an immediate consequence of C1.3), since clearly
CO
I Ф(к)
t=i
converges. П
Entry 32(i). Let h, a, fi > 0 with a/? = 2л:. Let <p be a continuous function of
bounded variation on [0, h]. Define
P
ф{г) — q>(x) sin(rx) dx.
Jo
Then, if n is real,
m(an/c) = ф(п) + ? {^(Д/с + и) - ф@к - и)}.
Proof. In the Poisson summation formula C1.2), put a = 0, b = h/oc, and
/(x) = ф(ах) sin(anx). Then
?' ф(а/с) sin(an/c)
О < * < fc/a
l l °° ГА
= -i/i'(n) + -^ ^(u){sin(n + pk)u + sin(n - Д/с)м} du,
% a k=i Jo
from which the proposed formula follows. ?
Entry 32(ii) is another version of the Euler-Maclaurin summation formula
A0.5).

236 13. Integrals and Asymptotic Expansions
Corollary. Let a, jS > 0 with ajS = я/2. Let cp(x) be continuous on @, сю),
integrable over @, d), of bounded variation on F, oo), and tend to 0 as x tends
to oo, where 0 < ё < я/2. Define
Ф(г) — ф(х) sin(rx) dx.
Jo
Then
This corollary gives the Poisson summation formula for Fourier sine
transforms (Titchmarsh [2, p. 66]).
Ramanujan concludes Section 32 by remarking that integrals such as
cos(nx) Г* sin(nx) Г* cos(«x)
(p(x)rfx p(x)rfx, and —; m(x)dx
J sinx
о cos х Jo cosx Jo sinx
may be determined. Ramanujan is evidently indicating that analogues of
Entry 31 (ii) exist.
Entries 33(i), (ii). Let n and I denote nonnegative integers with n > I. Let p > 0.
Then
/-_V2\/ C_iV
V A / V */
о Ll - x2" n(x2 - 1)
ж n/2~1 fnBl + l)k nk\
— Y, e~pcm{nkln) cos I p sin — j, if n is even,
П fc=l \ П П j
e + У
2n n fc=i
. Bk - lOl
p gln
In r In
if n is odd.
Proof. First observe that
"oo „V21 __ „2n-2 _ . . . _ 2 _ 1
— Yn cos(px) dx.
Let K(z0) denote the residue of
«"-**¦-'—^-lc,.
иA - z2")
at a pole z0. In the upper half-plane, /(z) has simple poles at z = exp(rcik/n),
1 < к < n — 1. Hence, by a familiar argument from the calculus of residues,

13. Integrals and Asymptotic Expansions 237
п-1
(-\fl = ni ? R(enikln)
7Г1 "~1
= __ у e«iBi+w« ехр(фе"*/в)
2n &=1
= — У е "»1П<«*/") sin I — — + p cos—
2nk=i \ n n
Observe that the terms with indices к and n — к are equal.
First, suppose that n is even. Singling out the term with к = n/2, we then
find that
(-1)'/ = ±(-Yfe-' + ~ T «-'-"» sin (ПЩ + m + p cos^
In n M. \ n n
= JL(_ 1Уе~, + (_ 1У « "I" e-Pcos
2П П k=l
where we have replaced /c by n/2 - /c in the former sum.
Second, suppose that n is odd. Then
and
n sin(fa) - (-
(П) ?
Corollary. // x is arbitrary, then
rccosh(fa) 1
Т-Г7 Г = ~2 + 2 2^ Гг
x sinh(nx) x2 t=i /c2 + x
and
n sinh(fa)
rccosh(fa) 1 (l^cos^fl)
W Т-Г7 Г = ~2 + 2 2^ Гг 2 ' ^ ^ я,
x sinh(nx) x2 /c2 + x2
00
(-1)'/ = 1T r>**™ sinf^±^ + p cos
It ("-1)/2
= (_!)'_. у e-pcos
/tiB/ + 1)BA - 1) . Bk - \)n
x cos p sin
\ 2n y 2n
where we have replaced к by (n + \)/2 — к in the former sum. ?
Entry 34. J/ x is arbitrary, then
n cos(fo) 1 « (-l)*+1cos(/cfl)

238 13. Integrals and Asymptotic Expansions
Proofs of Entry 34 and Corollary. First, Corollary (i) is proved in
Bromwich's text [1, p. 368, Eq. D.1)].
Entry 34(i) is easily obtained from Corollary (i) by replacing x by ix.
We next prove Entry 34(ii). Recall that the set of functions sin{Bfc + 1H},
0 < к < oo, is a complete orthogonal set on — n/2 < в < n/2. Calculating the
Fourier series of sin(fe), when x is real, with respect to this orthogonal set,
we readily deduce Entry 34(ii) for real x. By analytic continuation, Entry 34(ii)
holds for complex x as well.
Lastly, Corollary (ii) follows from Entry 34(ii) by replacing x by ix. ?
Entry 35. Let n denote a nonnegative integer, and let a, /? > 0 with a/? = n. Then
С 00
VaV + 2 L
(. *=i
where
и! "
Bn)!ktb (n-fc)!fc!
Proof. In the Poisson summation formula C1.2), set a — 0, b = oo, and
f{x) = 2A + a2*2)-". Thus,
I;(i+,W'=2 \o ot^+41Г^Ш^dx-C5л)
By Entry 26(i),
cosB?cfcx) _ 4n{nkloi)ne-2nkl* " (n + j)\
' " a2x2)"+1 "" 2оЫ
/Je** " (n ¦
(n - j)\j\
2e^n! - (n
[
a Г(и + 1) Bи)! А (и - j)lj\
F^^ C5.2)
¦y a F(n + 1)
Substituting C5.2) into C5.1), we complete the proof. ?
Entry 36. Let N be any positive integer. As m2 + n2 tends to oc,
m
L' "^—; m = tan \mln) +
IC = Q rrt т ^'* "T" Л.^ ft = l ^fc (,W1
+ 0((m2 + n2)-"),
w/iere Bj, 0 < j < oo, denotes the jth Bernoulli number.

13. Integrals and Asymptotic Expansions 239
Proof. Letting a = 0, b = со, and f(x) = {m2 + (n + xJ} in the Euler-
Maclaurin summation formula A0.5), we find that
First,
= Г f(x) dx + 1/@) - t ^/'"""(O) + *W C6-D
*=o Jo *=
Г°Л*)*с= r^^T = i^-tan-1^) = itan^. C6.2)
Jo Jn m + м m\2
Next, a straightforward calculation shows that
ilk - IV f i2* ( —iJ*
/•Bl-l)( \ _ V^11 '^ j ' I ^ *J
7 W 2m \(m + (и + x)iJk (m-(n + x)iJk j'
Thus, for /c > 1,
- шJк - (m + niJk
2mi { (m2 + n2J"
\kt~ih Ш
2f
n2
_ -2
m(m2 + и2)*
Hence, using C6.2) and C6.3) in C6.1), we deduce that
ш sinB/c tan ^m/n)). C6.3)
m y ' ' ' 2(m2 + n2)
1 Д B,tsinBfctan(m
w ([=i 2/c (m2 + n2)*
The remainder Лдг+1 is easily estimated, and the desired result readily follows.
?
Corollary. As n tends to со,
„ у + у
^o n2 + (n + kf 4 + h Bk + lJ2k+2n*k+2 ¦
Proof. Let m = n in Entry 36. ?

CHAPTER 14
Infinite Series
Since Ramanujan's death in 1920, there have perhaps been more published
papers establishing results in Chapter 14 than in any of the remaining 20
chapters. In many cases, the authors were unaware that their discoveries are
found in Ramanujan's notebooks. In [6] and [7], the author showed that
several results in Chapter 14, as well as many others as well, arise from a
general transformation formula for a large class of analytic Eisenstein series.
It should be emphasized, however, that Chapter 14 also contains many other
types of results.
Chapter 14 is primarily concerned with identities involving infinite series.
In Ramanujan's Collected Papers [16, p. xxv], Hardy remarked: "There is
always more in one of Ramanujan's formulae than meets the eye, as anyone
who sets to work to verify those which look the easiest will soon discover. In
some the interest lies very deep, in others comparatively near the surface; but
there is not one which is not curious and entertaining." There could not be a
more apt comment about Chapter 14 than this last sentence of Hardy. Some
of the formulas are fairly easy to prove; others require considerable effort. As
previously indicated, many of the formulas in Chapter 14 have their genesis
in elliptic modular functions. A large number of formulas arise from partial
fraction decompositions. Some formulas are instances of the Poisson summa-
summation formula. Six formulas lie in the realm of hypergeometric series. There are
also a few integral evaluations.
In the sequel, jR(/, z0) = R(z0) denotes the residue of / at a pole z0. Also,
x(n) always denotes the primitive character of modulus 4; that is,
0, if n = 0 (mod 2),
l{n)=-{ 1, if и si (mod 4), @.1)
-1, ifn = 3(mod4).

14. Infinite Series 241
Entry 1. For z2 ф —n{n+ l)/2, where n is a nonnegative integer, we have
\~1 00 l_\
Proof. From the partial fraction decomposition (Whittaker and Watson
[i, p. 1363)
sech x = 4я У -± ;2 I ' '' A.2)
„=o Bn + 1) n + 4x
we obtain, after some simplification,
—l (-
In sechGi;
From the product expansion (Gradshteyn and Ryzhik [1, p. 37])
4z2
cosh z =
and Wallis's product (Gradshteyn and Ryzhik [1, p. 12])
4 = „Ц BnTTF'
we find that
B=1 ^n(n + 1) V1 + Bn + \f
The result now follows. ?
Corollary. For Re z > 0,
» (~1)и+1B"+1) l? ,/я г^ г77\
У ~— —-—7=--= + - У sech -Jn2 - z2/4
"^^/ф+Ше2*^"'"""-!) Zn=l ^z /
= ^ + f-C, A.3)
27CZ 6
where
where the asterisk (*) on the summation sign above indicates that the terms must
be added in successive pairs in order for the series to converge.

242
14. Infinite Series
We first show that the series defining С converges. We have
2n+ I In + 3
1
и
1
5/
= 0(n),
and so С is well defined.
Formula A.3) does not agree with the corresponding entry in the notebooks
in that Ramanujan claims that С should be replaced by
1 n=i 2n
(-1)"
(-1)"
8
A.5)
2 „=i Bn + 1){2и + 1 + 2>/Ф+ I)}2
It is not difficult to prove the foregoing equality. Indeed, let С denote the left
side of A.5). Using Gregory's series for я/4, we find that
c = i-n-+j\ ()n + tl^!!
8 nt\ [2Bn +1) 2n + 1 + гУ^фГ+Т)
{2^/и(и + 1) - Bи + 1)}
8
and A.5) easily follows.
Calculations of J. Hill first demonstrated that the constant given by Ramanu-
Ramanujan is incorrect. In fact, С = 0.61144169•¦•, while С = 0.54661949¦¦¦. The
formula for С given in A.4) can be transformed into another formula that
exhibits Ramanujan's error. Letting an = Bn + 1)Д/и(п + 1), we have
с I ()
n=i
nodd
1 1 °°
x + x I
2. I n=i
n odd
{(an - 1) - (ая+1 - 1)}
:- 1

14. Infinite Series
243
Comparing the formula above with A.5), we find that Ramanujan neglected
a factor of 2^/n(n + 1) in the denominators of the summands on the left side
of A.5). The correct formula for С was first conjectured by R. Lamphere who
verified it numerically.
After stating the corollary of Entry 1, Ramanujan declares: "Similarly any
function whose denominator is in the form of a product can be expressed as
the sum of partial fractions and many other theorems may be deduced from
the result." But nonetheless, we have been unable to prove that A.3) is a
corollary of A.1). The following proof of A.3) is due to R. J. Evans.
Proof of Corollary to Entry 1. We prove the result for z = x > 0; the
more general result will then hold by analytic continuation.
For n > 1, let а„ be as defined above and put
/»W = -т—г-= l
Thus,
(-1ГЧ/„М=
1 » (_l)»+iBn+ 1)
2nx „=i n(n + 1)
By combining successive terms, we find after an elementary calculation that
1
n(n +1)
h \n n + 2
n odd
i
A.7)
Putting A.7) into A.6) and comparing the resulting equality with A.3), we find
that we must show that
00 1 GO
К-1ГЧ/я(*) + -1
n=l X n=i \X
nx
- — =-C, A.8)
where
1 1
ая-ап+1).
n=l
n odd
From Whittaker and Watson's text [1, p. 136],
1 1 . lx
2x
A.9)

244 14. Infinite Series
Using A.2) and A.9) in A.8) and then simplifying, we find that
n m=i n(n + l)x2 + m'
nx x » » (-l)mBm + 1)
L
" Л n^l m^O W('W + l)^2 + П2
n n=\ m=i m(m + l).x + n
Letting
we see that A.10) may be written as
- t t {в(т> и - 1) - B(m, и)}
7C
4 (l.ii)
^ Л m = l n = l
n even
A brief calculation gives
B(m, n - 1) - B(m, и) =
2n2x2 - 2m2
(m2 + n2x2 — nx2)(m2 + n2x2 + nx2)'
Replacing x by x/2, we see then that A.11) is equivalent to
n2x2 — m2 n " » n2x2 — m2
= -^+i x
n2x2J - n2x4/4 2x m^ „4i (m2 + n2x2J - n2x4/4'
A.12)
By a brief calculation,
GO GO
I E
n2x2 — m2 n2x2 — m2
% {(m2 + n2x2J - n2x4/4 (m2 + n2x2
is seen to be an absolutely convergent double series, and so an inversion in
order of summation is justified. Thus, A.12) is seen to be equivalent to
n2x2 -m2 n
2x - - (m2 4- n2x2J
oo oo vj2 V2
^Z ^ ^ /^2 , „2^-242' ()
zx „=1 m=1 (m + n x )
where on the right side we have replaced the indices m and n by n and m,
respectively. Let the left side of A.13) be denoted by F(x). Thus, A.13) may be

14. Infinite Series 245
rewritten as
F(x) + x~2F(l/x) = -tc/Bx). A.14)
Now return to A.9). Replace x by 27mx and differentiate the extremal sides
with respect to x. After some simplification, we find that
2л2 ! - У 2n2x2 » 1
(e"nx — e~nnxJ 2n2x2 „f=i (n2x2 + m2J m=i n2x2 + m2
Summing both sides of A.15) on n, 1 < n < oo, we deduce that
2
-я2 Y csch2Ginx) — ——2 = F(x).
2 „=i 12x
Thus, A.14) is seen to be equivalent to
oo jr oo 71 f л
nx Y csch2Ginx) 4- — Y csch2Gin/x) =—1+—lx + —
„ = l X „=i О \ X
If we put a = nx and /? = я/х, we find that for a, /? > 0 and a/? = я2,
GO GO
a X csch2(an) + p Y csch2(pn) = -1 + (a + p)/6. A.16)
n=l и=1
In summary, we have shown that A.3) is equivalent to A.16). But the author
[6, Proposition 2.25] has previously proved A.16), and hence the proof is
complete. ?
Observe that A.13) provides a beautiful example of a nonabsolutely con-
convergent double series whose order of summation cannot be inverted.
Entry 2. Let m, n, x, and у be complex numbers. Suppose that ГA + xz) and
ГA + yz) have no coincident poles and that z = 1 is not a pole of either. Then
if Re(m + n) > 0,
M Г(и! - к + 1)Г(и + 1 - ку/х)Г(к)(х + к)
(-1)*+1ГA -кх/у)
r'" — к т i;i yin т i — ал/уд 1лду т iv/
- hrVv -l- Л
B.1)
А Г(и - к + 1)Г(И1 + 1 - кх/у)Г(к)(у + к)
Г(х+
Г(х + т + 1)Г(у + и + 1)
Proof. Let
f, .__ ГA + Х2)ГA + yz)
J У2) — i~.
Y{m + xz + 1)Г(и + yz + \)(z -

246 14. Infinite Series
Then / has poles at z = 1, —j/x, and — k/y, where 1 < j, к < oo, and all poles
are simple by hypothesis. Routine calculations yield
Г(т + х+ 1)Г(п + у+ 1)'
f_ , = (-1)JTA-Jy/x)
( J/X) Г(т - j + 1)Г(п - jy/x + l)(j + x)T(j)'
and
T(m - kx/y + 1)Г(и - к + l)(ik + у)Г(к)'
Let CN be a positively oriented square centered at the origin and with
vertical and horizontal sides of length 2N. We shall let N tend to oo on some
countable subset of the positive real numbers chosen so that the sides of CN
never get closer than some fixed positive distance from the set of poles of /.
Using Stirling's formula, we find that
as \z\ tends to со. Hence, if Re(m + n) > 0, we deduce that
f(z)dz = o(l), B.2)
as N tends to со.
Now integrate / over CN and apply the residue theorem. Let N tend to oo
and use B.2). We then deduce B.1) immediately. ?
Corollary 1. Let m, n, and x be complex numbers such that x is not an integer
and that Re(m + ri) > -1. Then
f
khn (x + к)Г(т + 1 - к)Г(п + 1 + к)
n
~ sinGcx)r(m + x + 1)Г(п - x + 1)'
Corollary 2. Let a and fi be complex numbers with Re(a + /?) > 0. Then
oo С _ ] )k oo t—\)k
уL1+ у { '
B.3)
k% Bk + 1)Г(а - к)Г(/? + к + 1) k% Bk + 1)Г(/? - к)Г(а + к
п
2Г(а + ±
B.4)
Corollaries 1 and 2 are not really corollaries of Entry 2. Ramanujan
evidently means to imply that the proofs of the present results are very much
like the proof of the preceding theorem.

14. Infinite Series 247
Proof of Corollary 1. Let
77"
f(z) = ^~
x)T(m + 1 - z)Y(n + 1 + z)
Observe that / has a simple pole at z = — x and at each integer k. Routine
calculations give
R(-x)=--
sinGcx)r(w + 1 + х)Г(и + 1 - x)
and
(^
(к + х)Г(ш + 1 - к)Г(п + 1 + к)'
Let Сдг be the positively oriented square centered at the origin with vertical
and horizontal sides passing through ±(N + \) and ±(N + ^)i, respectively,
where N is a positive integer. By Stirling's formula,
f(z) = О(|гГКе(т+п)'2),
as \z\ tends to oo. Hence, for Re(m + n) > — 1,
%z)dz = o{\\ B.5)
as N tends to oo. Apply the residue theorem to the integral of/ over CN. Let
N tend to oo. Using B.5), we deduce B.3) at once. ?
Proof of Corollary 2. Integrate
л
sinGtz)(z - ^)Г(а + z)T{P - z
over the same square as in the foregoing proof. The present proof follows
along precisely the same lines, and we omit it.
A second proof can be given as follows. Let the left side of B.4) be denoted
by g(a, /?). After a little manipulation, we see that g(a, /?) may be written as
g{*. P) - -
which converges absolutely for Re(a + /?) > 0 by Stirling's formula. Now
apply DougalFs formula (Henrici [2, p. 52]) to the right side of B.6) to obtain
Corollary 1 may also be proved with the aid of Dougall's theorem. How-
However, Dougall's theorem is not applicable to Entry 2. The next entry is also an
instance of Dougall's theorem.

248 14. Infinite Series
Entry 3. Let a, /?, y, and 5 be complex numbers such that Re(a + P + у + д) >
-1. Then
k% Г(а - fc + \)T(p -к+ 1)Г(у + к
1
+ *ti Г(а + к + 1)Г(Р + к + 1)Г(у - к + 1)Г(<5 - к
Г(а + Р + у + д + \)
~ Г(а + у + 1)@ + у + 1)Г(а + с5 + 1)Г@ + 5+1)'
Proof. The left side of C.1) may be written as
sin(rox) sm(nP) " Г(/с — а)Г(/с — jS)
C.1)
n2 kJ±n Г(у + к + \)Г{5 + к + 1)'
which converges absolutely for Re(a + P + у + S)> — 1 by Stirling's formula.
A straightforward application of Dougall's theorem (Henrici [2, p. 52]) yields
C.1) immediately. ?
Entry 4. If z ф me**1'3, where m is a nonzero integer, then
у jл smhjnz^/3) - УЗ sin(^z)
M n2 + z2 + z4/n2 2z^/3 coshGtzy3) - cos(rcz)
Proof. Let f(z) denote the right side of D.1). We expand / into partial
fractions. Since
y — cosGtz) = 2 sinGcze'"/3) sin(nze~n'13),
f has simple poles at z = ne±ni/3 for each nonzero integer n. Now
^/ e-m/3\ _ -"""У""- у -v у -^ sinGrne *' )
Note that «(-ne""'73) = -R(ne""'73). The residues of the poles at ±ne"il3 are
obtained by replacing e~ml3 by e13 above. For each positive integer n, the
sum of the principal parts for the four poles + ne±n'13 is then
(-1)" je"'I'73{sinhGcne"iri/3y3) - Уз 8т(тшГя'/3)}
( sinGcne~2ltl/3)(z2 — n2e~2ni/3)
enil3 {sinh(nne"i/3y3) - Уз sin(nneni/3)} j
+ sinGcne2m/3)(z2 - n2e2nil3)
Elementary calculations give
/X if и = 2m,
sm(nne+-™>3) = |±гA) S
Щ ' l(-ir+1coshGmy3/2), ifn =

14. Infinite Series 249
and
sinGrne±Iti/3)}
_ J2(- 1)"' sinh(mrj,/3), if n = 2m,
~ |±2i(-l)m+1 cosh(jrnv/3/2), if n = 2m + 1.
Using the calculations above, we find that D.2) simplifies to n2/(z4 + n2z2 + n4)
for both n even and n odd. Hence,
OO jj2
/(z) = 5 ?T7^T^ + diz)>
where g is entire. However, as \z\ tends to oo, we clearly see that g(z) tends to
0. Thus, g is a bounded entire function. By Liouville's theorem, g(z) is constant,
and this constant is obviously 0. Hence, the proof is complete. ?
Corollary. For each nonzero integer n,
ffi к2 + BnJ + BnL/k2 12п2 2 й к2 + Зи2'
Proof. In the derivation below, we shall employ A.9) and the decomposition
(Whittaker and Watson [1, p. 136])
. . . 1 2z » (-1)"
csch(rcz) = — + — X —2 7j.
nz n k=i z + к
In Entry 4 let z = 2n to gel:
г1Т1 = —^T{co\hBnnyJl) + cschB7rny3)}
• BnJ +
v
1 . - 1 » (-1)"
12n2 „tin^ + fc^ktl^^ + fc2'
and the result follows. ?
Entry 5(i). Let 0 < x < n/(n + j), where n is a positive integer. Then
» sin2n+1(b) y/
h
h I ~ Tr(n + iy
Proof. Since (Gradshteyn and Ryzhik [1, p. 25])
sin2n+1 x = 2"" У {-\)n+j\ + ) sin{Bn + 1 - 2;»,
j^o V J )
we have
sin{BW+l-2j)fcx}
_(l)^ )U
2 0 \ J

250 14. Infinite Series
Using the familiar result (Gradshteyn and Ryzhik [1, p. 38])
" sin(kx) n — x
I —7— = -^—, 0 < x < In, E.2)
we find that, for 0 < x < n/{n + \),
2-,
k
sin
2n+1(
(fcx) _ (-1)" » /2и + 1\я-Bи + 1-2у)х
2-, у, т2л H
k=i к I j=o
E.3)
where we used the evaluation (Gradshteyn and Ryzhik [1, p. 3])
with m = n and к — In + 1.
We next show that the sum on the far right side of E.3) vanishes. We have
" /2n + 1\ 2n+1 fin + 1\
2 ? (-!)'( )Bn + 1 - 2Л = E (-DM )Bи + l ~ 2»
j=o \ J / j=o \ J J
+
J
2n + l /Ой 4.
-2 2 f-iy
j=o \ J
= 0,
where we have used E.4) and (Gradshteyn and Ryzhik [1, p. 4])
Hence, from E.3),
/c 22n+1 V и
which is easily seen to be equivalent to the desired result by the Legendre
duplication formula. ?
Entry 5(ii). Let 0 < x < n/(n + 1), where n is a positive integer. Then
^5, Sin [КЭС) */71 1 \П ~\~ ~t)
\ — _jl у (s >,\
k=i к 2 T(n + 1)
Proof. Let /(x) denote the left side of E.5). Since (Gradshteyn and Ryzhik

14. Infinite Series 251
sin2n+2 x
we find that
2, cos{2(n + 1 - j)kx]
Now (Gradshteyn and Ryzhik [1, p. 39])
л2 nx x2
Employing the formula above and E.4) with m = n and к = In + 2, we find
that E.6) becomes
f+i » /2n + 2\
1 ^ ' -<<2П + у{п+1-»1х2-п(п+\~Пх}
i + 2\tt2
where 0 < x < я/(и + 1). First,
2n+2
l(y )( ;) E (у(
2n + 2
= I (-iv
= 0. E.8)
Next, from two applications of E.4), we find that
Substituting E.8) and E.9) into E.7), we find that
/2n\

252 14. Infinite Series
which is again equivalent to the desired result by the Legendre duplication
formula. ?
Entry 6. For и > 0, let
\2~) -1
go r
= Д j1
Let a, /? > 0 with a/? = n. Then
Proof. Recall the Poisson summation formula. If / is a continuous function
of bounded variation on [a, b], then
Г f(k) = Г f(x) dx + 2^ Г f(x) cosBnkx) dx, F.1)
a<k<b Ja k = l Ja
where the prime on the summation sign at the left indicates that if a or b is
an integer, then only jf(a) or %f{b), respectively, is counted.
Now cp(x) was studied by Ramanujan in [8], [16, pp. 53-58]. On page 54
of [16] Ramanujan remarks that
T{n + ix)T(n - ix)
This is not too difficult to prove; use the Weierstrass product formula for
the quotient of Г-functions above, and after considerable simplification, the
desired equality follows. We shall apply F.1) with f(x) = <p{flx), a = 0, and
b = oo. By using Stirling's formula for \T(n + ix)T{n — ix)\, as x tends to oo,
we easily justify letting b tend to oo. Furthermore, for m real and n > 0, by
Entry 15 in Chapter 13,
<p(x) cosBmx) dx = ^ Г(" + z) seen2" m.
о 2 Г(и)
Hence, since <p@) = 1, F.1) yields
if00 2
() *
+ ? <P(№ я <?(*) <** + я Е ?(*) cosBnkx/P) dx
* * = 1 P Jo P k=l Jo
+
B
which is easily seen to be equivalent to the desired result. ?
Entry 7. Let a, j8 > 0 wit/i a/? = n and let z be an arbitrary complex number.
Then

14. Infinite Series 253
fi 00 ") ( 00 ~)
- + ? e-*42 cos(xzk)\ = Jfi \- + ? e"*2 cosh(jfeA:».
(.2 n=i J B t=i J
Proof. Apply the Poisson formula F.1) with f(x) — exp( — azx2) cos(azx),
a = 0, and b = oo. Now (Gradshteyn and Ryzhik [1, p. 480]),
e~cxl cos(rx) dx = l -е-'2*4*, G.1)
о 2V c
where Re с > 0 and r is arbitrary. With the use of the above evaluation, all
the calculations are quite routine, and the desired formula follows with no
difficulty. ?
Corollary. Let a., /? > 0 with afi = n. Then
Proof. Let z = 0 in Entry 7. ?
Note that the formula above is simply the functional equation for the
classical theta-function.
Entry 8(i). Let a, /?, n > 0 with tx.fl = n andO < fin < n. Then
» sinhBanfc) * sinB/M) ,
it=i e — l k=i e ' - l
Entry 8(i) arises from the transformation formulas of a function akin to the
logarithm of the Dedekind eta-function. The first proof of Entry 8(i) preceded
that by Ramanujan and was found by Schlomilch [1], [2, p. 156]. Later proofs
have been given by Rao and Ayyar [1], J. Lagrange [1], and the author [6,
Eq. C.31)], [2, Eq. A1.21)].
Entry 8(ii). Let a, /?, n > 0 with a.fi = к and 0 < аи < п. Then
coshBj8fifc) 2 u 2 2 f sin(an)
j^rrij =n -Ua - p) + LogШм
Entry 8(ii) arises from the transformation formulas of a function that
generalizes the logarithm of the Dedekind eta-function. Proofs have been
given by J. Lagrange [1] and the author [6, Proposition 3.4].
Entry 8(iii). Let a, ft n, r, t > 0 with a.fi = n,r = nfi, and t = n/fl2. Let С be
the positively oriented parallelogram with vertices ± i and ± t. Let q>(z) be entire.
Let mbea positive integer and put M = m + \. Define

254 14. Infinite Series
fm(z) = -2kMz 2niMzlt П'
and assume that fm(z) tends to 0 boundedly on С = С — { + i, ±t} as m tends
to oo. Then
+к
(p(ank) « (p(Pnki) + q>(-finki)
~к{е2*гк-\) +к~к к k(e2 - 1)
mcp@) aV
_ ^ + 6 2 + 2 4 ' (8Л)
provided that all series above converge.
The obviously very restrictive hypotheses on q> are of a technical nature.
We could state these hypotheses more specifically, but an even lengthier
statement of the theorem would be necessary.
Proof. We integrate /m(z) over C. On the interior of C, fm has simple poles
at z = ± ik/M and at z = + kt/M, 1 < к < т. Also, there is a triple pole at
z = 0. Straightforward calculations give
R{-ik/M)=^
2nik {e2nktt - 1
(p(-rki)
2nik(e2nklt - 1)'
2nik \е2жк'
and
Утл tri sy*'"-"'1 1 t
where 1 < к < т. Now,
/m(z) = B7rMJz3 {<?)@) + ^'^rMz + W'@)(rMzJ + ¦¦¦}
x {1 +7rMz + iGrMzJ + ---m
and so
4 + 12 12t + 4n + 4n +8^-
Applying the residue theorem and letting M tend to oo, we find that

14. Infinite Series 255
« <p{rki) + <p(
() iz e
k=i к(е — 1) n=i к l=i
яф(О)
~ &
rtcp'(O) <V@)
By our hypotheses and the bounded convergence theorem, the limit on the
left side of (8.2) is 0. Substituting r = nfi and t = n/p2 in (8.2) and rearranging,
we deduce (8.1). ?
We next show that Entry 8(ii) is a special instance of Entry 8(iii).
Let (p{z) = expBiz). Thus, q>{ank) + <p{ — ank) = 2 cosBanfc)and<p(/?nfci) +
(p(-Pnki) = 2 coshB(lnk). Since 0 < аи < п, by a standard result found in
Gradshteyn and Ryzhik's book [1, p. 38] and E.2), we have
I —j = E —r- = -Log{2 sin(an)} + i .
Second, an elementary calculation gives
jt=i K t=i K
Thus,
" cp(Pnki) " q>(a.nk) ni(p(Q) a.2(p@)
Ik * t=1 ~ * + ~2 6^
аиф'(О) Pni(p'(O) n2q>"@)
sin(an)
sin
Hence, formally, Entry 8(ii) follows readily from Entry 8(iii).
It remains to check the hypotheses concerning the parallelogram C. This
is easily done by parameterizing each side of C. In the first quadrant, /m(z)
trivially tends to 0 boundedly on С The same is true on С in the second
quadrant, but the hypothesis r > 0 is needed. Since 0 < аи < п, /m(z) tends to
0 boundedly on that part of С in the lower half-plane.
Corollary (i). Let a, p > 0 with a.p = n2. Then
This entry is really not a corollary of Entry 8(iii); however, a proof can be
given along somewhat the same lines.

256 14. Infinite Series
Formula (8.3) was first established by Schldmilch [1], [2, p. 157]. Other
proofs have been given by Malurkar [1], Rao and Ayyar [1], J. Lagrange [1],
Grosswald [2], Sitaramachandrarao [2], and the author [6, Proposition
2.11], [2, Eq. A1.7)]. In essence, (8.3) was also established by Hurwitz [1], [2]
and Guinand [1], although neither author explicitly states the formula.
Corollary (ii). Let a,P>0 with a.p = n2. Then
Proof. Let u, v > 0 with uv = n2. Write Corollary (i) in the form
? ke-2uk v » ke~2vk 1 v/u 1
и t4i J-
Integrate both sides of the equality above with respect to и over the interval
[л, a] to obtain
: e-2vk(v/u)
аи
7
— e
a - я If"
= -*T + 24 {V'
>/u) du+\ LogGt/a).
In the integrals that remain, make the change of variable и = n2/v. By the
hypothesis, the limits n and a are transformed into n and /?, respectively. Thus,
the last equality becomes
1 ^?, 1 — e ol — p
2 fc=i 1 — e 24
Multiplying both sides by 2 and then exponentiating both sides yields the
desired result. ?
(8.4)
This example is obtained from Corollary (i) by setting a = p = n. Ramanujan
stated (8.4) as a problem in [3], [16, p. 326]. He later gave a proof of (8.4) in
[7, p. 361], [16, p. 34] by using some formulas from the theory of elliptic
functions. But, as already indicated, (8.4) was first established by Schlomilch
[1], [2, p. 157]. Proofs of (8.4) have also been given by Krishnamachari [1],
Watson [1], Sandham [1], Lewittes [1], [2], and Ling [3], in addition to the
authors listed after Corollary (i).
Example.
We have
CO
z-
/c=i e
к
2жк _ j
1
24
1
8л

14. Infinite Series 257
Entry 9(i). Let a, ft > 0 with a/? = я/2. Let h > Obe chosen so that h/a > 1 and
h/a is not an odd integer. Let m be the greatest odd integer that is less than h/a.
Let n be an arbitrary real number. Let cp(x) be continuous and of bounded
variation on [0, h] and define
f*
ijj(i)= <p{x) cos(tx) dx. (9.1)
Jo
// x is defined by @.1), then
a X X(k) sm(ocnk)q>(ak) = - ? х(к){Ф(Рк - n) - фЦЗк + я)}.
k = l *¦ k = l
Proof. Let / be a continuous function of bounded variation on [a, b]. Then
the Poisson formula for sine transforms (Titchmarsh [2, p. 66])
I' X(k)№ = I X(k) f fix) sinGrb/2) dx (9.2)
a<k<b t=l Ja
is valid, where the prime on the summation sign on the left side has the same
meaning as in F.1). Let f(x) = sin(a.nx)q>(a.x), a = 0, and b = h/a. Then
со Г ty
/с)(p(afc) == ? xC^)
t=i Jo
sin(аи/с)(p(afc) == ? xC^) sin(anx)<p(ax) sin(nkx/2) dx. (9.3)
*=i t=i Jo
The integrals on the right side of (9.3) are easily calculated by (9.1) to complete
the proof. ?
Entry 9(ii). Let a, /?, h, m, n, and cp satisfy the same hypotheses as in Entry 9(i).
Define
f*
= (p(x) sin(tx) dx.
J
Then
m J go
a X X(k) соз{тк)фк) = - ^ Z(fc){^(j8* - в) + ф@к + я)}.
Proof. The proof is completely analogous to that for Entry 9(i). ?
Ramanujan stated Entries 9(i) and 9(ii) with the extra condition \n\ < ft,
but this hypothesis does not seem necessary. Entries 9(i) and 9(ii) are ana-
analogues of Entries 31(i) and 32(ii) in Chapter 13, respectively.
Entry 10. Let a, /J > 0 with afi = л/4, and let z be an arbitrary complex number.
Then
sin(azk) = Jp ? Z(*)e-'2*2 smh(^zk).

258 14. Infinite Series
Entry 10 should be compared with Entry 7.
Proof. Apply (9.2) with f(x) = e~*2 sin(azx), a = 0, and b = oo. By G.1),
о 4av
= yi^/oV*2-*2'4 sinh(j9z/c).
The entry now readily follows. ?
Entry 11. Let a, /? > 0 wif/z аД = л, and let n be real with \n\ < fi/2. Then
>• (П.1)
Proof. We shall use a transformation formula, Theorem 3(i), from the author's
paper [4]. Because the statements of the relevant theorem and notation from
our paper [4] would require considerable space, we kindly ask the reader to
refer to [4].
Let V(z) = - l/z, /-! = 0, and - 1 < r2 = r < 0. Then R1 =r,R2 = 0, and
p=0. Also, let s = -N = 1. By [4, Eq. D.5)], we find that
(_8z XV 4
where Bx (x) denotes the first Bernoulli polynomial. It follows that
t xW*(z,U0,r;l,n)=-ni. A1.2)
We next calculate, for Im z > 0,
oo oo oo oo
H2(z, 1; X; 0, r) = ?
Hence,
Jt=l e i
(- l/z, 1; X; 0, r) = - ? z(/c)^gg. A1.3)
Next, we calculate, for Im z > 0,
00 OO OO 00
= -2ni X Z l{m)e2nikim+r)z - 2ni ? ? l
m=l t=l m = l fc = l
= — 4tc* ? cosB7t/crz) Z Z(i) Z e2itik{4m+J)z
k = l j=0 m=0
Л J?, cosB7rferz)

14. Infinite Series 259
Lastly, we need to calculate J5f+A, x, r), where
Se+(s, x, r) = L(s, x, r) - e'"L(s, x, -r)
and where, for Re s > 0 and a real,
L{s,x,a)= ?
k> -a
Also define, for a, x real and Re s > 1,
Jc=O
where the prime on the summation sign indicates that the possible term
к = —a is omitted from the summation. The functions L(s, %, a), J?+{s, x, я),
and L(s, x, a, x) possess analytic continuations into the entire complex s-plane.
Now apply the functional equation for L(s, x, a, x) (Berndt [3, Theorem 5.1])
to get, for all s,
L(l -s,-r, 0, x) = r(s)BMs(i/2)e-"isl2^+(s, x, r).
Hence,
&+{l,X,r) = nL{0,-r,0,X). (П.5)
Now, for Re s > 0,
L(s, -гЛх)= ? e-«ik'l2X(k)k-s
k=l
= e-™12 ? e-2"ikrDk + l)"s - e'3"'^2 f e-2"ikrDk + 3)"s
t=0 k=0
= г-"вд4->(-г, |, s) - е-3"4->(-г, |, s), A1.6)
where, for x, a real and Re s > 1,
<p(x, a, s) = f' e2«ik*(k + ay*
k=0
denotes Lerch's zeta-function. By analytic continuation, the extreme left and
right sides of A1.6) are equal for all s. Now from Apostol's paper [2, p. 164],
(p(x,a,0) = ^cot(nx) + -. A1.7)
Hence, from A1.5)—A1.7),
f+a, X, r) = ^(e G \ cotftr)! <T U
-
= ^ sec(nr/2). A1.8)

260 14. Infinite Series
Substitute A1.2), A1.3), A1.4), and(l 1.8) into Eq. D.6) of our paper [4] to get
2тг » cos(nkr/2) n « cosBnkrz)
— L X fc) „ikK22) Г + T" secGrr/2 = -27i! X —¦ ,- ., , - ni,
z k=i e*lW2v - l 2z fe=i coshB7tifc)
where Im z > 0 and 0 < — r < 1. Now let z = in/Ba2) and r = 2n//J, where
a.p = n. Thus, 0 < - и < jS/2. Hence,
5 • v^ cos(an/c) ,, ~ . т
-4a i > ZC0)—Ji a ' sec(an) = — 2л( )
t=i ea * — 1 4=
Multiplying the last formula by i/Da) yields A1.1). Now note that both sides
of A1.1) are even functions of n. Thus, A1.1) is valid for 0 < \n\ < ft/2 and,
hence, by continuity, for \n\ < /f/2. ?
We remark that the differentiation of A1.1) with respect to n yields the last
formula in our paper [2] after suitable redefinitions of the parameters. How-
However, it appears to be difficult to deduce A1.1) from the latter formula.
Entry 12. Let a, fi > 0 with aft = я/2, and letO<n< тг/Bа). Then
- sin(anfe) » sinh(^nfc)
ай *k)z*u?k) = p& х(к)^щЩ- A2Л)
Proof. In our paper [6, Eq. D.23)] we showed that if 0 < r < 1 and a, ft > 0
with a/? = 7T2/16, then
sinhBj8rA;)
^} A12)
Replace a by a2/2 and ^ by P2j2; hence, in the new notation а]3 = л/2. Let
r = 2an/n. Thus, we need 0 < n < я/Bа). With these substitutions, we easily
find that A2.2) is transformed into A2.1). ?
Corollary. Let a, ft, t > 0 with а/? = тг/2 and t = a//J. Let С be the positively
oriented parallelogram with vertices +i and ±t. Let q>(z) be entire. For each
positive integer N, define
coshBnNz) coshBniNz/t)'
and assume that NfN(z) tends to 0 boundedly on С as N tends to oo. Then
cosh(a2fc) +#
The above entry is not a corollary of Entry 12. In fact, as we shall see later,
the converse is true. As with Entry 8(iii), at the expense of brevity, the
hypotheses on fN(z) can be made more explicit.

14. Infinite Series 261
Proof. We integrate /N(z) over C. On the interior of C, /N(z) has simple poles
at z = i{2k + 1)/DЛГ) and at z = Bk + l)t/DN), ~2N < к < 2N. Straight-
Straightforward calculations give
and
2nN c
Applying the residue theorem and letting N tend to oo, we find that
(-l)k{cp(iPBk + 1)) - V(-ij8Bfc + 1))}
ton N jcfN(z) dz = ^ cosh{Bfe+1)^2}
_ . » (- l)*{<P(foBfc + 1)) - cp(-ltBfc + 1))}
' bh cosh{BA:+ l)nt/2}
A2.4)
Putting t = a//? in A2.4) and using the fact that NfN(z) tends to 0 boundedly
on C, we readily deduce A2.3). ?
Next, we show that Entry 12 is a corollary of the preceding entry. Let
cp(z) = e'nz, where n > 0. We see at once that A2.3) then reduces to A2.1). It is
easily seen that the hypotheses on fN(z) are satisfied on the two sides of С in
the upper half-plane. In the lower half-plane on C, NfN(z) tends to 0 boundedly
if and only if n < я/Bа), which is precisely a hypothesis of Entry 12.
Entry 13. Let a, ft > 0 with а/J = n2, and let n be an integer greater than 1. Then
Entry 13 is stated without proof by Ramanujan in [13, p. 269], [16, p. 190].
The first published proof known to the author is by Rao and Ayyar [1].
Malurkar [1] and Hardy [3], [7, pp. 537-539] gave proofs shortly afterward.
Later proofs were found by Nanjundiah [1], J. Lagrange [1], Grosswald [2],
Sitaramachandrarao [2], and the author [2, Eq. A1.10)], [6, Proposition 2.6].
Corollary (i). | ^-^
00 k9 1
Corollary (ii). I ^37=264
Corollary (Hi). | ^r = l

262 14. Infinite Series
Corollary (iv). // n is a positive integer, then
qo t4" + l R
If a = p = n and n is odd, then Entry 13 reduces to A3.1) if и is replaced
by 2n + 1. Corollaries (i)-(iii) are special instances of Corollary (iv). Corollary
(iii) was communicated by Ramanujan in a letter to Hardy [16, p. xxvi].
Sandham [1] also proved this special case. M. V. Aiyar [1] and Ling [3]
established Corollaries (i)-(iii). The more general Corollary (iv) was actu-
actually first proved earlier by Glaisher [3] in 1889. In addition to the authors
who have proved Entry 13, Corollary (iv) has also been established by
Krishnamachari [1], Watson [1], Sandham [2], and Zucker [1].
As usual, let <rv(n) = ?d|ndv. It is easy to show that
I^K^I^, (B-2)
where у > 0. Thus, Entry 13 may be rewritten in terms of the left side of A3.2).
In this form, Entry 13 was established in Hurwitz's thesis [1], [2] in 1881 and
may be even older than 1881. Later proofs were found by Koshliakov [1],
Guinand [1], and Chandrasekharan and Narasimhan [1].
Entry 14. Let a, ft > 0 with aft — n2, and let nbe a positive integer. Then
oo k2nl °° к2
Entry 14 has been established by Malurkar [1], Nanjundiah [2], and the
author [6, Proposition 4.7].
Corollary of Entry 14. If n is a positive integer, then
» (-Щ2/С + 1Г"-1
к%со*ЦBк+l)n/2} " [ ' '
If a = P = n and n is even in A4.1), then A4.1) reduces to A4.2) upon the
replacement of n by 2и.
This corollary was, in fact, first established by Cauchy [1, pp. 313, 362].
Ramanujan stated A4.2) as a problem in [2]. In addition to the authors who
have proved Entry 14, A4.2) has been established by Rao and Ayyar [2],
Chowla [1], Sandham [2], Riesel [1], and Ling [3].
Entry 15. Let a, j3 > 0 with aj3 = тг2/4. Then
GO 00
2 X X(n) tan-1^-"") + 2 X X(n) tan-1^"")
n=l n=l
" sech(an) « sech@n) n
= I X(n)—-— + I Z(«)— = -. A5.1)
„=i П „=i n 4

14. Infinite Series 263
Proof. A proof of the rightmost equality in A5.1) has been given by Malurkar
[1], Nanjundiah [1], and the author [6, Proposition 4.5].
The leftmost equality in A5.1) follows from
= 2 I
„=i n k=b
where у > 0. П
Corollary. We have
? Z(«) tan" V""'2) = я/16.
n=l
The corollary follows trivially from A5.1) upon setting a = /? = л/2. Rao
and Ayyar [2] have also established this result. Chowla [1] has proved some
formulas similar in appearance to A5.1).
Entry 16(i). Let m and n be nonnegative integers. Then
- cos2m xdx =
sin2n+1x 2m л Г(т + ±)Г(п + i)
-Г
2Г(т + и + 1)
—z cos2m x dx.
Readers should compare the formulas for m = 0 with Entries 5(i), (ii).
Proof. The first equality can be found in Gradshteyn and Ryzhik's tables [1,
p. 457], but since the second is not in [1], we give a brief proof. (A proof of
the first equality can, in fact, be given along the same lines.) Let the integral
on the right side above be denoted by I(m, n). We induct on m. For m = 0,
by the tables of Gradshteyn and Ryzhik [1, p. 446]. Proceeding by induction,
we have
I{m, ri) = I(m — 1, и) — I(m — 1, n + 1)
-j)r(H+|) Г(т - $)Г(и + f)
2Г(т + и) 2Г{т + и + 1)
Г(т ± j
=
2Г(т + п + 1) '
and the proof is complete. ?

264 14. Infinite Series
The equalities below with p = 0 should be compared with Entries 5(i), (ii).
Entry 16(ii). Let n and p be nonnegative integers. Then
P
Jo
cosBpx)dx = (-!)
2 Г(и - p + 1)Г(и + p + 1)
sm2
v2
0 X
cosBpx) dx.
Proof. We prove the first equality; the proof of the second is virtually the
same. Let I(n, p) denote the integral on the left side above. For p — 0, the pro-
proposed formula is true by Entry 16(i). Thus, we assume that p > 0 for the
remainder of the proof. We induct on n. For n = 0, it is easy to show that
1@, p) *= 0 (Gradshteyn and Ryzhik [1, p. 414]), which agrees with the proposed
result. Using the identities 2 sin2 x = 1 — cosBx) and 2 cosBx) cosBpx) =
cos{2(p + l)x} + cos{2(p — l)x}, we find that, by the induction hypothesis,
l(n, p) = \l(n -Up)- il{n - 1, p + 1) - i/(« - 1, p - 1)
Г(и - р)Т(п + р)
1 1
2Г(и - р - 1)Г(и + р + 1) 2Г(и - р + 1)Г(и + р - 1)
2Г(и-р+ 1)Г(п + р + 1)
after several applications of the functional equation of F(z). ?
Entry 17(i). Let a, ft, n > 0 with a/3 = In. Suppose that я/Bа) is not аи integer,
and let m = [я/Bа)]. Let p be real. Then
fl ?
a<-+ 2j cos"(a/c) cos(apfc)
(.2 fc=i
тси! f 1
i(n - p + fik) + l}r{i(n + p -
+ F{i(n + p + Д) + l}F{i(n -p-fik)+
Proof. By Stirling's formula, the right side of A7.1) converges absolutely for
«>0.
Apply the Poisson formula F.1) with f(x) = cos"(ax) cos(apx), a = 0, and
b = 7t/Ba). After a simple change of variable, we find that

14. Infinite Series 265
\ m i Г"/2
- + У cos"{ock) cos(apfc) = - cos" t cos(pt) dt
2 *=i a Jo
2 °o C*/2
+ - X cos" tcos(pr) cos(pkt) dt. A7.2)
°U=i Jo
Now for v > 0 and arbitrary a (Gradshteyn and Ryzhik [1, p. 372]),
cos" x cos(ax) dx = _v.-r,fl. , * .y.,rfl,——-Ju"- <17-3)
о
If we calculate all the integrals in A7.2) with the aid of A7.3), we arrive at A7.1)
forthwith. ?
Entry 17(ii). Let a, P, n > 0 with a.p = я/2. Suppose that я/Bа) is not an odd
integer, and let m = [я/Bа)]. Let p be real. Then
m
¦x ? xik) cos"(a&) sin(apfc)
= 2^2 fcZ aw |р^(и _p + pq+ 1}г{^(и + р-Рк)
A7.4)
T{i(n + P + Pk) + l}T{i(n - p - pk) + 1}
Proof. As before, the series on the right side of A7.4) converges absolutely
for n > 0.
Apply the Poisson formula for sine transforms (9.2) with f{x) =
cos"(ax) sin(apx), a = 0, and b = я/Bа). After a simple change of variable, we
find that
m [ «j f*/2
X x(k) cos"(afc) sin(ap^) = - ^ z(^) cos" t sin(pf) sin(jSfct) dt. A7.5)
t=i «и Jo
If we calculate the integrals in A7.5) with the use of A7.3), we deduce A7.4)
immediately. ?
Corollary 1. Let а = я/(и + j), where n and j are positive integers of opposite
parity. Let m = [я/Bа)]. Then
Proof. In Entry 17(i) replace n by 2n and let p = 0. Let 2/„(а) denote the
infinite series on the right side of A7.1); that is,
00 1
Г(и + 1 + far/а)Г(п + 1 - kn/a)'

266 14. Infinite Series
Since fn(n/(n + ;)) = 0, we see that A7.1) reduces to
which can be transformed into the desired result by the use of Legendre's
duplication formula. ?
In fact, Ramanujan claimed that A7.6) is valid for 0 < a < n/(n + 1), that
is, /„(a) = 0, 0 < a < n/(n + 1), provided that n/Ba) is not an integer. (Of
course, for a = 0 the result is false.) In general, /„(a) does not vanish for all a
in @, n/(n + 1)), as the following counterexample shows.
Let n = 1 and put /(a) = Л (a). Let a = 2я/5 < л/2. Then
GO I
/Bti/5) = J] щ + 5/с/2)ГB - Sk/2)
& A - E^/2J)E/с/2)ГE/с/2)ГA - 5fc/2)
_ 2 " sinE7cfc/2)
= 5^ ttl A - Efc/2J)fc
_\^ » (-1)"
~ 5^ * A. A -{5Bfc + l)/2}2)Bfc + 1)'
The latter series can be evaluated by the residue theorem. Let
sec (яг)
(Z) = {l-EzJYz'
which has simple poles at z = 0, ±j, and Bk + l)/2, where /c is an integer.
Routine calculations give
K@) = 1, K(|) = -isecGc/5) = K(-i),
and
Я(B/с-
7t(l -{5B/c+ l)/2}2)Bfc + 1)'
Integrate h(z) over a positively oriented square Cn with center at the origin
and horizontal and vertical sides of length 2n, where n is a positive integer.
As n tends to oo,
f h(z)dz = o(\).
Hence, applying the residue theorem and then letting n tend to oo, we find that
Д2я/5) = ^A - secGr/5)) Ф 0,
which disproves Ramanujan's claim.

14. Infinite Series 267
Corolliary 2. Let a = n/(n — j), where n and j are integers of opposite parity
such that n > 0 and 0 < j < (n - l)/2. Let m = [>/Ba)]. Then
WJ
A7.7)
Proof. In Entry 17(i) replace n by 2и and let p = 0. After some manipulation,
we find that
where
1
(«) = E
^2 Г(и + 1 + кп/а)Г(п + 1 - /сл/a)'
For a = я/(и — j), 0 < j < (и — l)/2, gfn(a) = 0, and so the proof is complete.
?
Rarnanujan, in fact, claimed that A7.7) is true for n/n < a < 2n/(n + 1), that
is, #„(«) = 0, n/n < a. < 2n/(n + 1), provided that ;x/Ba) is not an integer.
Again, this claim is false, in general, and we give a counterexample.
Let и = 3 and put a = 2n/5; so л/3 < a < л/2. Then
g3Bn/5) = ?
t* ГD + 5/с/2)ГD - 5/с/2)
5Hti
1)'
where P(z) = (9 - 25z2)D - 25z2)(l - 25z2). This series can be evaluated by
the same method as used in the previous counterexample. Accordingly, we
find that
which disproves Ramanujan's claim.
Entry 18. Let an, bn, pn, qn, Pn, and Qn be complex numbers with а„Ь„ ф 0. Let
x and у be complex variables with xy ф 0. Let
<p(x) = Z—L. and ф(у) ? Qn

268 14. Infinite Series
Then
Tft-V \а„х/ »д„-Ьи;у \Ку]
where it is assumed that at least one of the two double series on the right side
o/A8.1) converges absolutely.
Proof. Without loss of generality, assume that the latter double series on the
right side of A8.1) converges absolutely. Inverting the order of summation
below by absolute convergence, we have
1
nPn
= у р» у 6ta** + у
nPn- anx Г anqkx - bkpny kqk- bky ? bkpny - anqkx
= у Рп у [ Qkanx . QAyiPn - anx)
кх - bkPny (qk - bky)(bkpny - anqkx)
Рп у Qk
y
nPn- а„х ^ qk- bky
= <р(х)Ф(у). п
Despite the simplicity of the result above, Ramanujan found many inter-
interesting applications of it, as we shall see in the sequel. However, each of the
following corollaries may be alternatively established by using partial fraction
decompositions directly and not employing Entry 18. The following entries
are valid except for obvious singularities which we shall not state.
Corollary 1. Let в and q> be real with \в\, \q>\ < n. Then for n, x, and у complex,
with x/y not purely imaginary,
2 2 cos(9nx) cosh(cpny) 2 » (—l)kk cos(k<p) cosh(kdx/y)
sin(Trnx) sinh(Ttny) (k2 + n2y2)
- 2nn2xy
Proof. For \в\ < n (Knopp [1, p. 377]),
(k2 + n2y2) sir
(- If к cos(kd) cosh(kq>y/x)
(к2 ~ n2x2) sinhGrfcj;/x) '
nnxcos{dnx) 22V (~~
sinGinx) kJ^oo Цпх - к)
кфО
Similarly, for | q> \ < n,
nny cosh((pny) inny cos(iq>ny) 2 2 v (~^fcos0{(p)
y) sin(inny) k^x, k(iny — k)
MO

14. Infinite Series 269
Define the functions cp, f, ф, and g by
and
nnxcos(dnx)
<P(x) = —; г 1 = f(x) - 1
sm(nnx)
nny cosh(cpny)
Ф У = г-гт : 1 = g(y) -
smh(nny)
Thus, in the notation of Entry 18, Pk = n2x2{-\f cos(fc0), pk = -k2, ak =
— kn, Qk = — n2y2{— If cos(kcp), qk = —k2, and bk = —ikn. Applying Entry
18, we find that, for |0|, \cp\ < я and y/x not purely imaginary,
ср(х)ф(у) = — ^-г-гт г f(x) - g(y) + 1
sin(^nx) sinh(^ny)
2 S (-lfkcos{ke)cosh(kcpy/x)
= -f(x) + 1 - g(y) + 1 + nn2xy X — 7
-nn2xy Y,
(кпх — к2) s\nh(nky/x)
(- If к cos(kcp) cosh(kdx/y)
(kiny — к2) sinh(nkx/y)
which yields the desired result after some simplification. ?
Corollary 2. Let 0 and cp be real with \в\,\<р\ < я/2. Let n, x, and у be complex
with y/x not purely imaginary. Then
я sin@nx) sinh((pny) 2 ™ #(fc) sin(kcp) smh(kQx/y)
An2 cos(nnx/2) cosh(^ny/2) y & k(k2 + n2y2) coshGib/By))
sinhjkcpy/x)
M k(k2 - n2x2)cosh(nky/Bx))' v ' ;
Proof. The set of functions sin{Bfe + 1H}, 0 <, к < со, is orthogonal and
complete on [ — я/2, я/2]. An elementary calculation gives the Fourier series
of sin@nx) with respect to this orthogonal set. Accordingly, we find that, for
|0| < n/2,
sin@nx) 2 « (-l)*+1 sin{Bfc + 1H}
cp(x) != = — X •
x cos(nnx/2) nx k^-co nx + 2/c + 1
Similarly, for \<p\ < n/2,
smh(cpny) sm(i(pny) 2i ™ (—If sin{Bk + I)cp}
'~ у cosh(nny/2) ~ iy cos(inny/2) ~ ny k=_aD iny + 2k + 1
Apply Entry 18 to cp(x) and \j/(y) as defined above. Then Pk = B/(ях)) х
(- l)k+1 sin{Bfc + 1H}, pk = 2k+ 1, ak= -n, Qk = Bi/(ny))(- if x
sin{Bfe + l)cp}, qk = 2k + 1, and bk = —in. A straightforward application of
Entry 18 yields A8.2) for |0|, |<p| < я/2. By continuity, A8.2) holds for |0|,
\(p\ < n/2. ?

270 14. Infinite Series
Corollary 3. Let в and cp be real with \в\,\ср\ < тс/2. Let n, x, and у be complex
with y/x not purely imaginary. Then
n cosFnx) smh(cpny)
4 sinGtnx/2) coshGtny/2)
_cpy 2 2 " (- l)k+1 sin{Bfe + l)cp} cosh{Bfc + 1)вх/у}
~Ъс+ПУ h Bk + l){Bk + IJ + n2y2} sinh{Bfc + \)nxl{2y)}
* (-It+1ca*Bke)sinhBk<py/x)
h 2k{BkJ - n2x2} cosh(nky/x)' { '
Proof. We first calculate the Fourier series of cos@nx) with respect to the
complete orthogonal set cosBfc0), 0 < к < oo, on [ — я/2, я/2]. Accordingly,
we find that
cos@nx) 2 « (-I)*cosBfc0)
x sin(nnx/2) nx k^aj nx + 2k
Define cp{x) = cos@nx)/(x sin(?cnx/2)) — #(x), where g(x) = 2/{nnx2). Thus, in
the notation of Entry 18, Pk = B/(ях))(- 1)* cosBfe0), pk = 2k, and ak = -n,
where к # 0. Let i/^(y) be as in the previous corollary. Thus, by Entry 18, for
|0|, | (p| < n/2 and y/x not purely imaginary,
xy sin^nx/2) coshGtny/2)
n « (-
2k) cosh(nkylx)
2in « (-l)t+1 sin{B/c + l)y} cosh{B/c + l)fa/y}
тех t^ao BЛ + l)(iny + 2k + 1) sinh{Bfc + 1)ях/Bу)}' l ' '
where
Ainy « (-l)*sin{Bfc+l)<j?}
n2x2 ц-^с (inj; + 2fe + l)Bfc + IJ
4iny « (-l)*sin{Bfc+l)y}[ 1 J_
' "' ' fe + IJ "*" n2v
^- A8.5)
In this last step, we have used the Fourier series of cp with respect to the
complete orthogonal set sin{Bfe + l)<p}, 0 < к < oo, on [-тс/2, я/2]. If we
substitute A8.5) into A8.4), we obtain A8.3) for |0|, \cp\ < я/2, after some
simplification. By continuity, A8.3) is valid for |0|, \cp\ <, я/2. ?

14. Infinite Series 271
Entry 19(i). We have
n2xy cot(Toc) coth(?iy)
ncoAinnx/y) « n coth(nny/x)
1^т?—2лхуЪ п>-*> ¦ A9Л)
We have stated Entry 19(i) with no hypotheses because, in general, the two
series on the right side of A9.1) do not converge. Ramanujan evidently used
Entry 18 to derive Entry 19(i), and so we formally derive Entry 19(i) in this
way. From A.9), we have
CO J
nx cot{nx) = 1 + x2 У f A9.2)
„ = -oo ПХ - П2
пфО
and
CO j
пу coth{ny) = 1 + y2 X
X 2•
„=-«, n* - my
Apply Entry 18 to q>(x) = nx cot(?cx) — 1 and ij/(y) = ny coth(Tty) — 1. Ig-
Ignoring the fact that the resulting two series on the right side of A8.1) diverge,
we arrive at A9.1) quite easily.
R. Sitaramachandrarao [1], [2] has found a corrected version of Entry
19(i), namely,
n2xy cot(Ttx) coth(Hj;) = 1 + — (y2 - x2) - 2nxy3
(y x) 2nxy ? J+
„tl n(n2 - x2)
We give Sitaramachandrarao's proof. From A.9),
n2xy cot(^x) coth(^y)
— n J \ m=i у + m
422V 1 / X2
ХУ
,f=1 m2x2 + n2y2 [x2 -n2 y2+m2
„^ \x2 -n2 y2 + n2
x2 (n coth(^ny/x) 1
2ny/x 2n2y2/x
S у2 Ы cothfomx/y) 1 \
m=i y2 + m2 \ 2mx/y 2m2x2/y2)

272 14. Infinite Series
00 / x2 X4 V2 V4 \
= l + 2 „5 (x1^2 ~ n2(x2 - n2) + ~y~2~V^ + n\y2 + n2))
27схз coth(^/x) _
' ti n(x2 - n2) J
ti n(x2 - n2) J „tl n(y2 + n2)
тс2 « cothGmx/y) » соЩппу/х)
= 1 H (y - x2) - Inxyr У r r-— 2пхъу > ——5 ^—,
3 ^ ^ ,4i n(n2 + y2) y h n(n2 - x2)
which completes the proof of A9.3).
Entry 19(ii). Let x and у be complex numbers such that x/y is not purely
imaginary. Then
n2xy csc(tcx) csch(rcy)
(-l)"ncschGmy/x)
(l)ncsch(^«x/y)
= 1 + 2nxy X -г—-2 2nxy X22
„=i n + у „=i n — x
Proof. From Whittaker and Watson's text [1, p. 136],
(—1)"
„=-«, ПХ - П2
cp(x) := тех csc(^x) — 1 = x2
and
ж (— 1)"
i/^(y) := ny csch(Tiy) - 1 = y2 ^ —
n^ao n* - inx
Apply Entry 18 with cp(x) and ij/(y) as defined above. Thus, Р„ = (— l)"x2,
pn = -n2, an= -n,Qn = (- l)"y2, ?„ = n2, and bn = in. Hence,
« (-1) \ninx fninx\
+ y > —: < CSC I —
„i^oo n — iny l у \ у J
The completion of the proof is straightforward, and we omit it. ?
Entry 19(iii). Let x and у be complex numbers such that y/x is not purely
imaginary. Then
j tan(?cx/2) tanh(?cy/2)
2 S tanh{Bn+ l)nx/{2y)} 2 « tanh{Bn + l)ny/Bx)}
~У ntt)B« + l){B« + lJ + y2} +X ЛBп+1){Bи + 1J-х2}'

14. Infinite Series 273
Proof. From Gradshteyn and Ryzhik's tables [1, p. 36],
1 2 °° 1
cp(x) := - tan(roc/2) = ?' - —
X ПХ n^-oo 2П + 1 + X
and
1 2i °° 1
i/^(y) := - tanhGiy/2) = — V' ,
У лу n=-<*> 2n + 1 + ty
where the prime on the summation sign on each right side above indicates
that the sum is to be interpreted as limjy-^^JL-jy. Apply Entry 18 to cp(x)
and ф(у) as defined above. Thus, Pn = —2/{nx), pn = 2n + l,an= —\,Qn —
2i/(ny), qn = 2n + 1, and bn — —i. Hence,
„^ Bn + l)Bn + 1 + x)
2i_ » tanh{Bn + l)nx/By)}
+ nx „ha, Bn + l)Bn + 1 + iy) '
and, after a little simplification, the desired result follows. ?
Entry 19(iv). Let x and у be complex numbers such that y/x is not purely
imaginary. Then
~ sec(nx/2) sechGty/2)
» x(n)n sech{7inx/By)} « x(n)n sech{nny/Bx)}
= к п^Т? + к п^х~2 •
Proof. From A.2),
2 °° (-1V
Л n = -oo 2.П + I + X
and
|JS±iL. A9.4)
Apply Entry 18 with cp and ф defined as above, and we readily obtaia the
desired result. ?
Entry 19(v). Let x and у be complex numbers such that y/x is not purely
imaginary. Then
— cot (nx/2) sechGiy/2)
_ 1 " ^(n)coth{^nx/By)} « sechGtny/x)
ZX „=i И + у „=1 (Zn) — X

274 14. Infinite Series
Proof. From A9.2),
q>(x) := cot(roc/2) = ^- ?
nx in
nx in „i^Qo nx/2 — n
Apply Entry 18 to cp(x) given above and to ф(у) given by A9.4). Hence,
пфО
2 « (-1)" ( JniBn
X 1cot
я л±±т In + 1 + iy { V 2У J Bn
Ах ?, sechGtny/x) \
Я „ = ! BпJ — X2 П.
Ay « (-l)"coth{Bn
Bn + IJ + y2
п2х „^ 2п + 1 + Су Г 2п
The last series above reduces to twice Gregory's series for я/4. Hence, after a
little simplification, the formula above reduces to the desired result. ?
After Entry 19(v), Ramanujan remarks that similar formulas can be derived
for tan(nx/2) cothGty/2) and sec(nx/2]
Entry 20(i). We have
" n coth(rcn)
я2^2 cot(rcz) coth(nz) = 1 —
L-. „4 .4
„=1 П — Z
Note that if we set x = у = z in A9.1), we obtain the equality above.
However, as previously observed, the two series on the right side of A9.1) do
not converge for x = y. A correct proof of Entry 20(i) is obtained from setting
x = у = z in A9.3).
Corollary. We have
cosGrz,/2) . « n соЛ(яп)
^ = 1 + 4nz* У 2 J~-
n* + zA
cosh(nZy/2) - cos(
Proof. In Entry 20(i) replace z by eKl/*z. We see that we must calculate
) соЩпе-^z) = coshMl-O/^coshMl + Q/^)
sinh{nz(l - i)/V2) зтЬ(я2A + OA/2)
The desired equality now follows. ?

14. Infinite Series 275
Entry 20(ii). We have
22 ,ч ut \ 1 л * V (-l)""csch(nn)
ttz csc(nz) csch(nz) = 1 — 4nz ^ 4 4 .
п = 1 П — Z
Proof. Let x = у = z in Entry 19(ii), and the result follows. D
Corollary. We have
2n2z2
cosh(rcz,/2) - cosinzy/l) »=i и4 +
Proof. In Entry 20(ii) replace z be e"'/4z. Use part of the calculation in the
proof of the corollary of Entry 20(i), and the desired result easily follows. ?
Entry 20 (iii). We have
71 ¦ / nw u/ /ox ^ Bn + 1) tanh{Bn
tanGtZ/2) tanhGtZ/2) = 5
8? „5o^MJ^T
Proof. Put x = у = z in Entry 19(iii), and the result readily follows. ?
Corollary. We have
n coshGtz/4/2) - cos(tcz/4/2) _ » B« + 1) tanh{Bn + 1)тг/2}
8z2 Cosh(nz/y2) + cos{nzj[/l) -=o Bn + 1)* + z*
Proof. Replace z by e^z in Entry 20(iii). The calculation that is needed is
precisely of the same type as that given in the proof of the corollary of Entry
20(i). ?
Entry 20 (iv). We have
n , ,„ч w ,* ? , ч  sechGtn/2)
- secGtz/2) sechGrz/2) = ^ l(n) 4 J '¦
о п = \ П — Z
Proof. Let x = у = z in Entry 19(iv), and the result follows forthwith. ?
Corollary. We have
n/4 « n3 sech(^n/2)
i
coshGcz/4/2) + cos
Proof. The corollary follows from Entry 20(iv) upon the replacement of z by
e"il4z and from the calculation in the proof of Entry 20(i). ?
Entry 21 (i). Let a, fi > 0 with afi = л2, and let n be any nonzero integer. Then

276 14. Infinite Series
«-JKCn + D + ^jL^J
1L-2H-1
"+1 R R
V С 14* D2k D2n+2-lk
-22" X (-1
where Bj denotes the jth Bernoulli number.
Entry 21 (i) is perhaps the most well-known result in Chapter 14. For
<x = /? = л and n odd and positive, the theorem is first due to Lerch [1]. A
proof of the more general Entry 21 (i) was first given by Malurkar [1]. Other
proofs of the aforementioned special case or of the full result have been given
by Grosswald [1], [2], Smart [1], Katayama [1], [4], Riesel [1], S. N. Rao
[1], N. Zhang [1] (see also the paper of N. Zhang and S. Zhang [1]),
Sitaramachandrarao [2], and the author [5], [6]. Several other authors have
established transformation formulas from which Entry 21(i) readily follows.
Thus, although Entry 21(i) was not explicitly stated by them, Guinand [1],
[2], Apostol [1], Mikolas [1], Iseki [1], Chandrasekharan and Narasimhan
[1], Glaeske [1], [2], Bodendiek [1], and Bodendiek and Halbritter [1] have
essentially proved Entry 21 (i). For a more detailed discussion of this formula,
see the author's expository paper [1]. Lastly, note that for n < — 1, Entry 21 (i)
yields Entry 13 (with n replaced by -n).
Many generalizations of Ramanujan's formula for ?Bn + 1) have been
given. First, analogues have been established for L-functions by Berndt [4],
Katayama [2], [3], and Toyoizumi [2], [3]. A special case is Entry 21(iii)
below. Other generalizations have been found by Katayama [3], [4], Goldstein
and Razar [1], and Nagasaka [1]. Some related formulas have been derived
by Terras [1].
Matsuoka instigated a series of papers by himself and Toyoizumi in a
different direction. Each [1], [1] first established formulas for C(s) at half-
integral arguments. Matsuoka [2] generalized his result for rational argu-
arguments. Toyoizumi [2], [3], [4] found some analogous results for L-functions
and Dedekind zeta functions attached to imaginary quadratic fields.
Interesting applications of Entry 21(i) and some of its corollaries have been
made by P. Kirschenhofer and H. Prodinger [1] to the analysis of special data
structures and algorithms.
Entry 21 (ii). Let a, ft > 0 with ocfi = n2/4. Let n be any integer. Then
_
sech(<x/c) « sech(^fe)
+(-/>) X X(fc)
l
k = l K- k = l
71 ^-n . . . d 'уъ S-j "уп -уъ
where Ej denotes the jth Euler number.

14. Infinite Series 277
Note that the latter equality in Entry 15 is the case n = 0 of Entry 21 (ii).
Also observe that Entry 21 (ii) reduces to Entry 14 when n < 0. (The parame-
parameters n, a, and /? must be replaced by — n, a/2, and /?/2, respectively, to obtain
Entry 14.)
Proofs of Entry 21(ii) have been given first by Malurkar [1] and then by
Nanjundiah [1] and the author [6, Proposition 4.5].
For Re s > 0, let
L(s) = t x(ri)n-\ B1.1)
n = l
Note Ihat L(s) is the Dirichlet L-function associated with the primitive charac-
character x and so can be analytically continued to an entire function.
Entry 21(iii). Let a, fi > 0 with afi = n2, and let n be any integer. Then
2
2n+1
i 1 V (~
B2n-
2k
2n-2k n-kOk+m
Bk)lBn-2k)\
The first published proof of Entry 21(iii) was given by Chowla [1, Eq. A.2)].
The author [6, Eq. C.20)] has also given a proof. (Unfortunately, formula
C.20) contains an error; replace (/?/8)к by pk+v22~Ak at the end of C.20).) Entry
21 (hi) also follows from results of Katayama [2], [3].
Entry 22(i). Let x and у be complex numbers with y/x not purely imaginary.
Then
2 c
= 2 + Anxf i nC°lhGrn:/y) + *nx>y I ^^. B2Л)
„=i n + у „=i n + x
Proof. Let
zf(z) = n2 cot(Ttzx) coth(^zy) and zg(z) = л2 cot(^zy) coth(^zx).
If we expand f(z) and g(z) into partial fractions, we obtain
xy{f(z) + g(z)} = -3 + 4rrx3yz ? -
¦& f2*-n*

278 14. Infinite Series
If z = 1, the equality above becomes
n2xy{cot(nx) coth(Tiy) + cot(ny) coth(nx)}
=2 - Anjy t nco;hGinf} - 4nXf t nc°^™xjy\. B2.2)
„=i n — x „=i n — у
Replace x by ел'/4х and у by eml*y in the formula above. The right side of
B2.2) then becomes the right side of B2.1). On the left side of B2.2) we have
2 fcosh(a — id) cosh(fe + ib) cosh(fc — ib) cosh(a + id)\
\ sinh(a — id) sinh(b + ib) sinh(b — ib) sinh(a + id) J
= 7i2xy{F(a, b) + F(b, a)}
G(a, b) ( ¦ '
where a = nx/yjl, b = ny/y/2,
F{a, b) = cosh(a — id) sinh(a + id) cosh(b + ib) sinh(fr — ib),
and
G(a, b) = sinh(a — id) sinh(a + id) sinh(b — ib) sinh(b + ib).
Now,
F(a, b) = i{sinhBa) + i sinBa)} {sinhBb) - i sinBb)},
and so
F(a, b) + F(b, a) = i{sinhBa) sinhBb) + sinBa) sinBb)}
= i(cosh{2(a + b)} - cosh{2(a - b)}
+ cos{2(a - b)} - cos{2(a + b)}). B2.4)
Also,
G(a, b) = |{coshBa) - cosBa)} {coshBb) - cosBb)}. B2.5)
If we substitute B2.4) and B2.5) into B2.3), we find that B2.3) is transformed
into the left side of B2.1). This completes the proof. ?
Entry 22(i) in the second notebook is slightly in error. Ramanujan has
replaced the numerator on the left side of B2.1) by
cosh{7t(x + y)y/2} cos{ti(x -
- cosh{7t(x - y)y/2) cos{7i(x
It also may be remarked that formally B2.2) can be derived from Entry 19(i).
Entry 22(ii). Let n > 0. Then
о coshft^/x) + cos(tc4/x) t=i cosh(nfc/2)

14. Infinite Series 279
Proof. Let
coshGcz) + cos(nz)'
We expand / into its partial fraction decomposition. There are simple poles
at z == Bk + 1)(± 1 + 0/2, - oo < к < oo. Since
R(Bk + 1)A + 0/2) = -:
27icosh{B/c
and
we readily find that
j <x> t_ ukBk + IK
/(Z) = ?
/() n t?0 c
where #(z) is entire. By the same argument as that used in the proof of Entry
4, g(z) = 0.
Letting z = 4/x, we multiply both sides of B2.7) by cosBnx) and integrate
with respect to x over [0, oo). Inverting the order of integration and summa-
summation by absolute convergence and using a result from Ramanujan's quarterly
reports (Part I [9, p. 322]),
00
cos(ax) dx к .
we find that
1 " /(fe)fc3 f °° cosBnx) dx
/(x) cosBnx) dx = - У —^ з—
О я fc = i СОлП^яК/Z^ Jo X -г К /*+
which completes the proof of B2.6). ?
Ramanujan claimed that the next entry is a corollary of Entry 22(ii). We
cannot show this and so proceed from scratch.
Corollary. Let a, /3 > 0 with а/З = жъ/А. Then
/3n2) 8' У '
Proof. Let N be an even positive integer. We shall let N tend to oo, but we
shall further restrict N by requiring that N2 remain at a bounded distance

280 14. Infinite Series
from the numbers Bn + 1)<х/я2, where n is a positive integer. Let
™ ~ z{coshGtNz) + cos(ttJVz)} cosB/?JV2z2) '
Elementary considerations show that /(z) has simple poles at z = 0, at z =
Bn + 1)(± 1 + i)/BN), where n is an integer, and at z = ±J{2k + l)a/{Nn),
where к is an integer. Straightforward calculations yield R@) = \,
R(Bn+l)(±l+i)/BA0) =
nBn + 1) cosh{Bn + 1)tc/2} cosh{Bn
and
R(±y/Bk
l){cosh УB^ + l)a + cos
Let С denote the positively oriented rhombus with vertices + 1 and + i. Hence,
employing the residue theorem and letting N tend to oo, we find that
lim —- fN(z) dz
2 л nJ~Lx Bn + 1) cosh{Bn + 1)я/2} cosh{Bn
2 ж ( — l)k+1
+ - E , ( > . B2.9)
я *=-oo Bk + l){cosh УBА: + l)a + cos ^BА + l)}
By the definition of fN and the choice of N, it is easily seen that the limit on
the left side of B2.9) is zero. A slight rearrangement of B2.9) yields B2.8), and
we are done. ?
Entry 22(iii). Let a, ft > 0 with afi = 4rc3, and let у denote Euler's constant.
Then
la 1 ? cos x/an
720 2 „=1 n(COsh yfim — cos ,/an)
_ у + LogB7r/,g) l 0 _ «> 1 l ^ coth(Trn)
4
In the notebooks, formula B2.11) contains a misprint; Log Г(|) is replaced
by Hog Щ).

14. Infinite Series 281
Proof. We first prove B2.11). A direct calculation gives
OO 1 00
У —= = -Log П A - Ч2"), B2.12)
„=! n(eiirn - 1) „=i
where q = e~n. Now from Whittaker and Watson's text [1, p. 488, problem
10],
where к, к', and К have their standard meanings in the theory of elliptic
functions. Here, к = к' = l/,/2 and К = я3/2/BГ2(|)). (See Zucker's paper
[1], for example.) Thus, B2.12) and B2.13) yield
12
= i ЬоёD/я) + Log Г(|) - тс/12,
as desired.
We now prove B2.10). Let N = n + %, where n is a positive integer. We
shall let JV tend to oo through a sequence such that N2n2/a remains at a
bounded distance away from the positive integers. Let
cot(nNz)
zie''2 - 1)
The function fN(z) has simple poles at z = ±^/ak{l + i)/BnN), at z = ik/N,
and at z = ik/N, where к is a nonzero integer. In addition, fN(z) has a quintuple
pole at z = 0. Using elementary trigonometric identities, we find, after some
calculation, that
Я(±^/ак{1 + i)/BnN)) = Я(±^/-оскA + i)/{2nN))
1 f 2cosV^ ., B214)
icosh y^ — cos
Easier calculations yield
coth{nk)
and
Observe that
^*^ + 5!^*. ,21,5,

282 14. Infinite Series
To calculate the residue at z = 0, write
If 1 nNz {nNzK ) f 1 nNz (nNzK
\2
1 1 1
After some simplification, we find that
/? 7a
Let С denote the positively oriented rhombus with vertices +1 and ± i. By
our choice of N, there are no poles of fN on С Applying the residue theorem
and employing B2.14)-B2.16), we find that
if,,, 2 „ cos ^/ak
c n i <k<n2N2/0L /c(cosh 4/afe — cos 4/a
_i у 1 + 1 у
B2.,7)
Next, we calculate directly the integral on the left side of B2.17). Let C,-
denote that part of С in the yth quadrant, 1 < j < 4. On Q set z = 1 — x + ix,
0 < x < 1, and on C3 set z = x — 1 — ix, 0 < x < 1. Then in either case,
{° ?<X<f B2.18)
On C2 set z = — x + A — x)i, 0 < x < 1, and on C4 set z = x + (x — l)i,
0 < x < 1. Then in either case,
{f/J ?<Х< B2.19)
By the choice of N, the convergence in B2.18) and B2.19) is bounded on С as
JV tends to oo. Hence, by the bounded convergence theorem,
If Iff1 f ( + 0/2 Г-i fd-0/2-)^
lU U(l+0/2 Ji J (-1-0/2 J-i )Z
= - Log 2. B2.20)
ж
Returning to B2.17), we examine
v coth(?c/c)
2
= 4 E ,, 2j n+2 ? 7- E ^ B2.21)

14. Infinite Series 283
Now from Ayoub's text [1, p. 43],
2 X z- X I
l^k<N к l^k<n2N2l<tk
= 2{Log N + у + 0A/N)} - {Log(n2N2/oc) + y + O(l/N2)}
= у - 2 Log n + Log a + O(l/N). B2.22)
Thus, letting N tend to oo in B2.17), using B2.20)-B2.22), and multiplying
both sides by л, we deduce that
Log 2 = -:
k(cosh ^/ofc - cos
which is equivalent to B2.10) after some elementary manipulation. ?
Entry 23(i). We have
Z^i + V V k coth(nk)(- l)"(fexL>Dn) = -^-{i<p(—2) + fc}, B3.1)
4л t=i „=o ix-4
where, the error h is nearly equal to
„=o x2n+lBn + 1)!
i/ x is small. (It is not clear whether the entry reads q>B) or <p( — 2) on the right
side of B3.1).)
It is not clear what interpretation should be given to Entry 23 (i). It is
surprising that a power series in x is to be approximated near x = 0 by a power
series in 1/x. Perhaps B3.2) is an asymptotic series for h. It seems quite certain
that Ramanujan derived Entry 23 (i) in a purely formal manner. We shall show
that perhaps Ramanujan made a mistake, because a formal argument seems
to produce a slightly different formula. For most of the discussion which
follows, we are very grateful to D. Zagier.
For each integer n, set cp(n) = ф(п + 2). Define
f(x\ _ у t_ iyv/f4n -i_ 2b-4" i2"\ 31
л=0
(Ramanujan seems to tacitly assume that F is entire.) Thus, Entry 23(i) may
be rewritten in the form
- У к coth(nk)F(kx) = ~ {\ф@) + h}, B3.4)
tvi fc=i 2x
when;

284 14. Infinite Series
л odd
In his theory of integral transforms, discussed by Hardy [9, pp. 188-193],
[4], [8, pp. 280-289] and the author [9], Ramanujan often writes
It is quite clear that Ramanujan is not assuming that G is an entire function;
he is simply indicating the form of the Taylor series of G about x = 0. Likewise,
in the setting at hand, Ramanujan is undoubtedly assuming that F has the
expansion given by B3.3), only for x sufficiently small.
As an example, let i/*(s) = k'\ where к > 0. Then F(x) = k'2{\ + x*/k*)~l.
Letting f(x) denote the left side of B3.4), we deduce that
J-2
The sum on the right side may be evaluated by letting z = e"mX/x in Entry
20(i). Temporarily putting и = nk/x, we then find that
Thus, in the notation B3.4), as x tends to 0,
2h = coshb/ги) + cos^u)
cosh(,/2u) - cos(y/2u)
_
cosh(,/2u)
0
= 2 у
n=o
= 4 «
n^o Dn)!

14. Infinite Series 285
Thus,
Dn + 3)! + \4* Dn)! '
B3.6)
According to B3.5), Ramanujan claims that
« cosC7in/4)Bu)"
n odd
_ _L v (-trc2"L"'1 . _L у (-i)"B»Ln+3 m7
Пк Dи+1)! \/2il Dn + 3)! ' l j
A comparison of B3.6) and B3.7) indicates that apparently the error h is twice
what Ramanujan claims. Furthermore, B3.6) contains an extra power series
Dn)! '
Observe that G(t) := e~kt is the inverse Mellin transform of F(s)A~s. Our
calculations above have shown that
iri/4
Because of the close proximity of Entry 23 (i) to Entry 20(i), we conjecture
that Ramanujan probably proceeded as we have above and then more gener-
generally considered those cp having the shape
00
<P(S) = Z Cj^'.
j=o
In regard to Entry 23(i), some series transformations of S. N. Aiyar [1],
published in 1913, might be mentioned. Recall that S. N. Aiyar was the
manager of the Madras Port Trust office when Ramanujan worked there as
a clerk for about 15 months during 1912-1913.
Entry 23(ii). We have
? t 1* l?-h, B3.8)
k=l n=0
where h is very nearly equal to
i/ x is small.
Comments similar to those made after Entry 23 (i) can be made about this
mysterious formula as well. However, as we shall shortly see, if we assume

286 14. Infinite Series
that the double series in B3.8) converges absolutely, then, in fact, B3.8) is
indeed true with h = 0. Of course, we are unable to make this hypothesis about
the double series in B3.1).
Proof. Assume that the double series in B3.8) converges absolutely. Then
inverting the order of summation and employing the corollary of Entry 14
and Entry 15, we find that
? ? Г^(/с)8есЬ(^/2)(-1)"(ЬL>Dп)
k=l n=0
= ? (-I)"x4>Dn) ? k^-'xik) sech(nk/2)
n=0 k=l
which establishes B3.8) with h = 0. ?
We are indebted to D. Zagier for the following very perceptive remarks on
Entry 23(ii).
Let G(t) be analytic at t = 0, and suppose that G(t) = О(ГС) as t tends to
oo for every с > 0. Define, for Re s > 0,
<p(s) = ^г,Г GWt*-1 dt.
T(s) J
Then cp is entire. Also, formally,
п=0 
In view of Ramanujan's work on Mellin transforms in his quarterly reports
(see Part I [9, p. 298]), we have determined the coefficients of G(t) from the
converse of Ramanujan's Master Theorem.
For x sufficiently small, suppose that
F(x)= ? (-l)VDn)x4n.
As in Entry 23(i), on the surface, it appears from B3.8) that Ramanujan is
assuming that F is entire, but this is not the case. With F and G defined above,
B3.8) can be rewritten in the form
" X(k)F(kx) _n
where
/ „ \
B3.10)
\xJ2
as x tends to 0.

14. Infinite Series 287
We now discuss certain cases.
Case 1. Suppose that G(t) is continuous for t > 0 and that G{t) = 0 for t > t0;
that is, G(t) has compact support on [0, t0]. It follows immediately from the
definition of q> that, for all s > 1,
r(s)cp(s)« ts0.
Then
. ^ <лоL" „.^
Hence, the left side of B3.9) converges absolutely as a double series for
xt0 < я/2. By our proof above, h = 0. Now
Thus, G(n/(xy/2)) = 0. Hence, our findings are consistent with Ramanujan's
claim B3.10).
On the other hand, suppose that the left side of B3.9) converges absolutely
as a double series for 0 < x < x0. It follows that
? f \(р{Щ\кАп-1е-~жк12хАп < oo, x < x0.
n=l k=l
Comparing the sum on к with the integral of t*n~le~nl2 over 0 < t < oo, we
deduce that
x < xo-
Hence,
|ГDи)ч»Dи)| < ( -^ + o(
^zx0
as n tends to oo. Moreover, if cp(s) is reasonably smooth,
\Г(з)ф)\ <(j^ + o(
as s tends to oo. By examining the inverse Mellin transform of T(s)(p(s) and
moving the line of integration to the right, we deduce that G(t) = 0 for
t > n/Bx0) =: t0. Again, this is consistent with Ramanujan's claim B3.10),
since
V2
and so G{n/{xy/2)) = 0.

288 14. Infinite Series
Case 2. Suppose that cp(s) = k~s, where X > 0. Then G{t) = e~kt and F(x) =
A + х4/Л4Г'. If f{x) denotes the left side of B3.9), we then find that
= „ AV^ sech(?t/c/2)
J ^ ' ^ h(\ i /,4V4 /14\
x(/c)/c3 sech(rc/c/2)
h к cosh(nfc/2) Й fc4 + Я4/х
Я4/х4
Applying Entry 25(vii) (or Entry 15) and the corollary to Entry 20(iv), and
letting nX/x = u, we deduce that
8 cosh(u/4/2)
Comparing this with B3.9), we see that
т
ft =
7
xJ2
B3.11)
as x tends to 0. Hence, B3.10) is established.
We have therefore shown that Entry 23(ii) is valid when (pis) — A~s.
Observe, from B3.11), that we may write h in the form
у g-njxHx^-Kikmxy/i)
j1
fc=j-l(mod2)
- «-At
where Git) = e я'. This suggests that, for more general functions q> and G,
A-If-D'-1 I c(^±^Y B3.12)
k=j-l(mod2) v
under suitable hypotheses. We now establish such a theorem.
It is clear that we now need to define G(z) in the quadrant Q := {z: |arg z\ <
тс/4}, instead of on just [0, oo). Thus, suppose that G is analytic on Q and that
Giz) = O(z~c) as z tends to oo in Q, for every constant с > 0. Define, as before,
for Re s > 0,
dt.
1 (S) Jo
Hence, if со = expGri/4),

14. Infinite Series 289
f00
= (p@) — \x G(t){a> sin(ctm) + ш 1 sin(co xxt)} dt
Jo
= c/>@) - \сохЩсох) - {ю^хЩсо^х),
by absolute convergence, where
f°°
H(u)= G(t) sin(fu) dt. B3.13)
Jo
Using Entry 25(vii) and the last expression for F(x), we find that
x(k)F(kx)
8 П ' h к coshGr/c/2)
oo vim)
J B3.14)
8 П ' h к coshGr/c/2)
oo vim)
x E J {
Jii 2 coshGrm/2)
Since
1 ?
> 0,
2 cosh у n^i
the right side of B3.14) may be written in the form
x E X(n)\co E х(т)е-«тп12Щсотх) + со'1 ?
w=l (_ m=l m=l
B3.15)
We now assume that the real and imaginary parts of H{coxu)e~nnul2 and
H(co"lxu)e~n'"'12, for each positive integer n, are integrable over @, 3) for some
3,0 < 3 < я/2, are of bounded variation over C, да), and tend to 0 as и tends
to oo. Then by Poisson's summation formula for Fourier sine transforms
(Titchmarsh [2, p. 66]) (see also (9.2) above), the expression within curly
brackets in B3.15) equals
OO ( C°°
У x(m) \ ы e~™"l2H(toxu) sm{nmul2) du
»=i I Jo
/«CO ¦)
+ Ю e"'""'/2H(co-1xu)sin(wmu/2)du>. B3.16)
Jo J
Next, replace u by co~1u and суц, respectively, in the two integrals above.
Assume that H(z) decays sufficiently rapidly in Q so that we may apply
Cauchy's theorem to replace the paths @, су») and @, суоо), respectively,
by @., oo). Collecting together the calculations from B3.14)—B3.16), we deduce
that

290 14. Infinite Series
_ « x(k)F(kx)
l fc coshGr/c/2)
= * ? ? *(ти) I Я(хи){е-ИВ8т(ятйГ1и/2)
n=l m=l Jo
du. B3.17)
Assume that the iterated sum above is equal to its double sum. Since the
coefficient x(mn) is symmetric in m and n, we may interchange the roles of m
and n in the second expression above. If we also employ the identity
2) = sin{n(con + йГхт)и/2},
we find, from B3.9) and B3.17), that
n °° f°°
^"'l=: E Z(m") Я(и) sin{тг(суп + cu 1m)u/Bx)} du,
2 m,n = l Jo
where x > 0. By the Fourier sine inversion of B3.13),
2 f00
G(t) = - Я(«) sin(tu) du.
к Jo
Hence, for x > 0,
ь v
m,n=l
AX
Setting j = (n + m)/2 and k = (n — m)/2, we find that the conditions m, n odd
and positive are transformed into the conditions j, k integral, j > \k\ + 1, and
j = k+l (mod 2). Also, x(m") - (-1)' and ши + ш-1т = (j + ifc^
Thus, for x > 0, we deduce B3.12).
As an example, let
G(z) = ze"cz2,
where с > 0. Then, initially for Re s > — 1,
2T(s) 2
where lastly s is any complex number, by analytic continuation. Also,
H(u) = Г te' sin(tu) it = ^e-/Dc),
Jo 4c3/2
which can be obtained from differentiating formula 3.896, No. 4, p. 480, in
Gradshteyn and Ryzhik's tables [1]. Lastly,

14. Infinite Series 291
F(x) = (p@) - ±
IX y/Tl ix2/Dc) IX V^ ix2/Dc)
" 8c3'2 8c3/2
x2v^ . /x2\
3, sin — 1.
4c3'2 \4c/
4сз/2
Thus, we have shown that, for x > 0,
sin(k2x2/Dc))
coshGtk/2)
k = ;-l(mod2)
The analysis above can be strengthened by beginning with the Fourier sine
transform B3.13), imposing conditions on H(u), and then defining F and G in
terms of H. Furthermore, an analogue of B3.12) undoubtedly holds for Entry
23(i) as well. However, in view of the limited applications that any more
rigorous and/or stronger versions of Entries 23(i), (ii) might have, it seems best
here to end our discussion of these entries.
Entry 24. For z complex,
ne~2nz
2z{coshB7rz) — cosB?tz)}
1
Snz*
1
4z2 +
OO
' „Si (e2m
л °°
4z „4l z2 -
n
- l)Dz4 + n
!
V(ZA
I4)'
hnJ
Proof. Let f(z) denote the left side above. We shall expand / by partial
fractions. The function / has a triple pole at z = 0 and simple poles at z =
±n(l ± i)/2, where n is a positive integer. By division of power series, it is
easily calculated that the principal part of/ about z = 0 is
8ttz3 4z2 4z '
Straightforward calculations show that
K(n(l + 0/2) = г = -«("(I - 0/2).
2in(e — 1)
Replacing n by — n above and manipulating slightly, we find that
K(-n(l + 0/2) = x-t-j1—tt + ^-= -R(-n(l- 0/2).
2ш(е2я" - 1) 2m
Now,

292 14. Infinite Series
1 f 1 1
2in (z + n(\ + J)/2 z + n(l - i)/2J z2 + (z
After much, but routine, simplification, we get
1 f 1 1
B4.2)
2in{e2nn - 1) (z - n(l + i)/2 z - n(l - 0/2
4nz
)'
z + иA + 0/2 z + иA - 0/2J (е2яя -
Using the principal parts in B4.1)-B4.3), we easily deduce the desired result
after employing an argument like that at the end of the proof of Entry 4. ?
Entry 24(i). For complex z we have
Inn
+ Zj
2z2 „t^z2 + n2 2z z(e2nz - 1)'
This result is just a reformulation of A.9).
Entry 24(ii). Let z be complex. Then
1 _ 1 _2 « 1
z(e"z + 1) ~ 2z ~ я „4 z2 + Bn + IJ '
A proof of Entry 24(ii) is easily obtained by expanding the function on the
left side above into partial fractions.
The next entry is complementary to Entry 24.
Entry 25. Let z be complex. Then
Д 1
4r{cosh(rcz) + cos(ttz)} 8z n=o z2 + (z + 2n + IJ
. ? 2n + 1
n% (eBn+1)« + l)Dz4 + Bи + IL)
Proof. Let /(z) denote the left side above. We expand / into partial fractions.
The function / has a simple pole at z = 0, and the principal part about 0 is
easily seen to be n/(Sz). Also, / has simple poles at z = ±{2n + 1)A + 0/2,
where n is a nonnegative integer. Routine calculations give
R(Bn + 1)A + 0/2) =
2Bn
and

14. Infinite Series 293
2Bn + 1)(е<2я+1)* + 1) 2Bn
= -R(-Bn + 1)A - i)/2).
The sum of the principal parts for the four poles +Bn + 1)A + i)/2 is thus
found to be
1 4zBn + 1)
z
2 + (z + 2n + lJ
lJ (е<2я+1>* + 1)Dг4 + Bn + IL)"
The theorem now readily follows. ?
Entries 25(i), (ii). We have
» coth(Ttk) 7tc3
k4 fc3 ~ 180 (
and
« cothfrfc) _ 19тг7
Й A:7 6,700" l '
Both B5.1) and B5.2) are special cases of the more general formula
- coth(nk)
_ ?
~2 h( ]
Bk)\Bn + 2-2k)!'
where n is an odd positive integer and Щ denotes the ;th Bernoulli number.
Ramanujan does not state the general formula B5.3) in his notebooks. How-
However, it does follow quite easily from Entry 21(i). (See our paper [6, p. 155].)
Formula B5.2) was communicated by Ramanujan in one of his letters to
Hardy [16, p. xxvi]. Entry 25 (i), in fact, was long ago established by Cauchy
[1, pp. 320,361]. Cauchy does not state the general formula B5.3), but he does
give a general method for evaluating the series on the left side of B5.3).
Preece [3] has established B5.1) and Sandham [1] has proved B5.2). The first
statement of B5.3) known to the author is by Lerch [1]. Later proofs of
B5.3) have been given by Watson [1], Sandham [2], Smart [1], Sayer [1],
Sitaramachandrarao [2], and the author [6, p. 155], [5].
Entries
and
25(iii),
(iv).
We have
» tanh{Bfe
+ 1)я/2}
4^0 B/c + IK
« tanh{B/c +
1O1/2}
л3
32
7я7
k% Bk + IO 23,040"
Both entries follow from the more general formula

294 14. Infinite Series
tanh{Bfc
tanh{Bfc + 1)я/2} = И
ti Bfc + 1Lп+3 Ъ Л
Bfc + 1Lп+3 Ъ Ло Bk + 1)! Dи + 1 - 2/c)!' l " '
where n is a nonnegative integer and Ej(x) denotes the jth Euler polynomial.
Formula B5.4) cannot be found in the notebooks. The first proof of B5.4) was
given by Phillips [1]. Later proofs have been given by Nanjundiah [1],
Sandham [2], Smart [1], Sayer [1], and the author [7, Corollary 4.10].
Formula B5.4) is, in fact, a special case of a more general formula. Let
a, p > 0 with aP = n2. Then
x tanh{Bfc+l)|
Bfc + lJ"+1 v " k% Bfc+lJ
' Ak%K ' Bk + 1)! (In - 2k - 1)! " '
where n is a positive integer. The first proof of this formula appears to be by
Nanjundiah [1]. The author [7, Corollary 4.9] has also given a proof. The
case n = 1 was established by Grosjean [3]. The case n = 2 was proved by de
Saint-Venant [1] in 1856 and occurs in the determination of the torsional
rigidity of a beam of rectangular cross section. This motivated a problem by
Boersma [1] who asked for an asymptotic expansion of the series on the left
side as a tends to 0. The identity above easily yields such a result.
The author [7, Theorem 4.11] has evaluated
^ tan|Bfe+ 1)-
k?o Bk + lJn+1
for a very general class of real quadratic irrationalities в.
Entries 25(v), (vi). We have
» (-l)*+1cschGrfc) _ я3
й к3 360
and
« (-l)/t+1cschGrk) Ш7
i?i k1 453,600"
Both entries follow from the more general formula
у в2к%
2 W
LQ I 1) Bfc)! D„ + 4 _ 2k)\'
where n is an integer and Bj(x) denotes the ;th Bernoulli polynomial. Formula
B5.5) is essentially due to Cauchy [1, pp. 311, 361] who gave a somewhat less

14. Infinite Series 295
explicit formulation. Otherwise, B5.5) was first established by Mellin [1].
Later proofs have been given by Malurkar [1], Phillips [1], Nanjundiah [1],
Sandham [2], Riesel [1], Sayer [1], and the author [7, Corollary 3.2]. The
general formula B5.5) does not appear in the notebooks.
Formula B5.5) is an immediate consequence of the following more general
result. Let a, ft > 0 with a/? = л2. Then, for any integer n,
= (-P)~" Z (-
fci
which has been proved by Mellin [1], Malurkar [1], Nanjundiah [1], and the
author [7, Theorem 3.1].
Entries 25(vii), (viii), (ix). We have
« sechGrfe/2) n
? Z(ft)
/t=i
B2k(l) g2n+2-2t(l) n+l-k
Bft)! Bn + 2-2k)!a
1
,I4sechG*/2)
[ ' к5
sechGrk/2)
8'
n5
768'
23тг9
and
V Л)«
к9 1,720,320"
All three entries follow from the general formula
Jc=l
Bfc), Dn _
which can be easily deduced from Entry 21(ii). Here n is any integer. Entry
25(vii) is a simple consequence of Entry 15 and was proved by Preece [3].
Zucker [2] has established Entry 25(vii) as well as some related results. Entry
25(vii) was also submitted as a problem to the Mathematical Gazette [1],
where several solutions are indicated and considerable discussion is found.
Entry 25(viii) appeared in one of Ramanujan's letters to Hardy [16, p. xxvi].
In addition to the proofs mentioned after Entry 21(ii), proofs of B5.6) have
been given by Watson [1], Sandham [2], Riesel [1], and Sayer [1].
Entry 25(x). We have
f *W 'у * ^2 Ч1—Лс B57)
k% fc V - 1) 8 h k2 cosh(nk) 96 " 2 Jo x { '

296 14. Infinite Series
Proof. Let
We shall integrate / over the positively oriented rectangle CN whose horizontal
sides pass through + (N + |)i and whose vertical sides pass through ±N,
where N is a positive integer. The function / has a triple pole at z = 0 and
simple poles at z = + Bk + l)/2, where it is a nonnegative integer, and at
z = + ki, where к is a positive integer. Routine calculations yield R@) = 5n/l2,
4(_l)*+i
R(Bk l ;
Bk + \Jn(e{2k+l)n - 1)'
Bk + 1)V
and
R(ki) = --——L-— = R(-ki).
2nk2cosh(nk)
Hence, applying the residue theorem and letting N tend to oo, we find that
0 = lim —
jv-oo 2ni JCjv
f(z)dz
_ 8 » X(k) 4 If 1 5" B58)
" rcilfcV-1) rr^^^^^cosh^ + n' (^'8)
where L(s) is defined by B1.1). A comparison of B5.8) with B5.7) indicates
that it remains to show that
B5.9)
о •*¦
Integrating termwise the Maclaurin expansion
tan-Jx_ » (-l)*x2*
x ~ Дь 2k + 1 '
we readily deduce B5.9), and the proof is complete. ?
Entry 25(xi). We have
1
у
k2 + (k + lJ}(cosh{Bk + 1)я} - cosh n)
—r^— |- + coth7r-Jtanh2Gr/2)t. B5.10)
2 sinh n \n 2

14. Infinite Series 297
Entry 25 (xi) is in error in the notebooks, for Ramanujan has written
sinh{Bk + 1)я} — sinh n instead of cosh{Bk + \)n} — cosh л on the left side
of B5.10). Ramanujan communicated B5.10), with the same error, in one of
his letters to Hardy [16, p. 349]. Watson [1] established B5.10) by contour
integration. Because Watson's proof contains a few errors, we briefly sketch
another proof by contour integration below. The calculations in both proofs
are extremely laborious. Sitaramachandrarao [1] has found another proof
based on Entry 20(i). Since his proof is very elegant and is likely the one which
Ramanujan had, we shall give this proof as well.
First Proof. Let
n sinh n
z{coshGrz) + cosh n) {cosGrz) + cosh n}'
which has a simple pole at z = 0 and poles at z == iBk + 1) + 1, if к is an
integer, and at z = 2n + 1 ± i, if n is an integer. These poles are simple except
when к = 0, — 1 and n = 0, — 1 when the two sets coalesce to give double
poles. Very lengthy calculations yield
Я@) =
R(iBk + !)+!) =
sinh n
±1) = {iBk+ l)±i}(ooSh{Bk+l)n}-cosh пУ
RBn + 1 ± i) = Bn + j ± -)(ch{-^ + 1W h ), и/0,-1,
and
coth n 1
±0= -
2 sinh n 2% sinh n
Integrate / over a square with vertical and horizontal sides passing through
±2N and ±2Ni, respectively, where N is an integer. Apply the residue
theorem and let JV tend to oo to deduce B5.10). ?
Second Proof. Let S denote the left side of B5.10). Since an elementary
calculation shows that
coth{kn) - coth{(/c + 1)я} 1
2 sinh n cosh{Bk + 1)я} — cosh n
we readily deduce that
1 " coth(k7r) - coth{(/c + 1)я}
~ 2 sinh n fc[ k2 + (k + if '
Transforming the right side by partial summation, we deduce that

298 14. Infinite Series
2(slnh „)S .^1 _ ?
coth я « к coth(Trk)
4 h
fc coth(Trk)
Setting z = A + i')/2 in Entry 20(i), we easily find that
« fe coth (як) 1
Using this in the foregoing equality, we complete the second proof of Entry
25(xi). ?
Entry 25(xii). We have
~ 5я/2). B5.11)
fco {25 + Bk + lL/100} (e<2k+1>* + 1) 11,890 8
This entry again was communicated by Ramanujan in one of his letters to
Hardy [16, p. 349]. The right side of B5.11), however, had the wrong sign on
both terms. This error is also made in the notebooks. Furthermore, the left
side of B5.11) is replaced by only the first three terms of the series in the
notebooks, and the second term contains another misprint. It may be of
interest to determine how well the first three terms on the left side of B5.11)
approximate the right side. We note that
- = 0.001665694154- ••,
25.01(e" + 1) 25.81(е3я + 1) 31.25(es
while on the other hand,
- I coth2E7r/2) = 0.001665694195 • • ¦.
о
Watson [1] has given a proof of B5.11) by contour integration. It will be
shown below that Entry 25(xii) is a corollary of Entry 25; hence, this is
probably the method used by Ramanujan to establish B5.11).
Proof. In Entry 25 put z = 5i. After some simplification and rearrangement,
we find that
S2k + 1» 1
L l L Bfe 1J
Lq {?{2к+1)п + Щ25 + BJfe + tf/щ L
B5.12)

14. Infinite Series 299
A comparison of B5.12) with B5.11) indicates that it remains to show that
1 4689
5- у l чооу B5 13)
,?<, Bk + IJ + 10Bk + \)i - 50 ' 11,890' l
or equivalently that
Д Bk + IL + 2500 " 11,890' (
since B5.12) obviously implies that the imaginary part of the left side of B5.13)
is zero. To show B5.14), write
k% Bk + IL + 2500
2 &, {Bk + IJ - 10Bk + 1) + 50 Bk + IJ + 10Bk + 1) + 50
1
2 [fa Bk + IJ - 10Bk + 1) + 50
1
-i
e5 Bk + 1 - 10J + 10Bk + l - 10) + 50)
-5y
2 k% Bk
4689
+
lJ-
1
10Bk H
-1)H
h 50
11,890'
and the proof of B5.14), and hence B5.11), is complete. ?
Infinite series involving the hyperbolic functions have attracted the atten-
attention of many authors. Our papers [6], [7] contain many such results as well
as numerous references. Readers may also wish to consult papers by Cauchy
[1], Zucker [1], [2], Ling [1], [2], [3], Forrester [1], and Bruckman [1] for
additional results not examined in the aforementioned papers. The papers of
Berndt [8] and Klusch [1] offer some hyperbolic series of different types.

CHAPTER 15
Asymptotic Expansions and
Modular Forms
The title of Chapter 15 does not entirely reflect its contents, because this
chapter contains several diverse topics. Of the 21 chapters in the second
notebook, Chapter 15 contains more disparate topics than the remaining
chapters. Ramanujan appears to have collected here several "odds and ends."
While much of the material is fascinating, a few parts have little substance.
The first seven sections are devoted primarily to asymptotic expansions of
series. For example, Ramanujan derives asymptotic series, as x tends to 0 +,
for
oo qo X Jcm~^
2>-Ч-1, I,-Log к, and lJVT-r^.
Frequently, theorems about such series are established by us in greater gener-
generality than indicated by Ramanujan. These generalizations not explicitly stated
by Ramanujan are labeled as "Theorems" in the sequel, in contrast to our
usual designations by "Entries" in relating Ramanujan's results. This has
necessitated some reordering of Ramanujan's findings in our description of
Sections 2-7 below.
Ramanujan's discourse in Section 1 seems to indicate that he used the
Euler-Maclaurin summation formula to establish his asymptotic expansions.
However, his comments are so cryptic and obscure that we have been un-
unable to find a proper interpretation for them. At any rate, it appears that
Ramanujan's use of the Euler-Maclaurin formula was formal and non-
nonrigorous. Despite the nature of his methods, Ramanujan's results are correct,
except for some minor errors. Our original proofs were likewise based on the
Euler-Maclaurin summation formula. These proofs were rather lengthy and
involved. We are extremely grateful to P. Flajolet for suggesting that con-
considerably shorter proofs could be achieved via the use of Mellin transforms.

15. Asymptotic Expansions and Modular Forms 301
In the second half of Chapter 15, modular forms, in particular, Eisenstein
series, are at center stage. However, as our summary below indicates, there
are many themes.
One of the most interesting theorems in Chapter 15 is found in (8.3) below.
This undoubtedly new result gives an inversion formula for a certain modified
theta-function. It may be surprising that an exact formula of this type exists.
Entry 11 is a beautiful and new reciprocity formula reminiscent of some of
the formulas in Chapter 14.
Section 12 contains several results found in Ramanujan's famous paper
[11], [16, pp. 136-162]. We mention, in particular, Entry 12(x) which is
equivalent to the very interesting identity
? o&k + 1)о3(п -к) = ~°5Bn +1), n > 0,
where av(m) = Yjd\mdv, m Ф 0, and (T3@) = jko- Ramanujan states this identity
without proof in [11], [16, p. 146] and indicates that he has two proofs, one
of which is elementary. We have not been able to find an elementary proof in
the literature nor can we produce one ourselves. All the results in Section 13
can also be found in [11].
Entry 14 offers a new recursion formula for Eisenstein series. It is quite
distinct from the most well-known recursion formula for Eisenstein series
which was discovered by Ramanujan in [11], [16, p. 140].
In Section 10, Ramanujan defines some terminology in the theory of infinite
series. His definitions are rather vague and do not seem to be important.
The motivation for most of the material in the last two sections of Chapter
15 is unclear. However, some of the work gains meaning when one realizes
that it is precursory to Ramanujan's profound work on modular equations in
Chapters 19-21. This will be described in Part III [11].
Most of Ramanujan's results in Section 2-7 are expressed in terms of
Bernoulli numbers В„. Recall that, for example, from Titchmarsh's book [3,
p. 19], for each positive integer n,
C(l-n) = (-l)"+1B>, @.1)
where ?(s) denotes the Riemann zeta-function. Since the values ?A — n), rather
than Bernoulli numbers, arise naturally in our proofs, and since the former
notation is more economical, we shall generally express Ramanujan's results
in terms of ?(s).
We quote precisely Ramanujan's cryptic formulation of Entry 1 and its
corollary.
Entry 1.
"ft f cp(kh) = Г (р(х) dx + F(h),
*=i Jo
where F{h) can be found by expanding the left and writing the constant instead
of a series and F@) = 0."

302 15. Asymptotic Expansions and Modular Forms
If we formally apply the Euler-Maclaurin formula, A0.5) of Chapter 13, to
f(x) = (p(hx) on [0, oo), we find that
F(h) = hf <p(kh) - Г q>{x) dx
*=i Jo
Z l + hRm, A.1)
where here
.m—l /*o
!~Jo
m
and where we have assumed that <p(k~^(co) = 0 for 1 < к < т. If <p is such that
A.1) is valid and if furthermore hRm tends to 0 as h tends to 0, then F@) = 0
as stated in Entry 1.
Corollary. "If
h<p(h) = ah" + bhq + chr + dhs + ¦¦¦,
then
aBph" bBqh"
v г _ 1 ..."
Apparently, Ramanujan assumes that p, q, ... are integers with 2 < p <
q < ¦ ¦ ¦ in his application of A.1). If A.1) holds for cp and Rm - O(hm) for each
m > 1, then this corollary yields a valid asymptotic formula as h tends to 0.
Ramanujan next observes that "if the expansion of (p(h) be an infinite series,
then that of F(h) also will be an infinite series; but if most of the numbers p,
q, r, s, t, etc., be odd integers F(h) appears to terminate. In this case the hidden
part of F(h) can't be expanded in ascending powers of h and is very rapidly
diminishing when h is slowly diminishing and consequently can be neglected
for practical purposes when h is small."
The first part of this observation refers to the fact that the coefficients Bp,
Bq, ... in the corollary vanish when p, q, ... are odd integers greater than 1.
The latter part about "the hidden part of F(h)" refers to situations as in the
following Example 1 (where <p(h) = A + h2)'1) and Example 2 (where (p(h) =
exp( — h2)), in which A.1) takes the form F(h) = — Л/2 + hRm for any positive
integer m. Ramanujan's claim is that F(h) + h/2 = hRm tends rapidly to 0 as
h tends to 0. Indeed, in the corrected versions of his Examples 1 and 2 which
we are about to present, it will be seen that F(h) + h/2 ~ exp( — 2n/h) and
F(h) + h/2 ~ exp(—n2/h2), respectively. It would be interesting to obtain
asymptotic estimates of F(h) for other classes of even meromorphic functions
m(h) as h tends to 0 (see Example (iv) of Section 3 and (8.3)).

15. Asymptotic Expansions and Modular Forms 303
Example 1. If <p(h) = 1/A + h2), then
and
Proof. By Entry 1,
dx
00 1 Г00 d
л л h л
= 2C°thh-2-2
_ n h
~ e2*1" - 1 2'
Ramanujan (p. 181) claims that F(h) = 2n/(e2"lh — 1), and so his value of
F(yo) is also incorrect.
Example 2. // q>{h) = exp( — h2), then
Proof. By Entry 1 and the well-known transformation formula for the classi-
classical theta-function B(z), found in the corollary to Entry 7 of Chapter 14,
F(h) = h ? e~h2k2 - Г e~x2 dx
k=l Jo
2 khj
The proposed approximation for F(yo) readily follows. П
In contrast, Ramanujan asserts that "F(jq) is very nearly
Entry 2 is the special case p — 1 of Entry 3 below.
Example (i). As x tends to 0 +,
Z e"ta Log к ~ "y~L°gX + i LogB^),
k = l X
у denotes Euler's constant.

304 15. Asymptotic Expansions and Modular Forms
Proof. This follows from the case p = m = 1 in Theorem 3.2, since ?'@) =
-i LogBrc) (Titchmarsh [3, p. 20]). ?
Example (ii). Let d(n) denote the number of positive integral divisors of the
positive integer n. Then as x tends to oo,
X d(n) ~ x Log x + By - l)x, B.1)
n<x
where у denotes Euler's constant.
This asymptotic formula is a well-known result in elementary number
theory. Let A(x) denote the difference of the left and right sides in B.1). By
elementary methods, A(x) = O(yfx), as x tends to oo. (See, for example, Hardy
and Wright's book [1, p. 264].) At present, the best O-estimate that we have is
A(x) = O(x7/22+c),
for each e > 0, which is due to Iwaniec and Mozzochi [1]. On the other hand,
Hafner [1] has shown that, for some constant с > 0,
A(x) = O+((x Log xI/4(Log Log x)<3+2Log2)/4 exp(_c(LOg Log Log xI/2)),
which is the best Q theorem at present. The problem of determining the order
of A(x) is known as the "divisor problem" and is one of the most difficult and
famous problems in the analytic theory of numbers. It is conjectured that
A(x) = 0(x1/4+e) for every e > 0. For a fuller discussion of the divisor problem
along with historical references, consult the books of Ivic [1, Chapter 10] and
Graham and Kolesnik [1].
Example (iii). // pk denotes the feth prime, then
" Log x
as x tends to 0 +.
Proof. From Landau's treatise [1, p. 215],
pk = к Log к + O{k Log Log k),
as к tends to oo. Thus, as t tends to oo,
F(t) := X pk = jt2 Log t + O(t2 Log Log t).
k<t
Therefore, by partial summation,
? e~kxpk = F(t)xetx dt
= x I it2(Log t + 0(Log Log t))e~tx dt
Jo

15. Asymptotic Expansions and Modular Forms 305
if00 / Гоэ \
= ~2 e'uu2 Log(u/x) du + Olx'2 e~uu2 Log Log(u/x) du
lx Jo V h* )
JO •*¦
as x tends to 0 +. ?
Example (iv). Write
= ? I(n)x", \x\ < 1.
n=0
Then "I(n) is of the order
If, /- smhn^/n]
— <cosh nJn ~—>.'
4«l я^А i
This declaration essentially appears in Ramanujan's first two letters to
Hardy [16, pp. xxvii, 352] and is not correctly stated. (This was pointed out
by Watson [2].) However, a corrected version can be found in the famous
paper of Hardy and Ramanujan [1] (Hardy [6, p. 334], Ramanujan [16,
p. 304]), wherein their asymptotic formula for the partition function p(ri) is
established. Example (iv) is an analogue of this theorem in that the generating
function for p(ri), the Dedekind eta-function, is essentially replaced by the
classical theta-function в(х), where x = — e~"ir. Littlewood [1] (see also
Andrews' book [2, pp. 68-69]) has written that in the collaboration with
Hardy on p(n), Ramanujan kept insisting that a highly accurate formula for
p(n) existed. This persistence especially pushed Hardy to the discovery of their
amazingly precise asymptotic formula. Example (iv) shows that the founda-
foundation for Ramanujan's confidence originated in India several years earlier.
In 1937, Rademacher [1] discovered an exact formula for p(n), which
yields, of course, Hardy and Ramanujan's asymptotic formula as a corollary.
(See also Ayoub's text [1, Chap. 3] for a proof of Rademacher's theorem.)
Zuckerman [1] shortly thereafter found an exact formula for I(n) in Example
(iv) as well as for the Fourier coefficients of the reciprocals of other modular
forms including all the classical theta-functions. Simpler formulas for the
Fourier coefficients of the reciprocals of the classical theta-functions, and
simpler proofs, have been derived by Goldberg [1].
Riesel [1] examined Entries 3-5 below and asserted that they were incor-
incorrect, because he interpreted them as exact formulas. As asymptotic formulas,
Entries 3-5 are, indeed, correct.
We begin Section 3 by stating Entry 3 and its corollary, as Ramanujan
probably intended them. Throughout, у denotes Euler's constant and В„ is the
nth Bernoulli number.
Entry 3. Suppose that m and p denote positive integers. Then as x approaches
0 + ,

306 15. Asymptotic Expansions and Modular Forms
Bm+pk
к px-'" h[ } m + рк k\ •
Corollary. Suppose that p is a positive integer. Then as x approaches 0 +,
k=i к- Р k=i рк- к:
By @.1), Entry 3 and its corollary follow from our Theorem 3.1 below.
Throughout the sequel, Z~ denotes the set of nonpositive integers and
a = Re s. The symbol ]T * indicates that those values of k yielding undefined
summands are excluded from the summation. The residue of a meromorphic
function / at a pole z0 is denoted by R(z0). (The identification of/ will always
be clear.)
Theorem 3.1. Let p > 0, let m denote a complex number, and define
f(x) = ? e'xkrkm-\
k = l
Then as x tends tо 0 +,
Г(т/р) ? (-xf
/M~^+XoCU-m-pfcL^ C-1)
if m/p ф Z~, while if m/p = —r e Z~,
' (~xY
f(x)~<-(Hr-y) + y--
r\
+ ^*f(l_m_pfe)L_L> C.2)
k=o k\
where у denotes Euler's constant and
H = Y {
' kk fc'
Proof. We shall assume that m is real; the more general result can be estab-
established by similar lines of reasoning.
Using the definition of/and inverting the order of summation and integra-
integration by absolute convergence, we easily find that
f(x)xs~l dx = Г(я)СA -m + ps),
provided that a > sup{0, m/p}. By Mellin's inversion formula (Titchmarsh
[3, p. 33]),
f(x) = -L Г(я)СA - m + ps)x~s ds, C.3)
•^7t! Ja-ioo
where a > sup{0, m/p}.

15. Asymptotic Expansions and Modular Forms 307
Consider now
1 Г
T(s)C(l - m + ps)x s ds,
where CM T is the positively oriented rectangle with vertices a ± iT and
— M ± iT, where Г > 0 and M = N + j. Here N is an integer chosen suffi-
sufficiently large to ensure that JV > \m\/p. The integrand has simple poles at
s = m/p and s = 0, — 1, — 2, ..., —N on the interior of CM T, unless m/p =
— r e Z~, in which case there exists a double pole at s = — r By the residue
theorem, if m/p ф Z~,
while if m/p = —r e Z~
Now,
Г(т/р) » (-xf
^г+1а1->"-^)—, C.4)
I* Ш-т-рк){—^. C.5)
k=o K!
and
x~s = xr - xr(Log x)(s + г) н . C.7)
Hence, if m/p = ~r e Z~,
fl
(H)
(ry) ygr C.8)
P p \ \
Putting C.8) into C.5), we see from C.3)-C.5) that, in order to establish
C.1) and C.2), it suffices to show that
J-M
- m + p(a + iT))-"+iT da = o{\) C.9)
as T tends to oo, and then that
T(-M + it)C(l -m+p(-M + it))xM-u dt « xM, C.10)
as x approaches 0 +.
Recall (Copson [2, p. 224]) the following form of Stirling's formula. Uni-
Uniformly for a in any finite interval, as \t\ tends to oo,
112. C.11)
Also (Titchmarsh [3, p. 81]), uniformly for a > cr0, there exists a constant

308 15. Asymptotic Expansions and Modular Forms
к = k(a0) > 0, such that
C(s) = O(| 114 C.12)
as \t\ tends to oo. Estimates C.9) and C.10) clearly follow from C.11) and C.12),
and the proof of Theorem 3.1 is complete. ?
The proofs that follow are similar to that of Theorem 3.1, and so we shall
not provide all the details. In particular, estimates analogous to C.9) and C.10)
are always needed, and they are always obtained in a manner very much like
that described above.
The case m = p = 1 of Theorem 3.2 below yields Example (i) of Section 2.
Observe that Theorem 3.2 follows formally from Theorem 3.1 via differentia-
differentiation with respect to m. This (nonrigorous) procedure may be what Ramanujan
used to deduce Example (i) of Section 2.
Theorem 3.2. Let p > 0 and define
00
g(x) = ? е-хкРкт-' Log к.
Then, if m/p ф Z~,as x tends toO + ,
T'(m/p) - T{m/p) Log x « <-xf
g(x)?C(impk)
Proof. As before, we may, without loss of generality, assume that m is real.
Inverting the order of summation and integration by absolute convergence,
we readily deduce that
g(x)xs~1 dx = -Г(я)С'A -m + ps),
if a > sup{0, m/p}. From Mellin's inversion formula,
9(x) =-J-.\ r(s)f'(l -m + ps)x~s ds,
2ra Ja-i00
where a > sup{0, m/p}.
Consider next
[ Г(я)С'A - m + ps)x~s ds,
M.T
where CM> T denotes the same rectangular contour as in the proof of Theorem
3.1. Since m/p ф Z~, by the residue theorem,
r(m/p)~T(m/p)Logx » (-xf
The remainder of the proof is similar to that of Theorem 3.1, but an estimate

15. Asymptotic Expansions and Modular Forms 309
just like C.12) is needed for {'(s) in place of {(«)• This can be obtained
by differentiating the formula for {(s) obtained from the Euler Maclaurin
formula, A0.5) in Chapter 13, with f(t) - t~s, a — 1, b = oo, and m sufficiently
large. D
Ramanujan now records several examples to illustrate his results. Example
(i) is the case p = 2 of the corollary following Entry 3. Example (ii) is the case
p = 4, m = 2 of Entry 3. Example (iii) is the case p = 3, m = 2 of Entry 3.
Example (iv). As x tends tо 0 +,
oo „-*2* 2
Proof. From A.2), as t tends to 0+,
у -*2< - * * f*
иhle ~ 2 2 v t
Integrating this equality over [0, x], we find that
OO C1 _ p~k2x\ Y
as x tends to 0+. Since {B) = n2/6, the proposed result follows. ?
Example (v) records the case m = 3, p — 6 of Entry 3.
Entry 4. Let p > 0 and let m and d be complex numbers with Re{pd — m) > 0.
Define
Then, ifm/p $ Z~, as x tends to 0 + ,
Г(т/р)Г(Л - m/p) » (-*)*
й(х;+ ?(d)CAm pk)
where, as usual, (d)k = F(d + k)/r(d).
Proof. As in previous proofs, we may assume without loss of generality that
m and d are real.
Inverting the order of summation and integration by absolute convergence,
we deduce that
Лоо oo Лео s-1
hMx-Ux^Zk"-1-1"] nT
Jo *=i Jo U +")
r{s)r(d-s)

310 15. Asymptotic Expansions and Modular Forms
provided that m/p < a < d and a > 0. By Mellin's inversion formula,
1 f+to r(s)T{d - s) _s
n(x) = -— — 4A — m + ps)x as,
2ni Ja-ic0 T{d)
where m/p < a < d and a > 0.
With the same oriented rectangle CM T as in previous proofs, we find that
1 Г ГE)Щ-5)_ ___ . __,___, ,_
Г(т/р)Г(«/ - m/p)
by the residue theorem. The remainder of the proof now parallels that of
Theorem 3.1. ?
Ramanujan concludes Section 4 with an example, which is the case p = 8,
m = 2, d = j of Entry 4.
Entry 5 is the case п = ц=\,тфро( Theorem 6.1 in Section 6. The
corollary of Entry 5 is the case n = q = 1, m = p of Theorem 6.1.
In the case m/p ф n/q, Theorem 6.1 is a somewhat more general version of
Entry 6, and in the case m/p = n/q, Theorem 6.1 generalizes a result marked
by "N.B." immediately following Entry 6.
Theorem 6.1. Let p, q > 0 and let m and n be complex numbers. Define
j,k=i
Then, if m/p, n/q ф Z~, as x tends to 0 + ,
/,,„(*) ~ ^«1 - n + qm/p) + ^«1 - m +
? C(\ - m - pk)C(l - n - qk)(~^, ifpn^qm,
+ ? C(\ - m - pk)C{l - n - qk){-^-, ifpn = qm.
In the case p = q = 1,

15. Asymptotic Expansions and Modular Forms 311
where d(r) denotes the number of positive divisors of r. Titchmarsh [3, p. 140]
used the asymptotic expansion for flA(x), which was first proved by Wigert
[1], and which is a special case of Theorem 6.1, in obtaining mean value
theorems for ?(s).
Proof. Without loss of generality, assume that m and n are real.
Inverting the order of summation and integration, we readily see that
fm n{x)xs l dx = F(s)?(l — m + ps)C(l — n + qs),
Jo
under the assumption that a > sup{0, m/p, n/q}. By Mellin's inversion formula,
i Ля + ioo
/m.nW = ^ T(s)C(l -m + ps)C(l - n + gs)x~s rfs,
provided that a > sup{0, m/p, n/q}.
Let CM T denote the rectangular contour given in the proof of Theorem
3.1. By the residue theorem,
~ f T(s)C(l -m + ps)C(l - n + qs)x~s ds
= <5m,n(R(m/p) + Я(п/д)) + X Ш - m - MKO - " - <^) ™Г"'
k=o /c!
where <5m „ = 1 if pn ф qm, and <5m „ = ^ if pn = дш. If pn ф qm, the residues
R(m/p) and R(n/q) are routinely calculated. In the case that pn = qm, the
integrand has a double pole at m/p = n/q instead of simple poles at m/p and
n/q. With the use of C.6) (and its analogue with q and n in place of p and m,
respectively), C.7), and the Taylor expansion of T(s) about s = m/p, we may
easily calculate R(m/p) for this double pole. The remainder of the proof is
almost identical to that of Theorem 3.1. ?
Theorems 6.2 and 6.3 below supplement Theorem 6.1 by providing asymp-
asymptotic expansions when m or n or both are equal to 0.
Theorem 6.2. Let p, q > 0 and let n be complex. Then, if n/q ф Z~,
ГЩ1 + pn/q) C(l - n) Log x
as x tends to0 + , where
A C'(l - n)q

312 15. Asymptotic Expansions and Modular Forms
Proof. The proof is identical to that of Theorem 6.1 except in one respect.
The former simple poles of the integrand at s = 0 and s = m/p now coalesce
to form a double pole at s = 0. An elementary calculation yields
and the desired result follows. ?
Theorem 6.3. Let p,q>0. Let
n2
where
" Log r Log2 и
Tften as x tends tо 0 +,
/• /..ч LoS2* (P + g ~ !)У L°g ^
РЧ
t
Proof. The proof is the same as for Theorem 6.1, except that the former
simple poles at s = 0, m/p, and n/q now coalesce to yield a triple pole at s = 0.
To calculate R@), we require the Laurent expansions
s
_ 1
PS ps У
1
and
x"s = 1 - s Log x + js2 Log2 x + • • •.
The Laurent expansion for F(s) about s = 0 may be calculated from the
Weierstrass product representation for F(s), while the Laurent expansion of
C(s) about s = 1 is found in Entry 13 of Chapter 7 (Part I [9]). It transpires that
2pq pq
The desired result now follows as in the proof of Theorem 6.1. ?

15. Asymptotic Expansions and Modular Forms 313
Note that An does not approach Ao as n tends to 0.
Ramanujan concludes Section 6 with two examples. Example (i) is the case
p = 2, n = q = 1 of Theorem 6.2, and Example (ii) is the case m = 3, p = q =
n = 1 of Theorem 6.1.
As customary, put ar(s) = ? dr, where the sum is over the positive integers
d which divide s.
We next offer a corrected version of Ramanujan's Entry 7. (Ramanujan
mistakenly indicates on^1(s) instead of <Tn_m(s) below.)
Entry 7. As x tends to0 + ,
» sm~4_m(s) r(m)
E "^31 rs-CN^1 + m - ")
s=l с 1 A
, rW r, mi , . , «2- m)CB - n)
+ —^C(n)C(l + и - m) +
X X
+ f СA-»я-Щ1-и-Щ-Л)Ц^, G.1)
provided that m ф п, т ф \,пф 1, and т,пф Z~.
Setting s = jk below, we see that the sum on the left side of G.1) equals
s=l i = l i,j,k-l
that is, the left side of G.1) equals
00
/„.».«(*):= I e-^/c-y-Y-1 G.2)
for the special choices p = q — r = l=l. In Theorem 7.1 below, we give an
asymptotic formula, as x tends to 0 + , for the triple sum in G.2), under
the general conditions m/p, n/q, l/r$Z~, qm ф np, rm ф pi, and rn Ф ql.
Ramanujan's formula G.1) then follows from Theorem 7.1 upon setting p =
Theorem 7.1. Let p, q, r > 0 and let m, n, and I denote complex numbers. Let
fm,n,i(x) be defined by G.2). Suppose that m/p, n/q, l/r ф Z~, qm Ф np, rm ф pi,
and rn ф ql. Then as x tends to 0 +,
/».«.«W ~ Г~^ CO - n + qm/p)t(l - I + rm/p)
+ ^rC(l - m + P«A?KA - ' + rn/q)
? Ц1 - m - pfc)C(l - n -
k=o

314 15. Asymptotic Expansions and Modular Forms
Proof. The proof follows precisely along the same lines as the proofs of
Theorems 3.1 and 6.1. ?
Ramanujan concludes Section 7 with an example, which is the case m — 3,
n = 5 of G.1). Again he mistakenly indicates <rn_i(s) in place of <rn_m(s) in the
example, and, moreover, he inadvertently omits the term — l/A440x) in the
asymptotic expansion.
Clearly, the theorems that we have proved can be generalized and extended
even further. In particular, restrictions imposed on the parameters can be
lifted. The computation of the residues would then be somewhat more difficult.
At the beginning of Section 8, Ramanujan remarks that "if F(h) in XV 1
terminates we do not know how far the result is true. But from the following
and similar ways we can calculate the error in such cases." To illustrate these
cryptic remarks, Ramanujan indicates a method for calculating the error in
the asymptotic expansion
|by> (8Л)
as x tends to 0 + , which is the case p = 2, q = m = n=lin Theorem 6.1. In
fact, he indicates that the equalities
? e~k2x cos(ax) dx =
Jo *=i i
_ n sinh^/^a) - sinGi
-j= j=
can be used to deduce the following exact formula extending (8.1).
Entry 8. If x> 0, then
-Л-In I— J - e-W**/* cos I -
cosh ( 2л /— I — cos I 2k
V V xj \ м x
(8.3)
This is truly a remarkable formula. The left side can be construed as a
modification of the theta-function
k=l
Thus, Entry 8 is an analogue of the inversion formula for в{х).
Before proving Entry 8, we first establish (8.2).

15. Asymptotic Expansions and Modular Forms 315
The first equality easily follows from inverting the order of summation
and integration on the left side and using a well-known integral evaluation
(Gradshteyn and Ryzhik [1, p. 477]).
To prove the second equality, we shall expand the right side into par-
partial fractions. An elementary calculation shows that the nonzero zeros of
cosh(n^2a) — cos(n*j2a) are at a = + k2i, 1 < к < со, and that they are
simple. Thus, if R(z0) denotes the residue of the function on the far right side
of (8.2) at a simple pole z0, we find that
R(±k2i)= T1-.
Thus, for some entire function g(a),
л
л sinh(n^2a) - sin(n^2a) _ i_ » f I 1 j
ijla cosh(n^/2a) - cos(n^/2a) 2 k=i \ a- k2i a + k2i\
Letting a tend to oo on both sides above, we find that g(a) tends to 0. Hence,
g(a) is a bounded entire function, and so by Liouville's theorem g(a) is
constant. Clearly, this constant is zero. Hence, the proof of the second equality
in (8.2) is complete.
A different proof of the second equality in (8.2) may be found in a paper of
Glaisher [1].
Proof of Entry 8. Setting x = ny, we restate (8.3) in the form
\6y 2j~y
where
_ 1 1
2 fci fk
2 fc=i ^fky I coshBK^/k/y) — cosBK^/lc/y)
2 » n f sinhGr4/2a) - sin (я ,y 2a) 1
= — ?-• —T= 1 T^ ~r=— Г'
пУ k=i 2J2a LcoshGrN/2a) — cosGrx/2a) J
where a := ak:= 2k/у.
For brevity, set
ф(и) = ? е~жк2и, и > 0.
k=l
Thus, by (8.2) and (8.5), with a = 2k/y,
oo ({"<*> du \
R = 2E ф{и) cosBnku/y) — >
о У 4^/ky)
= 2 ? j Г ф(иу) cosBnku) du - -^. (8.6)
*=i Uo

316 15. Asymptotic Expansions and Modular Forms
Now, for у > О,
GO
Z Ф(ку) =
fc l
= z z *-*
i = l k = l
(8-7)
By (8.6) and (8.7), the proposed formula (8.4) now becomes
1
cosB"'")* -m\-
о 4y/ky)
(8.8)
Let 0 < ? < 1. Applying the Poisson summation formula, F.1) of Chapter
14, we deduce that
У Ф(ку) = ф(иу) du + 2 Z ФШ cosB7iku) du. (8.9)
ф(иу) du = —.
by
From (8.2),
Г
Jo
Hence, by (8.9),
00 л =0 f<»
Z Ф(кУ)~т= Hm 2 Z ф(иу) cosBKku) du. (8.10)
t=i by ?^0+ k=i J?
By (8.8) and (8.10), it remains to prove that
= lim 2 Z \\ ф(иу) cosBnku) du 7=\. (8.11)
E^o+ k=i (.Jo
From the corollary to Entry 7 of Chapter 14, for и > О,
2 2,/цу Уцу W/
Therefore,
' cosBnku)
1 f1
(му) cosBtt/cm) du= ~-\ cosBnku) du + -
2 J 0 2
0 2 J 0 ^y Uy
uyj
The first term on the right side of (8.12) gives to (8.11) the contribution

15. Asymptotic Expansions and Modular Forms 317
= lim & - [г.] - |) = -i (8.13)
where we have used E.2) of Chapter 14. The third expression on the right side
of (8.12) contributes to (8.11)
W^'A — )du = O, (8.14)
E-o+ *=i Jo Juy \uyj
which can be seen after two integrations by parts. By (8.11)-(8.14), it remains
to prove that
(8.15)
2 -.T+AUo Ju 2у/кУ
Now (Gradshteyn and Ryzhik [1, p. 395]),
f °° COSB7lfcM) , IT f °° , , 1
—— аи = —~ cos и аи = ——.
Jo Ум V**Jo 2^Д
Using this in (8.15), we find that (8.15) becomes
?->0 +
We shall again apply the Poisson summation formula. Let 0 < ? < 1 < N
and suppose N is not an integer. Then
c<k<N Jk Je ^/U k = l J?
The left side of (8.17) may be written as
C
J.
N d{\u] - u) [u] - и
N I CN
I CN [и] -и
j
Using this in (8.17) and letting N tend to oo, we deduce that
Г , 1 Г" M - " л л ^ f °° cosBitM л
(8.18)
where letting N tend to oo inside the summation sign is justified by two
integrations by parts. Combining (8.16) and (8.18), we see that we must show
that
But this last formula follows immediately from a well-known representation
for ?(s) found in Titchmarsh's treatise [3, p. 14, Eq. B.1.5)]. Hence, the proof
of (8.3) is complete. ?

318 15. Asymptotic Expansions and Modular Forms
In the sequel, we shall set
CO
FB,,(x)= ? jmk"e-ikx, (9.1)
where x > 0 and m and n are nonnegative integers. Without loss of generality,
assume that m > n. In Theorem 6.1, an asymptotic expansion is given for
Fm „(x) as x tends to 0 + . Ramanujan begins Section 9 with the special case
p = q = \,тф n of Theorem 6.1. He then defines, for \q\ < 1,
,3 Qk
and
N = 1 - 504
k = l
The functions L, M, and iV were thoroughly studied in a famous paper [11],
[16, pp. 136-162] by Ramanujan, where L, M, and N are denoted by P, Q,
and K, respectively. We now show that L, M, and N are essentially the
Eisenstein series of weights 2,4, and 6, respectively, on the full modular group
ГA). To see this, first let q = ехрBл1т), where т is in the upper half-plane Ж,
and write
oo tv/7^ oo да oo
^Ц = I TvWe2™1, (9.2)
k-\ >¦ — 4 k = l j=\ r = l
where we put j/c = r and where <rv(r) = ?д|Ду. Next recall that the Fourier
expansions of the Eisenstein series En{%), where n is an even positive integer,
are given by (Rankin [2, p. 194])
?2(t)=l-24 ? ffl(fc)e2«-*_J_
= 1-24Ф! (<|)-— (9.3)
and
= 1-^Ф«-1(«), ">2, (9.4)
where у = Im i > 0 and where Bn denotes the nth Bernoulli number. Hence,
L = ?2(t) + 3/G1)/), M = EM and JV = ?6(т).

15. Asymptotic Expansions and Modular Forms 319
Ramanujan next claims that if m + n is an odd, positive integer, then the
function Fm „(x) of (9.1) can be evaluated exactly in terms of L, M, and N. First,
observe that, by setting jk = r in the definition of Fm „(x), we obtain
FmJx) = ? г-<тт.я(г)е-".
r=l
Thus, with x = —2nh, Fmn(x) is essentially an n-fold derivative of an Eisenstein
series of even weight if m — n is odd. If m — n = 1 and n > 1, then Fm „(x) is
clearly a multiple of an n-fold derivative of L. Suppose now that m — n is odd
and > 1. By a theorem in Rankin's text [2, p. 199], each modular form of even
positive weight can be expressed as a polynomial in Е4(т) and ?6(t). Thus,
can be so expressed, and since Fmn(x) is, up to a factor of ±1, an n-fold
derivative of the function above, then Fm n(x) can be represented as a poly-
polynomial in M, N, and their derivatives.
For further remarks and discussion, see Venkatachaliengar's monograph
[1, pp. 30, 31].
Entry 10(i) (First Part). For each positive integer n > 2,
can be expressed as a polynomial in M and N.
This statement was verified in Section 9 where we appealed to Rankin's
book [2, p. 199]. See also A4.2) and Entry 14 below.
Entry 10(i) (Second Part). For each positive integer n,
can be expressed as a polynomial in M and N. Here S^ = j and Sn = 1 if n > 2.
Proof. By (9.3) and (9.4),
so k2nnk 1 Л / R
V q - I 2" E*
where
if п > 1.

320 15. Asymptotic Expansions and Modular Forms
Thus, for и > 1,
An ) 6 An
By the aforementioned theorem in Rankin's treatise [2, p. 199], it suffices to
prove that Fn(x) is a modular form on ГA) of weight 2и + 2. We must therefore
show that (Serre [1, Eq. E), p. 80])
Fn(-l/x) = x2~+2Fn(x), хеЖ. A0.1)
Recall that for Vx = (ax + b)/(cx + d)e F(l) (Schoeneberg [1, pp. 50, 68])
f(ct + dJE*(x) - 6n-4c(cx + d), ifn=l,
(ct + dJKE2n(x), ifn>l.
4:
By A0.2), if n >1,
FA/)
F.A/t) &nx2«-lE2n{x) + х2»Е'2я(х))
B2n 2m
This proves A0.1) for n > 1. A similar argument can be used for the case
n = 1. ?
Alternatively, for и > 1, A0.1) follows from the theorem in Ogg's survey
[1, pp. 16, 17] that if f(x) is a modular form of weight k, then f'(x) —
{2nik/l2)Ef(x)f(x) is a modular form of weight к + 2.
Ramanujan did not consider the case и = 1 in Entry 10(i).
In the remainder of this long section, Ramanujan makes several definitions
and offers many examples to illustrate his definitions, which, for the most part,
are imprecise. For each definition, we quote from the notebooks (pp. 186,187).
Entry 10(ii). "The degree of a series is the sum of the highest powers of the nth
terms together with unity if the series contains all the powers of x or if the
powers of x be in A.P. (arithmetic progression).
If the coefficient of each nth term is homogeneous the series is said to be
pure and in other cases mixed.
The theory of indices holds good in terms of degrees of series.
If F(h) in XV 1. terminates the series is said to be perfect. // not it is said to
be imperfect.
// F(h) = 0 the series is said to be complete in other cases incomplete.

15. Asymptotic Expansions and Modular Forms 321
A series is said to be absolutely complete when it remains complete when
transformed or split up.
A linear series can only be expressed by linear, double by double, treble by
treble, pure by pure, perfect by perfect, imperfect by imperfect, and absolutely
complete by absolutely complete adhering to the laws of indices in all cases. But
a mixed series can be split up into a number of pure series of different degrees."
M. E. H. Ismail has suggested that the degree of a series is more properly
defined in terms of the order of a singularity on the boundary of convergence
of the series. Of course, this definition is possibly ambiguous if there is more
than one singularity on the boundary. However, for some of Ramanujan's
examples, Ismail's definition is more viable than Ramanujan's definition.
We do not know what is meant by "the theory of indices." The definition
of F(h) is given in Entry 1.
Example 1. Let
k=l
where и is a nonnegative integer. First, /t has degree и + 1 because the degree
of k" is n and xk contributes 1 to the degree. Since/, has a pole of order n + 1
at x = 1,/j has degree и + 1 by Ismail's definition as well. It is easily seen that
k" is homogeneous; that is, if g{k) = k", then g(jk) = (jk)" = j"g(k). Thus, /t is
pure. Here q>(t) = t"x'. Since ф'2*^) is not necessarily equal to 0 for each к
sufficiently large, F(h) does not terminate, and so /t is imperfect. It trivially
follows that /t is incomplete. It is uncertain what Ramanujan means by
"linear." But if he means that the series is not a multiple series, then it is clear
that /i is linear.
Example 2. For x real, let
™ sin(fcx)
Now sin(fex) probably has degree 1 in Ramanujan's definition. Since l/k has
degree —1, Ramanujan concludes that/2 has degree 0. The singularities at
x = 2пк, where n is an integer, are "jump" discontinuities, and so it is reason-
reasonable to say that they are of order 0. Hence, f2 has degree 0 by this interpretation
as well. The coefficients are equal to l/k, and so /2 is pure. It is clear that /2
is linear by the interpretation of "linear" given in Eixample 1. Now F(h) ф 0,
but cp(t) = sin(tx)/t is an even function oft, and so ф*2*^) = 0, k > 1. Hence,
/2 is perfect and incomplete.
Example 3. Consider Fm „(x), defined by (9.1). Clearly, Fmn is pure and is a
double series. Nowym, k", and e~x are of degrees m, n, and 1, respectively, and
so Fmn(x) has degree m + n + 1. By Theorem 6.1, the order of the singularity

322 15. Asymptotic Expansions and Modular Forms
at x = 0 is not equal to m + n + 1, however. Also by Theorem 6.1, Fm „ is
incomplete. The series on the right side of Theorem 6.1 consists of terms of
the form
Ax? Bm+k+1Bn+k+1(-xf
а-т-кШ-п-к)у-
k\ (m + к + 1)(и + к + l)fe!'
by @.1), if к > 1. If m + n is even, these terms will not be equal to 0 when m
and к are of opposite parity. Thus, Fm „ is imperfect if m + n is even. However,
if m + n is odd, then either Bm+k+l or Bn+k+i is equal to 0. Hence, if m + n is
odd, Fm „ is perfect.
Example 4. Let m and n denote positive integers with m ф п. Let
Thus, in the notation of Section 7, gm „(x) = /m+1 „+11. Ramanujan asserts
that gm „ is a treble, pure series of degree m + n + 1, which is clear. Note that,
by Theorem 7.1, the alternate definition of degree fails here. Also, by Theorem
7.1, gm „ is incomplete. A typical term in the asymptotic expansion for gmn(x),
by Theorem 7.1 and @.1), equals
(_1Г+„ в^к+
n + k+ l)(n + к + l)(/c +
Thus, if both m and n are even, we see that the asymptotic series does not
terminate, and so gm „ is imperfect. But if either m or n is odd, the expansion
does terminate, and so gm „ is perfect in these cases.
Example 5. Let m, n, and x denote real numbers with n > 0. Put
00 fcm
"m,nW — L, 7' kx o-kx\n "
lc=l ye -r e )
Ramanujan claims that hm „(x) is a double series, so that he evidently writes
hm „(x) in the form
*=i ./=0 ;!
The coefficients are not homogeneous, and so the series is mixed. Ramanujan
claims that hm „ has degree m + n, but it seems to us that the degree is equal
to m + n + 1, since km, e~kx", and e~2Jkx have degrees m, n, and 1, respectively.
Note that hmn has an essential singularity at x = 0. It is easy to see that hm „
is incomplete.
Example 6. Consider the theta-function
f(x) = Utxk2 = \ I xk\ \x\<\.

15. Asymptotic Expansions and Modular Forms 323
Clearly, / is pure. Since x'2 is an even function of t, it is trivial that F(h)
terminates, and so / is perfect. Ramanujan also claims that / is a pure, double
series. This is enigmatic, for if we expand x*2 in a power series in k, f is no
longer pure. However, possibly, in this instance, Ramanujan intends "double"
to mean "bilateral," in which case, Ramanujan's assertion is correct. Lastly,
he asserts that / has degree \. We are unable to justify this claim by using
Ramanujan's definition of degree. Now / is analytic at x = 0. However,
if we set x = ек", т е Ж, then, by the theta-transformation formula A.2), it
may be loosely construed that f(emz) has a "singularity of order \ at т = 0."
Of course, this is not really the case, since the real axis is a natural boundary
for f(en"). Thus, a fuzzy interpretation of Ismail's definition has a modicum
of viability.
Example 7. Ramanujan remarks that L, M, and N are perfect, pure double
series of degrees 2,4, and 6, respectively. By expanding A — qk)~l in a geometric
series, we readily see that L, M, and N are pure double series of degrees 2, 4,
and 6, respectively, since q>k, 1 <j,k< oo, is of degree 1. Now apply Theorem
6.1 with m = p = q = 1, e~x = q, and n = 2, 4, and 6, respectively. Since
B\+A+k = 0, /c > 1, L, M, and N are perfect. Lastly, Ramanujan assets that
M and N are complete, but L is incomplete. It appears to us, however, that
all three series are incomplete, for in Theorem 6.1,
f B -ri) Г(н)Ци)
I ^ u.
Entry 11. Ifa,p>0 and af} = л2, then
1 S 1\_ •
4 tti k2 sinh2(a/c) 4 »ti k2 si
ao
- 2a X fc2 Log(l - е~2ак) - 2p ? /c2 Log(l -
k=l k=l
120 72'
Proof. By an elementary calculation,
f /c2 Log(l - e~2xk) = - ? "г
~i\
A - e-
cosh(«/)
(li.i)
With A1.1) as motivation, we define

324 15. Asymptotic Expansions and Modular Forms
/ 1 2a cosh(az)
/(z) = л cotGtz) . + —. з
yz'* sinh (az) z sinh (az)
We shall integrate / over a suitable rectangle, to be described later, and apply
the residue theorem. We let R{z) denote the residue of a specified function at z.
First, / has simple poles at each nonzero integer к with
= 1 , 2ot cosh(afc)
W k2 sinh2(afc)
k2 sinh2(afc) к
By A1.1), the sum of all such residues is equal to
2 1 r2 ¦ I 2, ,v - 16a t k2 Log(l - е~2л). A1.2)
t=i к-1 sinh (ак) »=i
Second, let/t(z) = p(z)/q(z), where p(z) = n cot(rcz) and q(z) = z2 sinh2(az).
The function /t(z) has double poles at z = i/ся/а, for each nonzero integer k.
To calculate the residue at ikn/a, we shall use a formula from Churchill's text
[1, p. 160] for the residue of a double pole. Accordingly,
„,,M 2p'(ink/a) 2p{mk/<x)q'"{iiik/0L)
R№) AU)
Elementary calculations yield
p{ink/a) = n cot{pki), p'(ink/tx) = -я2 csc2(Pki),
q"(mk/tt)= -2n2k2, and q'"(ink/oi) = I2a.nki.
Using these values in A1.3), we find that
„,. , , , 1 2 coth(W
Thus, the sum of all such residues is
4»coth№)
Consider a function F(z) = p(z)/q(z), where p and q are analytic at z0,
p(z0) / 0, and q has a zero of order 3 at z0. Then a somewhat lengthy, but
routine, exercise shows that
3p(zo)q<s»(zo) 3p(zo){^»(zo)}2
z)}2 8{q"'(z)}3 ' l ' '
Now set /2(z) = p(z)/q(z), where p(z) = 2яа cot(rcz) cosh(az) and q(z) =
z sinh3(az). The function /2(z) has triple poles at z = ink/a., for each nonzero
integer k. Elementary calculations yield

15. Asymptotic Expansions and Modular Forms 325
p(ink/a)= -2(-\)knu.icoth(pk),
p'(mk/a) = 2(- If л2 a csch2(Pk),
p"(mk/a) = 4{-lfn3ai csch2(?/c) coth(^/c) - 2(- 1)*ла3
q"'{ink/a) = 6(- 1)*'ла.2 ki,
qw(mk/oC) = 24(-l)"a3,
and
qE)(ink/a) = 60(- If mx4fci.
Using these values in A1.5), we find, after much simplification, that
OR 0 0
R{mk/a) = — csch2(/3k) coth(^) + -r csch2(/3fc) + —r coth(
к к pk
Thus, the sum of all such residues, by A1.1), is equal to
». ,„.6,
Lastly, / has a pole of order 5 at the origin. We have
7
/м = * ~-^f-
360
Г 3z2 15
Hence,
15л 9 15a2/ 9 15
Consider next
1
In ¦= f
where Cw is a positively oriented rectangle with sides parallel to the coordinate
axes and passing through the points ±([,/jV] + \) and m(N + ^)/a, where
N is a positive integer. Note that CN is free of poles of /. Estimating the
integrand on the vertical and horizontal sides separately, we find that
^ o(l), A1.8)
as N tends to oo.

326 15. Asymptotic Expansions and Modular Forms
Apply the residue theorem to IN and then let N tend to oo. Using A1.2),
A1.4), A1.6), A1.7), and A1.8), we deduce that
°=2 x ,2. \2ГГ,-l6a ?fc2Log(l -eo"')
*=! /c2 sinh2(a/c) n=i
00 1 OO
tti /c2 sinh2(^fc) r ^l
a? a2 +
which is readily seen to be equivalent to the proposed identity. ?
Another proof of Entry 11 may be constructed from results in Berndt's
paper [6, Theorems 2.2, 2.16] together with A1.1).
Entry 12. Let L, M, and N be as defined in Section 9, and recall that ?„(т),
n > 2, and Фп(а) are defined by (9.4) and (9.2), respectively. Define the dis-
discriminant function A(t) by
A(t) = q ft A - qkJ\ q = e2*", re/.
Then, for \q\ < 1,
(i) M3- N2 = 1728Д(т),
(ii) ?8(t) = M2,
(iii) ?10(t) = MN,
(iv) ?u(t) = M2N,
. . S /cV M - L2
lyj \ I rr:
^ ' L-i /1 л*\2 TOO '
k=l A — q j /50
v fc4<5ffI LM ~ N
720 '
2 ~LN
=i A - qkJ 1008 '
.. » k8qk _LM2-MN
Ш) к A - qkJ ~ 720 '
00 00
fe = 0 ft = 0
oo (Off 11/7^ °° C2.h, 1 }^ Q^
W M Z 1 _ 2fc-l = Z ^_ 2t-l ¦
Proofs of (i)-(viii). Formulas (i)—(iv) are very well known and are special
cases of the general theorem in Rankin's book [2, p. 199] which we applied
in Section 9. In particular, (i)—(iv) can be found in [2, pp. 195,197, Eqs. F.1.8),
F.1.9), and F.1.14)]. These formulas were also derived by Ramanujan in [11],
[16, p. 141].

15. Asymptotic Expansions and Modular Forms 327
Formulas (v)-(viii) are originally due to Ramanujan, and proofs can be
found in his paper [11], [16, pp. 141, 142]. ?
First Proof of (ix). This formula is a special case of a general formula
established by Ramanujan in Chapter 16. See Part III [11, Chap. 16, Entry
35(i)]. ?
Second Proof of (ix). Rearranging in (ix), we find that
Jy (_l)Bfc+ l)9«*+D/*_ f (-lfBk+ l)Vj-
24 [k% *=o J
Equating coefficients of q", n > 0, on both sides, we find that
a(n) - Ъа(п - 1) + 5a(n - 3) - la{n - 6) + • • • = 0, A2.1)
if n is not a triangular number, while if n = r(r + l)/2 is a triangular number,
a{n) - Ъа(п - 1) + Sa(n - 3) - la{n - 6) + • • •
= (-1Г1Е^2- A2.2)
Thus, formula (ix) is equivalent to the arithmetic identities evinced in A2.1) and
A2.2). These identities are due to Glaisher [2] in 1884, although they are really
consequences of a formula proved seven years earlier by Halphen [1]. Hence,
appealing to the theorem of Glaisher and Halphen, we have shown (ix). ?
For generalizations of Entry 12(ix), see two additional papers of Glaisher
[4], [5]. For further references to the literature, consult Dickson's history
[1, P- 289].
Proof of (x). If
/(т) = ? anq\ q = e2*<\
define functions /„, /0, and _/\ by
/oo(t) = /Bi) = t anq2\ /0(т) = /(т/2) = ? anq»*,
л=0 n=0
and
n=0

328 15. Asymptotic Expansions and Modular Forms
Then Entry 12(x) may be rewritten in the form
N,-N0 = 2lM{L1 - L0). A2.3)
If w = (x + l)/2, observe that
1 - 1/t w - 1
2 2w- 1
Thus, from A0.2), we readily find that
LJX + 1) = LJt), L0(t + 1) = LM L1(t+1) = L0(t),
b.(- 1/t) = it2L0(t) + ^, Lo(- l/i) = 4t2L0O(t) + ^,
1() ,
t), N0(t + 1) = Л^т), iV^x + 1) = iV0(t),
(- V*) = Z^^oM. #„(- 1/t) = 64t6JV0O(tX
64
and
Next, define
Xx = L1 — Lo, Xo = <*Lm — L1( Xl = Lo — 4^00,
Za> = N1-N0, Zo = 64Nao-N1, Zl=N0-64Nx.
Then the foregoing equalities readily imply that
XJx + 1) = -JUt), ^o(t + 1) = -XM X,{x + 1) = -*0(t),
Z00(-1/t)=-t2Z0(t), Хо(-1/т)=-т2Х0О(т), ^(-1/т)= -т2
A2.4)
and
Ze(T + 1) = -Ze(r), Z0(t + 1) = -Z^t), Zt(T + 1) = -Z0(t),
^(-l/t) = -t6Z0(t), Z0(-1/t) = -xbZJz\ Z^-1/т) = -x6Z
A2.5)
Let Mt denote the space of modular forms of weight к on the modular
subgroup ГB). If S(x) = x + 1 and Г(т) = — l/т, then, by a paper by Frasch
[1, p. 245], generators of ГB) are
S2(x) and TS2T{x)= T
Using these generators and A2.4), we may easily verify that Xo, Xxe M2.
Suppose that к is even. Then from Rankin's text [2, pp. 104, 105], dim Mk =

15. Asymptotic Expansions and Modular Forms 329
1 + \k. Moreover, since
•¦• A2.6)
and
--- A2.7)
are obviously linearly independent in M2, we conclude that ХЦ2, Xf/2~1Xao,
..., XoX**2'1, XkJ,2 form a basis for Mk. Now suppose that fe Mk and that
/(t) = o(qk'*), as g tends to 0. Then from A2.6) and A2.7), /(т) = О.
In our situation, we take к = 6. Clearly, MXX e M6, and, from A2.5), we
may verify that Zm e M6. From the expansion
Zx = 1008^1/2 + 24595243/2 + ¦•¦,
A2.7), and the definition of M, we find that 2\MXao - Zx = o(q3/1) as q tends
to 0. Hence, 2\MXaD -Z.,50, and A2.3) is proved. D
We are very grateful to D. W. Masser for supplying us with the proof above.
Another proof of Entry 12(x) based on the theory of modular forms on ГоB)
was constructed for us by A. O. L. Atkin.
Entry 12(x) was stated by Ramanujan in [11], [16, p. 146] without proof.
Ramanujan indicated that he had two proofs, one of which was elementary,
while the other used elliptic functions. However, he provided no hints to either
proof. It is very unlikely that the proofs of Masser and Atkin are the same as
either of Ramanujan's proofs. In her thesis, Ramamani [1, p. 59] has given a
proof of Entry 12(x) that uses the theory of elliptic functions. Entry 12(x) is
equivalent to the elegant identity
? a, Bk + 1)<x3(ii -k) = ~osBn +1), n > 0,
*=o ^4U
where G3@) = jio. It would be interesting to have an elementary proof of this
identity and hence of Entry 12(x) as well.
In his paper [11], [16, pp. 136-162], Ramanujan studies
n
Ег,Л«)= Z or{k)os{n - k),
where r and s are odd, positive integers and am@) = j?( —m). He establishes
an asymptotic formula for ?r,s(") as n tends to oo with an error term.
He, however, conjectured a better error term [11], [16, p. 136, Eq. C)].
This conjecture remained unproved until 1978 when Levitt [1] proved
Ramanujan's conjecture in his thesis. In some instances, Ramanujan showed
that the error term is identically equal to 0. Levitt [1] established necessary
and sufficient conditions for the vanishing of the error term and so showed
that the instances of such found by Ramanujan are exhaustive. Such a theorem
was also found by Grosjean [1], [2] who has made a systematic study of
recursion formulas connected with ?rs(")-

330 15. Asymptotic Expansions and Modular Forms
An informative survey paper on convolutions involving ak(n) has been
written by Lehmer [1]. For other papers in this area, consult [1, Sect. A30],
edited by LeVeque.
Entry 13. Let Фч(<?) be defined as in Entry 12. Then, for\q\ < 1,
(i) 691 + 65,520 ФИМ = 441М3 + 250JV2,
(ii) 3617 + 16,320 Ф15(<?) = 1617М4 + 2000МАГ2,
(iii) 43,867 - 28,728 <D17(q) = 38,367M3iV + 5500JV3,
(iv) 174,611 + 13,200 <D19(q) = 53,361M5 + 121,250M2iV2,
(v) 77,683 - 552 Ф21(д) = 57,183M4iV + 20,500MiV3,
(vi) 236,364,091 + 131,040 Ф23(д) = 49,679,091Мб + 176,400,000M3iV2 +
10,285,000JV4,
(vii) 657,931 - 24 Ф25(д) = 392.931М5 AT + 265,000M2N3,
(viii) 3,392,780,147 + 6960 Ф27(д) = 489,693,897M7 + 2,507,636,250M4N2 +
395,450,000MiV4,
(ix) 1,723,168,255,201 - 171,864 <f>29{q) = 815,806,500,201M6iV +
881,340,705,00OM3iV3 + 26,021,050,000iV5,
(x) 7,709,321,041,217 + 32,640 Ф31(д) = 764,412,173,217M8 +
5,323,905,468,000M5N2 + l,621,003,400,000M2JV4.
Note.
dL L2 - M dM LM - N J dN LN - M2
and
ql T^' q^T =т' and q^T т
dq 12 dq 3 dq 2
Examples. Define, for \q\ < 1,
(Thus, Ф„» = *,(*)¦) Then
(i) 20,736 Ф4,М = 15LM2 + 10L3M - 20L2iV - 4MN - L5,
(ii) 1728 Ф2>7'(д) = 2LM2 - MN - L2N,
(iii) 3456 ФЪ,М) = L3M -3L2N + ZLM2 - MN.
All of the foregoing results may be found in Ramanujan's paper [11],
[16, pp. 141, 142], where the method of proof is indicated.
Let <x>l and co2 denote two complex numbers linearly independent over the
real numbers. Put о = mco1 + noJ, where m and n are integers. Recall that
the Weierstrass 0> function SP{z) is defined by
where the sum is over all pairs of integers (m, n) ф @, 0).
In order to prove Entry 14, we shall need the following facts about
and Eisenstein series taken from Apostol's text [3, pp. 12, 13], as well as a
lemma.

15. Asymptotic Expansions and Modular Forms 331
Pror n > 1, put
b(n) = 2Bn + l)CBn + 2)?2n+2(r), A4.1)
where ?„(т) is defined by (9.4). Then, for n > 3,
Bn + 3)(n - 2)b(n) = 3 "X b(fc)b(n - 1 - k). A4.2)
t=i
(This is a more explicit version of the first part of Entry 10(i).) Furthermore,
for | z | sufficiently small,
where a^ = 1 and w2 = т, with те/. Lastly, ^(z) satisfies the two differential
equations
{SP\z)Y = 40>3(z) - 20ft(l)^(z) - 28bB) A4.4)
and
^"'(z) = 6^>2(z) - 10b(l). A4.5)
In fact, A4.2) follows immediately from A4.5).
Lemma. We have
Proof. Differentiating A4.5) twice, we find that
^D)(z) = 12.^'(zJ + \23P(zH"\z\ A4.6)
Also, by A4.5),
12^(z)^"(z) = 72^3(z) - 120b(l)^(z), A4.7)
and by A4.4),
72^3(z) = 18^'(zJ + 360b(l)^»(z) + 504feB). A4.8)
Substituting A4.8) into A4.7), we find that
12^(z)^"(z) = 18^'(zJ + 240b(l)^(z) + 504feB). A4.9)
Substituting A4.9) into A4.6), we complete the proof. ?
If n is an even positive integer, Ramanujan now defines
5„ = ^ + (-1)хЛ>
In k=i 1 — q
where \q\ < 1 and В„ denotes the nth Bernoulli number. If и > 1 and
q = ехрBтпт), with x e Ж, then, by (9.4),
(-1)" lB
ln

332 15. Asymptotic Expansions and Modular Forms
Furthermore, from A4.1),
Yb(n), л > 1. A4.10)
In Entry 14, Ramanujan provides a recursion formula for S2n+2 which is
different from A4.2). It should be remarked that in his paper [11], [16, p. 140,
Eq. B2)], where a different definition of Sn is used, Ramanujan gives a very
ingenious proof of A4.2). Rankin [1] has given an elementary proof of A4.2)
as well as some other recursion formulas for S2n- His paper also contains other
references to the literature. However, the recursion formula of Entry 14, which
is incompletely stated by Ramanujan in his notebooks (p. 191), does not
appear to have been given elsewhere in the literature.
Entry 14. If n is an even integer exceeding 4, then
(n + 2)(n + 3) _
2^1Г"+2
n + 3 ~ 5k){n ~ 8 ~ 5k)
- 5{k - 2)(* + 3)}S2t+2SB_2Jt,
where the prime on the summation sign indicates that if (n — 2)/4 is an integer,
then the last term of the sum is to be multiplied by \.
Proof. First, rewrite Entry 14 in the form
where n is even and at least 6. With n = 2(m + 1), where m > 2, the last
equality may be rewritten as
(m + 2)Bm + 5)b{m + 1) = l0b(l)b(m - 1) + 10 ^ k(m - кЩкЩт - к)
- Bm2 - m) 'X' b(k)b(m - k), A4.11)
where A4.10) has been employed. Now A4.2) can be written in the form
[m/2]
Bm + 5)(m - l)b(m +1) = б?' b(k)b{m - k), m> 2, A4.12)
where the prime on the summation sign indicates that if m is even, the last
summand is to be multiplied by \. Using A4.12) in A4.11), we find that

15. Asymptotic Expansions and Modular Forms 333
(m + 2){2m + 5)b(m + 1) = №(l)b(m - 1) + 10 ']Г' k{m - k)b(k)b(m - k)
,t=i
- m)Bm + 5)(m - l)b(m + 1).
Thus, it remains to show that, for m>2,
^Bm + 5){m + l)Bm2 - 5m + \2)b(m + 1)
= 2b(\)b(m -\) + \ Цт - k)b{k)b(m - к). A4.13)
Subtracting 2(m + \)b{m + 1) from both sides of A4.13), we see that A4.13) is
equivalent to
iotn(m + l)Bm - l)Bm + V)b{m + 1)
= 2b(l)b{m - 1) + \ k(m - k)b(k)b(m - k) - 2{m + \)b(m + 1),
A4.14)
for m > 2.
Now observe that the first expression 2b(l)b(m - 1) on the right side of
A4.14) is the coefficient of z2m in the power series for 2b{\)SP{z\ by A4.3).
Also, by A4.3), the latter two expressions on the right side of A4.14) constitute
the coefficient of z2m in the power series expansion for ^'(zJ/4. Lastly, the
left side of A4.14) is the coefficient of z2m in the expansion of ^<4)(z)/120.
Thus, A4.14) follows from the lemma above, and this completes the proof.
?
Differentiating A4.5), we find that &>'"{z) = \20>(z)@'{z\ which yields
another recursion formula for b(n) midway in complexity between A4.2) and
A4.14).
At first glance, the material in the next two sections appears uninteresting.
However, it is a precursive introduction to Ramanujan's work in Chapters
18-21 on modular equations. The definition of "modular equation" given
below is Ramanujan's personal one and is different from the standard defini-
definition which he used later and which can be found in Hardy's book [9, p. 214],
for example. See the author's paper [10] for a discussion of the analogies
between these two definitions.
With F(x) = A — x)/2, Ramanujan begins Section 15(i) with the trivial
identity
= A + t)F (t2), A5.1)
written in terms of binomial series. If we set a = 2t/( I + t) and ft = <x2/B — aJ,
then A5.1) may be written as
= M2(«)F(p), A5.2)
where M2{<x) = 2/B - <x). Ramanujan says that Д -- oc2/B - <xJ is a modular

334 15. Asymptotic Expansions and Modular Forms
equation of the second degree. The factor M2(a) appearing in A5.2) is the
"multiplier." Ramanujan also records the following representations for M2(a):
MM =
Each of these formulas for М2(я) is easily verified.
Consider now a more general equation
A5.3)
where /? = К„(а) is a function of "degree n" and F(x) is not necessarily equal
to A — x)~m. The factor М„(а) is the "multiplier" of "degree n." The meaning
of "degree" is not clear. In the sequel, modular equations and multipliers of
degree 2m will be obtained by iteration. We emphasize that in standard
definitions of modular equations, the meaning of "degree" is precise.
Returning to the penultimate paragraph, we derive further modular equa-
equations by iteration. To obtain a modular equation for n = 4, iterate A5.2) to
find that
{«7B-aJ}2
2 - a 2 - a2/B - aJ \{2 - a2/B - aJ}2
4B - a) / ~4
a2 - 8a + 8 \(a2 - 8a + 8J,
Thus, P = a4/(a2 — 8a + 8J is a modular equation of degree 4. This procedure
only yields modular equations when n is a positive power of 2. However,
Ramanujan claims that the modular equation of degree n, for any positive
integer n, is given by
P = 4a" -== . A5.4)
{A + УГ^с)" + A - JT^af}2
Possibly Ramanujan established A5.4) by induction when n = 2m and then
"interpolated" to obtain a general formula for each positive integer n. Note
that when m = 0, 1, 2, A5.4) is in agreement with our previous calculations.
The inductive proof of A5.4) for n = 2m is straightforward, but rather tedious,
and so we shall omit it.
We next calculate the function М„(а) corresponding to A5.4). For brevity,
set
Pn = A +
and
1+ ,/1-a"- 1- Jl-aY
a -a
where n > 1. Then, by A5.3) and A5.4),

15. Asymptotic Expansions and Modular Forms 335
«f м = ^M = -J^L_ = (p-2 - 4a"I/2 = a.
"W F@) FDa"/Pn2) Р„A - aI/2 Р„'
after a straightforward calculation. Observe that М„(а) is a rational function
of a. In particular, if n = 3,
, . 4-a
JV ' 4 -3a
Ramanujan asserts that
M3(a) = 1+2 /? =
and both equalities are readily verified.
In a corollary, Ramanujan claims that "if 2nd be a2 + 2a = /J, then the nth
is y5 = (a + 1)" — 1." We have not been able to discern any connection between
this statement and the original function F. It appears that Ramanujan is
claiming that "modular equations" of degree 2m can be obtained from the
given "modular equation" of degree 2 by iteration. Since
{(x + 1)" - I}2 + 2{(x + If - 1} = (jc + IJ* - 1,
for each positive integer k, Ramanujan's assertion is easily established when
n = 2m, m > 0. As above, Ramanujan evidently used an "interpolative" argu-
argument to establish his corollary for general n. It should be remarked that in his
quarterly reports, Ramanujan defines the nth iterate of a function, for any real
number n, by the same type of interpolative argument. (See Part I [9,
pp. 324-326, 328-329].)
Ramanujan commences Section 15(ii) with the following theorem and
corollary.
Entry 15(ii). "// pth and qth be <p(x) and ф(х) and rth be f(x), then if pth and
qth be q>F(x) and \j/F(x), then rth is fF(x). And also if pth and qth be Fq>{x) and
Ftj/(x) then rth is F/(x)."
Corollary. "Thus we may add or subtract any constant and multiply or divide
by any constant to x in each function or to each function."
What can be said? It appears that Ramanujan is simply attempting to make
some elementary remarks about the composition of functions.
Define, for n = 2m, where m is any nonnegative integer,
F(n)(x) = FF-F(x), A5.5)
where F occurs m times on the right side. In particular, FA)(x) = x.
Corollary (i). lffA\x) = x andfi2)(x) = x2 + Ax, then
/<»>(*) =

336 15. Asymptotic Expansions and Modular Forms
Corollary (ii). IffA)(x) = x and fB\x) = x2 - 2, then
/r^Y (x - ^HF- 4N
/<">(*) =
As above, these two corollaries are statements about the iterates of func-
functions when n = 2m. Ramanujan then presumably is assuming that his formulas
are valid for all positive integers n by interpolation. For both corollaries, the
inductive proofs are completely straightforward.
Entry 15(iii). "///(x) and F(x) be of the pth and qth degree, find <p(x) such that
JyfW=J<pF{x) = x(x) A5.6)
suppose, then the function for the rth degree = cp^1 {x(x)Y and the self-repeating
series is ^/(р(х)/(ф(х)ср'(х)), where n is any quantity and ф(х) any suitable
function. Supposing the series to be S(x) we have
sf(x) V «<№)/'(*)"
¦
We have quoted Ramanujan (p. 192) for Entry 15(iii), which is very enig-
enigmatic indeed. There is no guarantee that the function <p exists. It also is not
clear what a self-repeating series is.
We offer a proof under several assumptions.
Proof. We shall assume that a function cp exists so that A5.6) holds. Without
loss of generality, we assume that p = 1; thus f(x) = x. We furthermore
suppose that q and r are nonnegative powers of 2 with 2 < q <2r. Since F is
of "degree q\ we put F(x) = G(q\x), where G(q)(x) is defined by A5.5). With
our assumptions, A5.6) now takes the form
(p^x) = cp(F(x)), A5.8)
and we are required to prove that
G(r)(x) = (р'ЧФП A5-9)
We shall establish A5.9) by induction on r. For r = 1, A5.9) clearly holds.
We shall now assume that A5.9) holds up to a fixed integer r > 1 and show
that A5.9) is valid with r replaced by 2r. Using A5.8), A5.9) with x replaced
by F(x), and A5.5), we deduce that
= Gi2rlq){F (x))
= GBrlq\Giq){x))
= Gi2r)(x).

15. Asymptotic Expansions and Modular Forms 337
This concludes the proof of A5.9) and Ramanujan's first assertion in Entry
We next prove A5.7). There is now no need to make any restrictions on p
and q, except that pq # 0. We do need to assume that /, F, and <p are
differentiable.
Using the chain rule and A5.6), we find that
SF(x) =
Sf(x) \tF(x)d(<pF)/dx cpf(x)f'(x)J
_ fi//f(x)F'(x) (pF(x)d((pFflqldxVln
yl/F(x)f'(x) (pf(x)d((pF)/dx
p
\l/f(x)F'{x) q
•№)/'(*) <pf(x)
(
\i/,F(x)f'(x)q)
which completes the proof. П
For the example below, which closes Section 15, we again quote Ramanu-
jan(p. 192).
Example. "If 1 = x and II = x2 + 2nx, then if x is great
. 3n(n + 1) n(n - l)(n - 2)x
As in the examples above, we interpret this statement as an example in the
iteration of functions. First, observe that, in the corollary in Section 15(i), the
third function is equal to x3 + 3x2 + 3x, which agrees with A5.10) when n = 1.
Second, by Corollary (i) in Section 15(ii), /<3)(x) = x3 + 6x2 + 9x, which is in
agreement with A5.10) in the case n = 2.
In accordance with our comments made earlier in Section 15, Ramanujan
probably derived a representation for the rth iterate when r = 2m and then
replaced r by an arbitrary positive integer. He then evidently derived a type
of asymptotic formula for the third function and terminated the series to
obtain the given approximation. Thus, for r = 2m, m > 0, define a sequence of
polynomials Pr(x) by Px{x) = x and
P2r(x) = Pr2(x) + 2nPr(x).
We can prove by induction that for r = 2m > 2,
Pr(x) = xr + rnx'-1 + irn(l + (r - 2)n)xr~2
+ \r{r - 2)n2(l 4- n(r - 4)/?)xr-3 + ¦¦¦. A5.11)
If we interpolate by setting r = 3 in A5.11), we find that

338
15. Asymptotic Expansions and Modular Forms
P3(x) = x3 + 3nx2 + \n{n + l)x + f n2(l - n/3) + ¦ • •. A5.12)
The first three terms on the right side of A5.12) agree with those in A5.10).
However, the last term in A5.10) approaches, as x tends to oo, —n{n — 1) x
(n - 2)/2, which differs from f n2(l - n/3) in A5.12).
We do not know how to find a general closed formula for the coefficient
ofx* in A5.11).
Entry 16. If the modular equation of degree n — 1 is
then the modular equation of degree (n — IJ is
f "/«/1 P\ n/R(\ fY^\" ( П П П/R\" ~\~ { П/\ /? /1 ftl"
Proof. For brevity, set
C = ^y, a = </l - a,
b = ^/i _^, and с = УП^у.
The modular equation of degree n — 1 for у as a function of a is
/4C + ac=l,
and the modular equation of degree n — 1 for /? as a function of у is
A6.1)
BC + bc=\. A6.2)
Thus, /? is of degree (n — IJ in a, and we can determine the modular equation
of degree (n — IJ by eliminating у from A6.1) and A6.2). After subtracting
A6.2) from A6.1), we readily find that
A - В /1 \1/n
or
b-a \y
У =
-BY
b-a
Substituting in A6.1), we arrive at
A
'А-В
1 \ 1/n A - В
— \ + a
b~a \ A - B\"
+
l/n
= 1.
— a j I \\b — a
Multiplying both sides by {(A — B)" + (b — a)"}1/n and simplifying, we deduce
that
Ab~aB = {(A - B)" + (b- a)"}1/n,
from which the identity that we sought follows. ?

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Index
Abel-Plana summation formula 221
Abel's formula 88
Absolutely complete series 320
Achuthan, P. 126, 130
Aiyar, M.V. see Ayyar
Aiyar, S.N. 285
Allen, E.J. 109
Andoyer, H. 137
Andrews, G.E. 1, 6, 15, 127, 305
Apery, R. 155, 162
Apostol, T.M. 259, 276, 330
Appell, P.E. 8
Appledorn, C.R. 216
Askey, R.A. 6, 8-9, 36, 41, 86-87,
137, 219, 225
Asymptotic expansions 5, 54-56, 168,
181-184, 185-218, 238-239, 283-
291, 300-314, 337
Atkin, A.O.L. 329
Ayoub, R. 73, 283, 305
Ayyar (Aiyar), M.V. 253, 256, 261-263
В
Bailey, W.N. 42, 48, 59
Barnes, E.W. 214
Batut, С 162
Bauer, G. 24, 145
Bauer-Muir transformation 159, 162
Belevitch, V. 134
Berndt, B.C. 240, 245, 253, 256, 258-
261, 263, 276-277, 293-295, 299,
326, 333
Bernoulli number 36, 301
Bessel functions 51, 58-59, 133, 186,
214
Binet, J. 206, 221
Binomial coefficient identities 69
Birthday surprise problem 182
Blaum, M. 182
Boas, R.P. 226
Bodendiek, R. 276
Boersma, J. 294
Bowman, K.O. 181
Bromwich, T.J.I'A. 238
Brouncker, Lord 145
Brackman, P.S. 299
Buckholtz, J.D. 182
Biihler, W.K. 8
Buhring, W. 77
Burkhardt, H. 206
Carlitz, L. 182
Carlson, B.C. 15
Carr, G.S. 102, 134

356
Index
Catalan's constant 40, 45, 151, 153
Cauchy, A. 187, 262, 293-294,
299
C-fraction 175
Chandrasekharan, K. 262, 276
Character (mod 4) 240
Chowla, S. 262-263, 277
Chrystal, G. 106
Churchill, R.V. 324
Clausen, T. 58
Coddington, E.A. 88
Cohen, H. 6, 151, 162, 175
Complete series 320
Continued fractions 4, 103-107, 112—
166, 168-172, 175-183
of Bessel functions 133
even part 121
of gamma functions 132, 140-164
general convergence 123
of hypergeometric functions, see
Hypergeometric functions
Mi numerator and denominator 105
notation 104-105
periodic 116
tails 105, 107, 112, 115, 119, 120
Copson, E.T. 34, 66, 182, 203,
307
Coulomb approximation 40
E
Edwards, J. 102, 162, 192
Eisenstein series 5, 240, 301, 318-320,
326-333
Elliptic integral 38, 79-80, 281
Erdelyi, A. 44
Euler, L. 103, 106, 114, 129, 132-133,
137, 141, 168, 185
Euler-Maclaurin summation formula 208,
300-302
Evans, R.J. 5-6, 12, 40, 48, 65, 70, 77,
186, 202, 216, 243
Exponential integral 184
Exponential series 181
Fichtenholz, G.M. 232
Fields, J. 6, 12
Flajolet, P. 6, 130, 300
Fletcher, A. 96
Forrester, P.J. 226, 299
Fourier transform 223-224
Frank, E. 126
Fransen, A. 96
Frasch, H. 328
Frullani's theorem 162
D
Darling, H.B.C. 46-47
Dedekind eta-function 253, 281,
305
Dedekind zeta-function 276
Degree of a modular equation 334
Degree of a series 320-321
De Morgan, A. 127, 129
de Saint-Venant, M. 294
Dickson, L.E. 327
Discriminant function 326
Divisor problem 304
Divisor sums 301, 313-314, 319, 327,
329-330
Dixon, A.C. 4, 7, 15, 22, 24-25, 46,
60-62
Dougall, J. 4, 7-9, 11, 247-248
Dutka, J. 8, 40, 43, 47, 145
Dyson-Gunson-Wilson identity 15
Gamma function 5, 96, 172-175
continued fractions, see Continued
fractions
logarithmic derivative 8, 43, 104
Gauss, C.F. 4, 7, 34, 36, 42, 50, 57, 92
Gauss's continued fraction 103, 134
Gauss's theorem 17, 25, 26, 36, 56, 84,
120
Generalized hypergeometric series 8
Gessel, I. 7, 15
Glaeske, H.-J. 276
Glaisher, J.W.L. 192, 262, 315, 327
Glasser, M.L. 6, 226
Goldberg, L. 305
Goldstein, L.J. 276
Gosper, R.W. 7
Gould, H.W. 69
Goulden, I.P. 126, 130
Graham, S. 304

Index
357
Grosjean, С.С. 294, 329
Grosswald, E. 256, 261, 276
Guinand, A.P. 256, 262, 276
I vie, A. 304
Iwaniec, H. 304
H
Hafner, J.L. 304
Halbritter, U. 276
Halphen, M. 327
Hansen, E.R. 69
Hardy, G.H. 2, 4, 7-9, 11, 15, 18, 20-
22, 24, 39, 41, 50, 57-61, 63, 86,
102-103, 145-146, 156, 168, 181,
185, 190, 192, 214, 225, 240, 261-
262, 284, 293, 295, 298, 304-305,
333
Henriei, P. 221, 247-248
Herschfeld, A. 109
Hill. J. 242
Hill, M.J.M. 34
Hurwitz, A. 256, 262
Hypergeometric series 4, 7-102, 186,
196, 245-248
asymptotic formulas 12-15, 40-43,
52, 70-77, 193-205
balanced 13, 48, 70-77, 80
confluent 186, 192-193, 202-205
continued fractions 103, 112, 115-117,
120, 131, 134-137, 139-140, 142-
145, 164-165, 180
differential equations 49, 87-92
partial sums 11-14, 39, 41-42, 45-47
products 48, 58-64
quadratic transformations 48-51, 58,
64, 92-99
Jackson, D.M. 126, 130, 133
Jacobi, C.G.J. 9, 185
Jacobsen, L. 6, 104, 107, 109, 121-124,
132, 137, 141, 146, 148, 156-157,
159, 162, 164-165, 169
Jogdeo, K. .182
Jones, W.B. 105, 116, 121-122, 129,
134
Jordan, W.B. 145
К
Kampe de Feriet, J. 8
Karlsson, P.W. 7
Katayama, K. 276-277
Khovanskii, A.N. 105, 134, 170
Kirschenhofer, P. 276
Klamkin, M.S. 182
Klein, F. 1, 8
Klusch, D. 299
Knopp, K. 268
Knuth, D.E. 182
Kolesnik, G. 304
Koshliakov, N.S. 262
Krishnamachari, C. 256, 262
Krishnan, K.S. 226
Kummer, E.E. 4, 7, 36-37, 42, 48-50,
64, 92-93, 96, 117
Kummer's theorem 17, 21, 24
I
Imperfect series 320
Inclusion-exclusion principle 28
Incomplete series 320
Infinite products 230-231, 241
Infinite radicals 108-112
Integrals 31-32, 54-56, 79-80, 166-
171, 178-182, 185-207, 216-217,
219-227, 229-237, 263-264, 278-
279
Iseki, S. 276
Ismail, M.E.H. 6, 321, 323
Iteration of function 335-338
Lagrange, J. 253, 256, 261
Lagrange, J.L. 133
Lagrange inversion formula 100-102,
198, 202
Laguerre, E. 132
Lambert, J.H. 133
Lamm, G. 40
Lamphere, R.L. 5-6, 243
Landau, E. 304
Landau's constant 40
Lavoie, J.L. 19
Lawden, D.F. 182

358
Index
Lebedev, N.N. 186, 193, 202
Legendre, A.M. 165
Legendre polynomials 65-69
Lehmer, D.H. 330
Lerch, M. 276, 293
Lerch zeta-function 259
LeVeque, W.J. 330
Levitt, J. 329
Lewittes, J. 256
/.-function 259, 276-277
Lindelof, E. 221
Linear series 321
Ling, C.-B. 256, 262, 299
Littlewood, J.E. 305
M
Macmahon, P.A. 28
Madhava, K.B. 150
Malmsten's integral representation for
Log T(z) 162
Malurkar, S.L. 256, 261-263, 276-277,
295
Marsaglia, J.C.W. 181
Masser, D.W. 6, 329
Masson, D. 141, 142, 145
Matala-Aho, T. 38
Mathematical Gazette 182, 295
Matsuoka, Y. 276
McCabe, J.H. 130
McKay, J.H. 108
Mellin, H. 295
Mikolas, M. 276
Mixed series 320
Modified theta-function 5, 314-317
Modular equation 301, 333-335, 338
Modular form 327-329
Modular group 318
Morley, F. 25
Mozzochi, C.J. 304
Muir, T. 129
Multiplier for modular equation 334
N
Nagasaka, C. 276
Nanjundiah, T.S. 261-263, 277, 294-
295
Narasimhan, R. 262, 276
Newman, D.J. 182
Newton, I. 28
Nielsen, N. 32, 165, 181, 184, 206
Norlund, N.E. 135, 156
О
Ogg, A. 320
Olivier, M. 162
Olver, F.W.J. 6, 104, 168, 184, 208,
212-214, 216
Paris, R.B. 182
Parseval's theorem 186, 207, 224-225
Partial fraction decompositions 237, 241,
248-249, 267-275, 277-279, 291-
293, 314-315
Partition function 305
Perfect series 320
Perron, O. 105-106, 112, 120, 132-135,
137, 141, 146, 148, 156, 159, 170
Pfaff, J.F. 9, 36, 93
Phillips, E.G. 294-295
Pincherle, S. 142-143
Poisson distribution, 182
Poisson summation formula 5, 225, 233,
235, 238, 240, 252-253, 264, 316-
317
Poisson summation formula for sine
transforms 236, 257, 265, 289
Pollak, H.O. 217
Pollard, H. 226
Polya, G. 109
Ponnuswamy, S. 126, 130
Preece, C.T. 121, 146, 156, 293, 295
Primes 304
Prodinger, H. 276
Pure series 320
Putnam, William Lowell 108
R
Rademacher, H. 305
Ramamani, V. 329
Ramanathan, K.G. 134, 137, 141, 146
Rankin, R.A. 318-320, 326, 328, 332
Rao, M.B. 253, 256, 261-263

Index
359
Rao, S.N. 276
Razar, M.J. 276
Residue for double pole 324
Residue for triple pole 325
Riemann zeta-function 30-31, 150, 153,
155, 173,276, 301, 306-314
Riesel, H. 262, 276, 295, 305
Riordan, J. 34
Rogers, LJ. 11, 121, 125, 127, 129,
148, 150, 151, 163
Roy, R. 7, 225
Saalschiitz, L. 4, 7, 9, 15, 99, 187
St. Paul 1
Samuels, S.M. 182
Sandham, H.F. 256, 262, 293-295
Sayer, F.P. 293-295
Schlafli, L. 59
Schlomilch, O. 253, 256
Schoeneberg, B. 320
Self-repeating series 336
Serre, J.-P. 320
Shepp, L. 217
Sitaramachandrarao, R. 6, 256, 261, 271,
276, 293, 297
Sizer, W.S. 109
Slater, L. 8
Smart, J.R. 276, 293-294
Srivastava, H.M. 61
Stanton, D. 7, 15, 48, 70, 77, 102
Stieltjes, T.J. 141, 149-151, 154, 156,
163
Szabo, A. 40
Szego, G. 109, 181-182, 184
Terras, A. 276
Theory of indices 320
Theta-function 253, 301, 314, 322-323
Theta-function reciprocal 305
Thomae, J. 7, 39
Thron, W.J. 105, 116, 121-122, 129,
134
Titchmarsh, E.C. 189, 191, 223-225,
235, 257, 289, 301, 304, 306-307,
311,317
Toyoizumi, M. 276
Tricomi, F.G. 44
Trinity College, 5
Tschebyscheff, P. 169
Tschebyscheff polynomials 99
U
University of Illinois 1
Vaananen, K. 38
Vandermonde's theorem 36-37
Vaughn, J. 6
Venkatachaliengar, K. 319
Vijayaraghavan, T. 108-109
Vivekananda 1
W
Waadeland, H. 105, 107
Wall, H.S. 105-106, 112, 134
Wallis, J. 106
Wallis's product 241
Wang, Т.Н. 232
Watson, G.N. 5, 39^1, 43, 47, 51-52,
59, 139, 157, 163-164, 181, 184,
193-195, 214, 256, 262, 293, 295,
297-298, 305
Watson's lemma 203-204, 212-213
Weierstrass P-function 330-333
Whipple, F.J.W. 7, 12, 16, 41
Wigert, S. 311
Williamson, B. 102
Wilson, B.M. 5
Wilson, J.A. 11, 86-87, 219-220, 225
Worpitzky, J. 112
Wrigge, S. 96
Wright, E.M. 304
Zagier, D. 6', 283, 286
Zhang, N. 276
Zhang, S. 276
Zippin, L. 109
Zucker, I.J. 262, 295, 299
Zuckerman, H.S. 305