Infinite Series: Ramifications (Pocket Mathematical Library)

Infinite Series: Ramifications G. M. Fichtenholz

Revised English Edition Translated and Freely Adapted by Richard A. Silverman

Course 4

L!! POCKET MATHEMATICAL LIBRARY

INFINITE SERIES: RAMIFICATIONS

THE POCKET MATHEMATICAL LIBRARY JACOB T. SCHWARTZ and RICHARD A. SILVERMAN, Editors

PRIMERS: 1. THE COORDINATE METHOD by I. M. Gelfand et al.

2. FUNCTIONS AND GRAPHS by I. M. Gelfand et al.

WORKBOOKS:

1. SEQUENCES AND COMBINATORIAL PROBLEMS: by S. I. Gelfand et al.

2 LEARN LIMITS THROUGH PROBLEMS! by S. I. Gelfand et al.

3. MATHEMATICAL PROBLEMS: AN ANTHOLOGY by E. B. Dynkin et al.

COURSES:

1. LIMITS AND CONTINUITY by P. P. Korovkin

2. DIFFERENTIATION by P. P. Korovkin

3. INFINITE SERIES: RUDIMENTS by G. M. Fichtenholz

4. INFINITE SERIES: RAMIFICATIONS by G. M. Fichtenholz

INFINITE SERIES: RAMIFICATIONS BY

G. M. FICHTENHOLZ

Revised English Edition Translated and Freely Adapted by RICHARD A. SILVERMAN

GORDON AND BREACH SCIENCE PUBLISHERS

NEW YORK LONDON PARIS

Copyright © 1970 by Gordon and Breach, Science Publishers, Inc. 150 Fifth Avenue New York, N.Y. 10011 Editorial office for the United Kingdom Gordon and Breach, Science Publishers Ltd. 12 Bloomsbury Way London W. C. I

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All rights reserved. No part of this book may be reproduced or utilized in any form or by any means, electronic or mechanical, including photocopying, recording, or by any information storage and retrieval system, without permission in writing from the publishers. Printed in east Germany.

Preface The present volume of The Pocket Mathematical Library continues the study of infinite series begun in its companion volume Infinite Series: Rudiments, by the same author. Together the two volumes give a detailed treatment of the theory of numerical series, i.e., infinite series whose terms are numbers. The picture is then completed by a third volume, entitled Functional

Series, which, as its name implies, is devoted to the study of infinite series whose terms are functions. The set of three volumes makes up a comprehensive treatise on all aspects of a key topic of pure and applied mathematics. As in the companion volume, the problems appearing at the end of each section constitute an important part of the course, and should not be neglected by the serious student.

v

Contents Chapter 1. Operations on Series

1

1. Associativity of Convergent Series 2. Commutativity of Absolutely Convergent Series 3. Riemann's Theorem 4. Multiplication of Series 5. Toeplitz's Theorem 6. The Theorems of Mertens and Abel

1

5

7 13 21

26

Chapter 2. Iterated and Double Series

30

7. Iterated Series 8. Double Series 9. Examples 10. Power Series in Two Variables

30 35 43

Chapter 3. Computations Involving Series

54 61

11. General Remarks

61

12. Examples 13. Euler's Transformation

63

14. The Transformations of Kummer and Markov Chapter 4. Summation of Divergent Series

15. Introduction 16. The Method of Power Series 17. The Method of Arithmetic Means 18. Application of Generalized Summation to Multiplication of Series 19. Other Methods of Generalized Summation 20. The Methods of Borel and Euler Index

70 78 87 87

90 98

110 113 121 129

vii

CHAPTER 1

Operations on Series 1. Associativity of Convergent Series

The concept of the sum of an infinite series differs substantially from the concept of the sum of a finite number of terms (considered in arithmetic and algebra) in that it involves taking a limit. In certain cases, properties of ordinary sums carry over to sums of infinite series, but usually only under certain conditions to be determined below. In other cases, the ordinary properties of finite series break down in a striking fashion. Thus, in general, we must be very careful when attempting to extend properties of finite sums to the case of infinite series. Our first result along these lines expresses the associativity of

convergent series, analogous to the similar property of finite sums: THEOREM 1. Let (1)

a convergent series, and let'

al +a2 + ... +ak + ... = (al + ... +

(1)

1. Here [nk} is some subsequence of the sequence of positive integers 1,

2, ... such that n1 < n2
Ak,

Ak-1 < An 5 Ak

(4)

for the same values of n. But Ak -+ A as k - oc, where A is the sum of (1) and hence, by (4), An , A as n , oc. Example. Prove the convergence of the series (5)

k

k=1

where [x] is the integral part of x, i.e., the largest integer <x. Solution. Grouping together terms of identical sign in (5), we get k=1

(-1)k

1

+

kZ

(-1)k Ck, k=1

1

k2 +1

+ ... +

1

(k+1)2

-1 (6)

Infinite Series: Ramifications

4 But

k terms

1

2

2k+1

+... +

1

k2+1

k2

r

1

k2+k-1

k+1 terms

+ k + ...

1

1

+ k2

1

+ (k + l )2 -


0, there is an m such that E

Ixnl
m,

and hence, by (2)

Ixnl '< Itnlxl + ' + tnmxmI + Itn.m+lxm+l + ... + tnnxnl < Itnlxl + ... + tnmxmI + (Itn.m+1l + ... + Itnnl)

E

2K E

+ tnmxmI + - if n > m.

Itn1x1 +

(3)

2

But m is fixed in (3), and hence, by (1), there is an N > m such that

Itnlxl + "' + tnmxmI
N.

Combining (3) and (4), we find that Ixn,l < E

i.e.,x;,-0asn-+ oc.

if n > N,

(4)

Operations on Series

23

COROLLARY 1. Suppose the coefficients tnm satisfy the condition8

T. = tn1 + tn2 + ... + tnn --i

I

as n - oc, in addition to the conditions (1) and (2), and suppose

lint xn=a

n-x (a finite). Then

lira XIl = lira (tnlxl + tn2x2 + -'- + tnnxn) = a.

n-.x

n-+m

Proof. Clearly

xn = tnl (xl - a) + tn2 (x2 - a) + ... + tnn (X. - a) + Tna, where, by hypothesis, xn - a-+ 0 as n - co. Hence, by Toeplitz's theorem,

lim x;, = lim Ta = a.

n-.x

n-.x

COROLLARY 2. If

limxn=a,

nix then

lim

xl + x2 + '' + xn

= a.

17

Proof. Choose 1

tnl = tn2 = ... = tnn = n

in Corollary 1. COROLLARY 3. Suppose

limxn=limyn=0,

n-.m

n-,m

where 1Y1

+ 1y21 +

+ Iynl < K

(n = 1, 2, ...)

8. In the applications, we usually have Ti, -- I.

Infinite Series: Ramifications

24

for a suitable constant K, and let Zn = X 1 Yn + X2Yn - I + ... + x,Y 1 Then

limz,,=0.

n- W

Proof. Choose tnm = Yn-m+i in Toeplitz's theorem.

I

COROLLARY 4. Suppose

limx,, = a, limy,, = b, n-ao

n-+w

and let Z,, _

XiYn + X2Yn-i + ... + XnYl n

Then

Jim z,, = ab. n -,

Proof. First let a = 0. Then z,, - 0, by Corollary 3, if we replace yn by y,,1n, noting that 1Yn1 < K (why does K exist?) implies

If a

Yi

Y2

n

n

+

Yn

+

n

5 nK = K. n

0, w e wr ite

Zn -

(x 1 - a) Y1 + (x2 - a) y.- + ... + (xn - a) y, n

+a

+Y2+...+Yn

Y1

n

Then the first term on the right approaches zero, for the reason just given, while the second approaches ab, by Corollary 2. 0 PROBLEMS

1. Given two sequences {x,,} and {y,,}, suppose that a) {y,,} is an increasing sequence with limit + cc ; b)

lim X,. - Xn-

n-- Y.

1

Yn-1

=a

(x0 = Yo = 0) .

Operations on Series

25

Prove that

- = a.

lira X.

n- m Y.

Hint. Choose tnm =

Ym-Ym-1 Yn

2. Given a sequence xo, x1i x2 , ..., xn , ..., let xn =

Coxo + C;x1 + C2"x2 +

+ C,",xn f

2"

where Cm is the binomial coefficient n!

Cm =

(m = 0, 1, ..., n).

m! (n-m)!

Prove that xn -a a implies x;, -, a. Hint. Let C"m

tun,

= 2n

noting that Cm < nm and hence tnn, -- 0 as n --b oo, while n

M=0

Cm = 2n.

3. Given a sequence xo , x1, x2 , . . . , x , x n

= Coxo + Cix1z + Cx2z2 +

... ,

let

+ Cnxnz"

(1 + Z)"

xn"

Coxoz" + CI,xlzn 1 +

C2x2zn_2

+ ... + CIxn

where z > 0. Prove that xn - a implies x -+ a, x' -i a.

Infinite Series: Ramifications

26

6. The Theorems of Mertens and Abel

We are now in a position to prove a theorem generalizing Cauchy's theorem on the multiplication of series: THEOREM 1 (Merten's theorem). Given two convergent series X

n=1

an=a1 +a2 +... +an+

(1)

and M

I

(2)

b,,

n=1

with sums A and B, respectively, suppose at least one of the series is absolutely convergent. Then the product of (1) and (2) in Cauchy's form is convergent, with sum AB, i.e.,

AB = a1b1 + (a1b2 + a2b1) + (alb, + a2b2 + a3b1) + . (3)

Proof. Suppose the first series (say) is absolutely convergent, i.e., suppose the series M

n=1

(4)

lanl =jail

is convergent. Combining terms on the nth diagonal of the matrix (3), p. 13 we write

C = a1b +

anb1,

so that the series on the right in (3) is just c1+c2+....+cn+...,

with partial sums C = c1 + c2 +

+ C

(n = 1, 2, ...).

Thus our aim is to show that C - AB as n -+ co. First of all, it is easy to see that + C. = a1Bn + a2Bn-1 + anB1,

Operations on Series

27

where B. is the nth partial sum of the series (2). Setting B. = B - P. (where the remainder fi -. 0 as n --' cc), we can write C. in the form where An is the nth partial sum of the series (1) and(

yn = a1#n + a2fn-1 + "' + an-1r2 + anf1 Since An--+ A, everything reduces to showing that n- x

But this follows at once from Corollary 3, p. 23 (with x if we bear in mind that

y=

(n = 1, 2, ...),

Ia1I + Ia2I + ... + IanI < A*

where A* is the sum of the series (4). Example 1. It cannot be asserted that the series on the right in (3) converges if both series (1) and (2) are only conditionally convergent. For example, suppose we try to multiply the conditionally convergent series

(-1)n-1

=1-

n

1_

2 +73

+ ... + (-

l)n_1

n

(Ruds., Prob. la, p. 76) by itself. This gives

C. _ (- l)n-1

+

1

-

+ ...

1

1

1

1

+ But

1, since 1

(k = 1, ...,

> n

rt),

+ ...

Infinite Series: Ramifications

28 and hence

M

Y_Cn

n=1

diverges.

Example 2. On the other hand, suppose we multiply the conditionally convergent series

ln2=Y n=1

n

2

+ (-1)n-1

1

3

+ ...

n

(Ruds., p. 117) by itself. This gives

Cn = (-1)n-1 x 1

+

X 1

-n

1

2(n - 1 ) k=0

(-1)n-1

1

k(n -k+ l)

+

+

1

n(n-k+l)

n

n+1 k= =

+

I

(-1)e-1 1 on- I

+

2

n+1

k

n-k+1 1+

1

+ ... +

2

1

n)

Thus {1cJ} is a decreasing sequence converging to zero (why?), and hence M

I cn

(5)

n=1

converges by Leibniz's test (Ruds., p. 73). It is now natural to ask whether the sum of (5) is (In 2)1. The answer is affirmative, as

shown by the next theorem.

Operations on Series

29

THEOREM 2 (Abel's theorem). Given two convergent series (1) and (2), with sums A and B, respectively, suppose their product m

I C = c1 + C2 + ... + Cn + ...

n=1

is convergent, with sum C. Then

C = AB. Proof. With the same notation as before, we easily find that

C1 + C2 + +Cn n

lim

A1Bn

+ AZBn-1 + ...

n-.m

+

n

But the left-hand side equals C, by Corollary 2, p. 23, while the right-hand side equals AB, by Corollary 4, p. 24 (with xn = An, yn = B.).

I

PROBLEM

Suppose the power series x I anxn n=0

has the interval of convergence (-R, R), where R < 1, and suppose the series converges (possibly only conditionally) at the end points x = ±R. Prove that the identity in Problem 3, p. 17

continues to hold for x = +R.

CHAPTER 2

Iterated and Double Series 7. Iterated Series

Given infinitely many real numbers aj'k)

(j, k = 1, 2, ...)

indexed by two integers j and k, imagine these numbers arranged in an infinite rectangular matrix a(1) a2 a(1) a3 a(1) "' 1 (2)

a1

a2(2) a3(2)

a

i

(1)

... aj(2)

j

a (3) a (3) a (3) ... a (3) 1

2

3

(1)

(k) a2(k) a3(k)

a1

... aj(k) ...

Suppose we first sum the elements in each row of (1) separately, obtaining an infinite sequence of series of the form Go

ask)

(k = 1, 2,...),

(2)

j=1

and then sum the terms of this sequence in turn, obtaining the expression

Y > a(k)

k=1j=1

30

(3)

Iterated and Double Series

31

called an iterated series. Alternatively, if we sum the matrix (1) first over columns and then over rows, we get another kind of iterated series, of the form M

I>

acjk>

i=lk=1

(3')

DEFINITION 1. Suppose summation of the elements in the kth row of (1) gives a convergent series, with sum Ack>, for every k 1, 2, ..., and suppose the series A(k) k=1

(4)

is convergent, with sum A. Then the iterated series (3) is said to be convergent, with sum A. DEFINITION 1'. Suppose summation of the elements in the jth column of (1) gives a convergent series, with sum Aj, for every

j = 1, 2, ..., and suppose the series

Y A,

(4')

J=1

is convergent, with sum A'. Then the iterated series (3') is said to be convergent, with sum A'.

The elements of the matrix (1) can be represented as an ordinary sequence U1, u2, ..., ur, ... (5)

in infinitely many ways, and there is an ordinary series ur r= 1

(6)

corresponding to each such sequence.' Conversely, given any ordinary sequence (5), the terms of the sequence can be arranged in infinitely many ways (without regard for order) to make up a 1. A similar situation has already been encountered in connection with the matrix of the special form (3), p. 13.

Infinite Series: Ramifications

32

matrix of the form (1), and there is an inerated series of the form (3), say, corresponding to each such arrangement. Thus it is natural to ask about the relation between the series (3) and (6). This is done in the following two theorems: THEOREM 1. If the series (6) is absolutely convergent, with sum

U, then, no matter how its terms are arranged as a matrix of the form (1), the corresponding iterated series (3) is convergent, with the same sum U. Proof. By hypothesis, the series M

r= 1

(6*)

l url

is convergent. Let U* be its sum. Then, given any k and n,

j=1

al < U*. lik)

Therefore (k)

l aj

j=1

l

converges, and hence so does

aj(k) J=1

since an absolutely convergent series is always convergent (Ruds., Theorem 1, p. 63). Moreover, given any E > 0, there is an integer ro such that r=r0+ 1

l url < E,

(7)

and hence, afortiori, ro

cc

I

r=r0+1

ur

U - Y ur < E.

(8)

F=1

The terms u1i u2 , ..., uro of the series (6) all belong to the first m

columns and the first n rows of the matrix (1) if m and n are

Iterated and Double Series

33

sufficiently large, say if m > m0, n > no. Therefore, by (7), n

m

r0

(m > mo , n > no),

L+ L ajk) - I ur < E

k=1j=1

r=1

since the expression whose absolute value appears on the left is a group of terms ur with indices greater than ro. Taking the limit asm --), cc, we get n

ro

k=1

A(k)_ r=1 u r < E

(n > no),

which, together with (8), implies n

YA(k) - U

< 2E,

k=1 and hence

n

m

A(k) = lim > A(k) = U. k=1

k=1

n-

Remark. Theorem I obviously continues to hold in the case where some of the rows of the matrix (1) contain only finitely many terms. DEFINITION 2. The iterated series (3) is said to be absolutely

convergent if the iterated series (k7

k=1j=1

laj

(9) I

whose terms are the absolute values of those of (3), is convergent. Remark. Similarly, the iterated series (3') is said to be absolutely convergent if the series cc

w

L

J=1k=1

l ajk 7

is convergent. THEOREM 2. Suppose the iterated series (3) is absolutely convergent. Then (3) is convergent, and the series (6) made up of the

elements of (1) arranged in any order whatsoever is absolutely convergent, with the same suns as (3). 3

Fichtenholz(2094)

Infinite Series: Ramifications

34

By hypothesis, the series (9) is convergent. Let A* be its sum. Then, given any m and n, we have [n

``m

L L.

k=1j=1

(10)

A

Iajk)I

Consider any partial sum

U,* =lull

+Iu21+...+lurl

of the series (6*). The numbers u1, u2 , ..., u, all belong to the first m columns and the first n rows of the matrix (1) if m and n are sufficiently large. Hence (10) implies

U* < A* (r = 1, 2, ...), so that (6*) is convergent, i.e., (6) is absolutely convergent. The

rest of the proof is now an immediate consequence of Theorem 1.

Remark. Clearly, Theorems 1 and 2 remain true if (3) is replaced by (3'). THEOREM 3. Given a matrix (1), suppose the iterated series (3) is

absolutely convergent. Then the iterated series (3') converges and has the same sum as (3) : m

n

j=1k=1

n

ajk'

m

k=1j=1

ask )

Proof. By Theorem 2, the series made up of the elements of (1)

arranged in any order whatsoever is absolutely convergent. Hence, by Theorem 1, the iterated series (3) corresponding to the matrix (1) as it stands and the iterated series (3'), which is of the form (3) if we interchange rows and columns in the matrix (1), both converge and have the same sum. 0 Remark. Clearly Theorem 3 remains true if we interchange the roles of (3) and (3').

Iterated and Double Series

35

PROBLEM

Interpret Theorem 1 as a generalization of both Theorem 1, p. 1 and the theorem on p. 5. 8. Double Series

There is another kind of series associated with the infinite rectangular matrix (1), p. 30, namely the series of the form ail) + aZl) + ... + aj.l) + ... +aiz)+aZZ)+....+;Z)+...

+a2)+... +a" + ...

+ al +

called a double series and denoted by x

a(k)

(1)

J.k= 1

(with only one summation sign). By a partial sum of (1), we mean a finite sum of the form 111

n

An.

J=1k=1

aJ

made up of the terms in the first m columns and the first n rows of the given matrix. Let m and n approach infinity independently. Then the limit'

A = lim

(2)

nt-.Ft N -M

2. The limit in (2) is a double limit. Thus if A is finite, (2) means that, given any e > 0, there is an integer N such that Ann)

Whenever m and n both exceed N.

- AI < e

36

Infinite Series: Ramifications

(finite or infinite) is called the sum of the double series (1), and we write

A=

a(k) j,k= 1

J

'

The series (1) is said to be convergent if it has a finite sum, and divergent otherwise.

It is now natural to compare the double series (1) with the iterated series a(k)

L k=1j=1

(3)

and rc

.0

(3') J=1k=1 Since m

n

(k)

(n)

An' - k=1j=1 L aJ we have n

lim

A(k) k=1

m-.cc

after taking the limit as m -+ co for fixed n. Here A(k) = > ask)

(k = 1, 2,...),

J=1

(4)

and every "row series" on the right is assumed to converge. Thus the sum of the iterated series (3) is just the iterated limit

lim lim A(?n).

n-.ac) nl-r

Similarly, the sum of (3') is the other iterated limit lim lim Am(n) m-.ao n-.oc

(give the details). THEOREM 1. If the double series (1) converges and if every "row series" (4) converges, then the iterated series (3) also converges and

Iterated and Double Series

37

has the same sum as the double series:

Lj

j

k=1j=1

- j.k=1 l aj

aj(k) _

(k)

(5)

Proof. Given any E > 0, there is an integer N such that Al < E, provided that m, n > N. Holding n > N fixed, we take the limit as In - cc, obtaining

AM-A I k=1

C. I

Taking the limit as n - oo now gives

Y A(k)_A k=1

which implies (5).

1

Remark. There is an obvious analogue of Theorem 1 for the case of the iterated series (3'). DEFINITION 1. A double series (1) such that afk)

0

(j, k = 1, 2, ...)

is said to be positive. For such series we have the exact analogue of a familiar result for ordinary positive series (Ruds., Theorem, p. 9): THEOREM 2. A positive double series always has a sum. The sum is finite, and hence the series converges, if the partial sums of the series are bounded from above. Otherwise the sum is infinite, and hence the series diverges.

Proof. The last assertion is obvious. To prove that a positive

double series (1) converges if its partial sums are bounded, suppose

0, there is a partial sum AM(o) such that

Anno'>A-E, by the very definition of A. Choosing m > mo, n > no, we have

A;n"'>A-E, a fortiori, since A;"' obviously increases with both m and n. Hence

Al < E

if m, n > N = max imo, no}, since no partial sum exceeds A. But then A = llm A,,,"', m-.y

n-.m

as asserted.

Next we consider arbitrary double series, i.e., double series whose terms can have values of either sign. Just as in the case of ordinary series, we need not consider the case where all the terms of the series are negative or the case where there are only a finite number of positive or negative terms, since these cases immediately reduce to the case of positive series (cf. Ruds., p. 63). Thus we will assume that the double series (1) has both infinitely many positive terms and infinitely many negative terms. The following definition and theorem are the exact analogues of those for the case of ordinary series (Ruds., pp. 63, 65): DEFINITION 2. A double series (1) is said to be absolutely convergent if the series

Z a/'

(1*)

made up of the absolute values of its terms is convergent. A double series which is convergent but not absolutely convergent is said to be conditionally convergent.

Iterated and Double Series

39

Remark. Thus (1) is absolutely convergent if and only if the double series corresponding to the infinite rectangular matrix la(1')I lazl'1 ... Iail)l ... 1a12)I lazz)I ...

1a(2 k)l

Ia(1k)I

Iajz)I ...

... Iaik)I ...

is convergent. Similarly, the interated series (3) and (3') are absolutely convergent if and only if the iterated series corresponding to this "absolute value matrix" are convergent (cf. Definition 2, p. 33). THEOREM 3. If the series (1*) converges, then so does (1).

Proof Let (k)

(k)

aj

(k)

= Pj

- 4j

where (k) _ Iai I + aj = 2

(k)

(k)

Pj

(k)

(k) _

l aj

=

91

(k)

I - aj 2

Since

Pj< lajIk' , 9jS k) k' Iaik' 1, the convergence of (1*) implies that of the double series (k)

ilk= 1

/

r

ao

pi = P,

(k)

1 7j

_

j,k=1

^ Q

(see Problem 5). But then the series (k)

J,k=1

aJ

=Y

J,k=1

(Pjk)

- 9j

,

)

also converges (see Problem 3), and in fact has the sum A =

P-Q. Finally we prove a result analogous to Theorems 1 and 2 of the preceding section:

40

Infinite Series: Ramifications

THEOREM 4. Given a double series (1), let ur

(6)

r=1

be an ordinary series consisting of the same terms written in any order whatsoever. Suppose either of the series (1) and (6) is absolutely convergent. Then so is the other series, and both series have the same sum.

Proof. Suppose (1) is absolutely convergent, so that (1*) is convergent with sum A* (say). Then, just as in the proof of Theorem 2, p. 33, it is easy to see that

U* 1) which are

Infinite Series: Ramifications

44

not powers, then

G=I

1

_

+mm3-1 +... 1

mm2-1 1

+

1

m-1 3

m4

+

+ ...

m6

+(m3 +m6

+...)

in9

- m { \m2 + m3

+

+\m6

+...1

...) + (J4 +

1

m6

+...1

m9...)

I J(;W(M-

I

1

+ m2 (m2 ` 1) + m3 (m3 - 1) + 1)

(justify the various rearrangements). It follows that

- n=2 n (n

G_

°°

1

where n now ranges over all positive integers starting from 2. Therefore

OD

G2: ( n=2 n - I

1

1

n

Example 3. Consider the matrix with general term (j -_ 1)! (k - 1)! ck> aj =

j(j+ 1)...(j+k) k(k+ 1)...(k+j)

(where 0! = 1). Setting a = 0, p = k in the formula 00

1

Y.

n=i

(a + p)

(a + p)

(1)

Iterated and Double Series

45

(Ruds., Prob. 2, p. 4), we can easily sum the terms of the kth row :

OD

a(k)

= (k

k - k!

k2

Hence the sum of one iterated series is 00

k

00

CO

J=1

,

1

()

k

k

Because of the symmetry of a() with respect to j and k, the other iterated series is identical with the first, and nothing new can be deduced by equating the sums of the two iterated series. Suppose we now modify the matrix as follows: We retain the first k -- 1 terms in the kth row, but replace the kth term by the sum of all the terms of the kth row starting from the kth, dropping the remaining terms altogether. This gives the new matrix

rl a(2) 1

r2

a(3) 1

a2

(3)

r3

(k) a2(k) a3(k) a1 a1(k+1) a2(k+1) a3(k+1)

(k) ak-1 rk

(k+1) ...ak-1

(k+1) rk+1 ak

with the same "row sums" as the original matrix. Hence the sum of the first iterated series has the same value (2) as before. To sum the matrix "by columns," we first calculate 00

rk

(k)

00

(k - 1).

a( k)

CO

-v

n=1 (k - 1

(k - 1)(

k2(k+1)...(2k- 1)

1 + n) (3)

46

Infinite Series: Ramifications

using formula (1) with a = k - 1, p = k. The sum of the remaining terms of the kth column equals `L

ak`)

(k - 1)!

_

i=k+1 i (i + 1)

i=k+1

11

(i + k) (k - 1)!

= y

n=1 (k+n)(k+n+1).-(2k+n) (k- 1)!

_

(4)

k(k+ 1)...2k

where we seta = p = k in (1). It follows from (3) and (4) that the sum of all the terms in the kth column is just (k - 1)!

1

k (k +

=3

1

k +

1)

2k)

(k - 1)!

= 3 [(k - 1)!]2

k (k + 1) . . . (2k - 1) 2k

(2k) !

Using Theorem 3, p. 34 to equate the sums of the two iterated series, we arrive at the interesting formula Y

1

= 3 > [(k - 1)!]2

(5)

k=1

k2

k=1

(2k)!

The series on the right converges very rapidly, thereby facilitat-

ing the approximate calculation of the sum of the important series on the left.4 Example 4. Investigate the convergence of the double series a)

1

(o > 0).

(6)

.%.k=1 (% + k)°

4. An argument which will not be given here shows that this sum equals :72/6. See G.M. Fichtenholz, Functional Series (in the Pocket Mathematical Library), Sec. 7, Problem 7b.

Iterated and Double Series

47

Solution. Arrange the terms of (6) as an ordinary series, com-

bining terms on each diagonal of the corresponding matrix. Since the terms on each diagonal are equal, this gives (7)

n=2

t1

But clearly

1n

< n - 1 < n,

2

and hence, dividing by n°, we get

2 n °1

1

5 (n - 1)


2 and

diverges if a < 2. Therefore, by Theorem, 4 p. 40, the same is true of the double series (6). PROBLEMS

1. By applying Theorem 3, p. 34 to the positive matrix

1.2

2.3

3.4

4.5

1

1

1

2.3

3.4

4.5

1

1

3.4

4.5 1

4.5

Infinite Series: Ramifications

48

(the missing elements can be replaced by zeros), prove that the harmonic series

1+

+...+

1

2

+...

1

(8)

n

must diverge.

Hint. If (8) had a finite sum s, then one iterated sum would equals, while the other would equal s - 1, contrary to the theorem.

2. Prove that =

2 k=2 mk

(the same power can occur more than once in the left-hand side).

3. If Jxi < 1, the Lambert series M

q9(x) =

x" an

n=1

(9)

1 - x"

converges for precisely the same values of x as the power series M

J(x) = > anx"

(10)

n=i

(Ruds., Prob. 3, p. 86). Let R > 0 be the radius of convergence of the series (10) and suppose lxi < R as well as Ixi < 1. Prove that M

9 (x) _

IXnxn,

n=1

where IXn = I ak kIn

and kin means that the sum is restricted to those positive integers which are divisors of n.

Hint. Noting that x"

1 -x"

= xn + x2n + ... + xkn + ...

(Ixl < 1),

Iterated and Double Series

49

consider the matrix a1x a1x2 a1x3 a1x4 a1x5 alx6 a1x7 a1x6 a1x9 a1x10 ... a2x6

a2x4

a2x2

a2x8

a3x6

a3x3

a2x10...

a3x9

a4x4

... ..

a4x8

asx.lo ...

asx5 a6x6 a7X7

aex8

a9x9

a10Xto ...

where identical powers of x appear in the same column and empty spaces can be replaced by zeros. Now equate the corresponding iterated series (why is this justified?).

4. Prove that W

xn

I r(n) xn,

YT

n= 1

nxn n=1 1 - Xn

=

`'

n=1

l J o(n)

x",,

where z(n) is the number of divisors of n and a(n) is the sum of the divisors of n. Hint. Use the result of the preceding problem. Comment. It is easily verified that R = 1 in both cases, so that we need only assume that Jxj < 1. 4

Fichtenholz (2094)

Infinite Series: Ramifications

50

5. Prove that co

49(x) _ >.f(x"),

(12)

n=1

where the functions c and f are the same as in Problem 3. Hint. Arrange the elements of the matrix (11) without "gaps," in the form a1X a1x2 a1x3 a1X4 ... a2x2 a2x4 a2x6 a2x8 ...

a3x3 a3x6 a3x9 a3x12 ... a4x4 a4x8 a4x12 a4x16 ...

6. Prove that (aX)"

n=1 1 - x"

-r n=i

ax"

1 - ax"

(lal < 1, Ixl < 1).

Hint. Use (12), choosing an = an. 7. Given two power series

f(x) = Y

amx"',

m=1

g(x) = > bnx", n=1

prove that M

GC

I amg (xm) =

m=1

n=1

bn.f (x")

(Ixl < 1),

where x is such that both series converge absolutely. Hint. Consider the matrix with elements ambnx"", noting that Iambnxmnl < l amxml Ibnx"I

,

since mn > m + n (m > 1, n > 1), from which it follows that the corresponding double series is absolutely convergent.

Iterated and Double Series

51

Comment. The choice b = 1 (n = 1, 2,...) gives (12), since then

x 1 - x

8. Prove that the "power series in two variables" xJyk

(13)

J,k=O

is absolutely convergent if Ixi < 1, Iyj < 1 and divergent otherwise.

Hint. Use Cauchy's theorem (p. 14), noting that (13) is the product of the two ordinary power series 00

yk k=O

Comment. The study of power series in two variables is pursued in the next section.

9. Prove that the double series

(a>0,P>0)

1

J.k=1 j ak

is convergent if a > 1, 8 > 1 and divergent otherwise. 10. Consider the double series M

J. k 1

ask) _

1

J. k i (Aj2 + 2Bjk + Ck2)°

(e

> 0),

)

(14)

where the quadratic form Axe + 2Bxy + Cy2 is positive definite, so that A = AC - B2 > 0, as well as A > 0, C > 0. Prove that (14) converges if o > 1 and diverges otherwise, thereby generalizing Example 4. Hint. If M is the largest of the numbers JAI, IBS and ICI, then

a(k) >

1

1

M° (j + k)2p

Infinite Series: Ramifications

52 Moreover

aI(k)

= C [(AC - B2)j2 + (Bj + Ck)2] > C j2,

and hence aj(k)

dC

A \Q

(k)

,

aJ

j29

d

)Q

I

1

kzp

which implies

(,IAC)' ,4

1

j °kQ

11 . Give an example showing that in Theorem 1, p. 36, one

cannot drop the condition that every row series be convergent. Hint. The double series corresponding to the matrix

is convergent, with sum zero, but its row series diverge. 12. Sum the following double series: °°

a) C)

1

m, n=2 (p +

n)m

(P > -1);

m

1

m,n= 1

(4n - 1) 2m+1

rm

e) m, n` 1 (4n

1

2) 2m

b) m=2, n=1 (2n)m °°

1

d) m.n=1 (4n - 1) 2m

Iterated and Double Series

53

Hint. Resort to iterated series, first summing over m. Use the expansions

In2 = 1 -

-

+ 2

4

+ 4

3

3

7

5

(Ruds., pp. 115, 117). Ans.

a)

1

;

p+l

b) In 2; c) -7' 8

12

In 2;

d)

14 In 2;

e)

.

8

13. Consider the function of two variables 99 (x, z) = es'2 (z -z-1)

(z 0 0).

Prove that5 M

9' (X, Z) = I JJ(X) Z"' where

(-1)k

k X

J"(x)

= o k! (k + n)!

Z

(15) 2k+n

(2)

if n >, 0 and (- 1)k

J(C = x

k=-n k! (k + n)!

X

2k+n

(2)

if n < 0. Show that J-.(x) _ (- 1)" Jn(X) Hint. Multiply the absolutely convergent series e

(x'2)z _ W

X , ZJ

j= 2)

j!

e

X k (- 1)kZ-k

(xj2)z-1

k!

k=O

(Rods., p. 113) to get the absolutely convergent double series X,

5

n=-w

X

+k (-I)k f-k z

an is shorthand for the sum _7 an + n=0

a-n. n=1

54

Infinite Series: Ramifications

is called the Bessel function of Comment. The function order n, and plays an important role in mathematical physics, celestial mechanics, etc. Because of the expansion (15), the func-

tion ip (x, z) is called the "generating function" of the Bessel functions. 10. Power Series in Two Variables

By a power series in two variables x and y is meant a double series of the form M Y_

j, k=0

(1)

ajkxJYk,

involving nonnegative integral powers of the variables x and y multiplied by numerical coefficients aJk. Just as in the case of ordinary power series in a single variable (Ruds., Sec. 11), we now consider the problem of finding the "region of convergence" of (1), i.e., the set .,t of all points M = (x, y) for which (1) converges. LEMMA. If the power series (1) converges at a point M = (z, y) whose coordinates are both nonzero, then it converges absolutely at every point M = (x, y) satisfying the inequalities IxI < IxJ, !YI < IYI, i.e., in the whole open rectangle with center at the origin

and vertex at M. Proof. The proof is virtually the same as in the single-variable case (Ruds., p. 68). Suppose the series W

Y

ajkxJYk

j. k=0

converges. Then its general term approaches zero (see Problem 4, p. 42), and hence is bounded, i.e.,

Iajkxjykl < C

(j, k = 0, 1, 2, ...)

for some constant C. Therefore J

I ajkxJYk) = I aJkxjykl

X rI

y

R(O). 6. The distance from 0 to M is denoted by OM.

56

Infinite Series: Ramifications

Note that if R(O) = + oo for even a single ray, then, by the lemma, the series (1) converges (absolutely) in the whole xyplane, i.e., the "region of convergence" reduces to the whole xyplane. Excluding this case of an "everywhere convergent" series, we now assume that R(O) is finite for all O. Let MB be the point of the ray OL such that OMB = R(O).

Then the "boundary point" MB separates the points of the ray for which (1) converges (absolutely) from the points for which (1) diverges. At the point MB itself, the series (1) may or may not

converge, depending on the particular series under consideration. Drawing the vertical line PP' and the horizontal line QQ' through MB, as shown in the figure, we can assert (using the lemma) that the series certainly converges inside the rectangle OPMBQ and diverges inside the (right) angular sector Q'MBP'. Therefore on a new ray OL', corresponding to another angle 0', the series must converge at every point of OA and diverge at every point of BL'. Hence the boundary point MB, must lie between A and B on OL'. This makes it clear (why?) that R(O) varies continuously as 0 varies from 0 to nr/2. In other words, the point MB describes a continuous "boundary curve" as 0 varies

from 0 to 7r/2. Since the abscissa xe of the point M. is nondecreasing as 0 - 0 while the ordinate ye of M. is nonincreasing, xe and y9 must both approach limits as 0 -+ 0. Hence R(O) must also approach a limit as 0 -> 0. Suppose this limit

Ro = lim R(O)

(2)

B-o

is finite. Then, as 0 - 0, MB approaches a limiting point MO* =(R,, 0) on the x-axis.' If (2) is infinite, the boundary curve must 7. The limiting point Mo need not coincide with the boundary point Ma on the x-axis itself. In fact, MO may lie to the right of MM (or even lie at infinity). This possibility does not contradict the lemma, which applies only to points off the coordinate axes.

Iterated and Double Series

57

have an asymptote parallel to the x-axis, which may coincide with the x-axis itself. The same considerations apply to the case where 0 - n/2, with the roles of the x and y-axes interchanged. The above construction gives the boundary curve in the first quadrant. To complete the construction, we now reflect this portion of the boundary curve in both axes and in the origin. This gives the full boundary curve, determining the "region of convergence" .&. In fact, the series (1) converges in the part of the plane lying inside the boundary curve and diverges in the part of the plane lying outside the curve.8 On the boundary curve itself, the series may or may not converge, depending on circumstances. PROBLEMS

1. Sketch the region of convergence for the series xlyk

(3)

j,k=O

(already considered in Problem 8, p. 51). Ans. The square shown in Figure 2. Y

Figure 2

8. Except possibly on the parts of the coordinate axes lying outside the curve, where it may turn out that the series converge (see footnote 7).

Infinite Series: Ramifications

58

2. What is the sum of the series (3) in its region of convergence? Ans.

1

1

1-x 1-y

3. Find the region of convergence of the series Go

J.k=1

xJY"

differing from (3) in that the summation indices begin with 1 rather than 0. Ans. The same square as in Problem 1, together with the coordinate axes themselves. Comment. In this case, even though the boundary point M9 approaches the point MO* = (1, 0) as 0 - 0, the series converges on the whole x-axis (cf. footnotes 7 and 8). 4. Find the region of convergence of the series xJyk

J.k o j! k! Ans. The whole xy-plane.

5. Find the region of convergence of the series

J.k=o

j! k!

(4)

XJY

Hint. A necessary and sufficient condition for the absolute convergence of the series (4), i.e., for the convergence of the series

+ k)! J.k=o

j! k!

IxIJ IYI",

is the convergence of the series

"=o

(Ixl + IYI)" _

I

I - IxI - IYI

(4)

Iterated and Double Series

59

obtained by summing (4') "by diagonals." This leads to the condition IxI + IyI < 1. Ans. The square shown in Figure 3. Y

Figure 3

6. Find the region of convergence of the series

IXJy1,

j>k

+ X2y2 + X3y2 + X4y2 + ...

.

(5)

Hint. A necessary and sufficient condition for the absolute convergence of the series (5), i.e., for the convergence of the series J>k

Ix1J Iylk,

(5')

is the convergence of the series

(1 .+ IxI + IX12 + ...) (1 + Ixyl + Ixyi2 + ...) 1

1

1 - IxI 1 - Ixyl obtained by summing (5') "by rows." This leads to the conditions IxI < 1, Ixyl < 1.

Infinite Series: Ramifications

60

Ans. The region shown in Figure 4, bounded in part by the equilateral hyperbolas xy = ± 1.

Figure 4

7. Generalizing double series, discuss multiple series of the form9 Y,

ajk...{

(6)

j, k,..., 1=1

and power series in several variables, of the form W

ajk...1Xjy k ... Z I

How should the sum of a series like (6) be defined? 9. In discussing iterated and double series, we favored the notation a( k) over the simpler notation ajk, merely to emphasize the distinction between , rows and columns in the infinite matrices figuring in Sees. 7-9.

CHAPTER 3

Computations Involving Series 11. General Remarks We now consider the problem of using infinite series to make

approximate computations. Suppose we want to calculate a number A which is known to have a series expansion of thu form

A=al+a2+... +an+...,

(1)

where the terms a1, a2, ... are readily computable.' Then we have the approximation

A : Anal+a2+...+an, where An is the nth partial sum of the series (1). The error committed in making this approximation is obviously just the sum of the omitted terms, i.e., the remainder

an =an+1 +an+2 +...

(after n terms). Clearly, an can be made arbitrarily small for sufficiently large n. Our job is to find simple estimates for an. We will then be in a position to determine the value of n for which An approximates A to within any given accuracy. If (1) is an alternating series whose successive terms decrease in absolute value, i.e., if (1) is a series of the Leibniz type, then the remainder an has the same sign as its first term and a smaller

absolute value than its first term (Ruds., p. 74). Thus, in this 1. In fact, the terms a1, a2, ... are usually rational numbers. 61

62

Infinite Series: Ramifications

case, we already have at our disposal a technique for estimating an, which leaves nothing to be desired from the standpoint of

simplicity. The situation is more complicated for a positive series. To estimate the error a,,, we look for another, easily summable positive series whose terms are larger than the corresponding terms of an. For example, in the case of the series 1

?n=1 Y mz

we have the estimate2

a_

Y

1

1

m=n+1 m (m - 1)

m=n+1 m2

_m=n+l i (_1 m-1

1

_

m)

1

n

Remark. One is ordinarily interested in decimal approximations to the number A, although the terms a1, a2, ... rarely turn

out to be terminating decimals. Thus the "round-off error" committed in writing a1, a2, ... in decimal form must be taken into account, as well as the "truncation error" a,,. Even when the series (1) has simple terms and a remainder which is easily estimated, it often fails to be a practical way of calculating A, due to its having too slow a rate of convergence. In other words, the partial sums A,, A 2, ... may approach their limit A too slowly as n -> oc. For example, the series

1-+ 2 3- 4+ ..., 1

1

1

1

1

(2)

1-

3

+

5

-

1 +... 7

converge to In 2 and .7r/4, respectively (cf. Problem 12, p. 52). However, to calculate these numbers to within 10-5, we need 100,000 terms in the first case and 50,000 terms in the second 2. The same estimate has already been found in Ruds., formula (6), p. 45, in connection with the integral test.

Computations Involving Series

63

case! Obviously, such calculations are feasible only with the help of a high-speed calculating machine. As will be shown in the next section, it is not very hard to find series expansions of a and In 2 which converge much more rapidly than (2). PROBLEM

Let an be the remainder after n terms of the series 1 + Prove that 1

an < - .

n! n

Hint. Note that

11-

I n! m=n+1 (n + 1) ... m 1

m=n+1 m!

I

1

1

n! m-n+1 (n +

1)m-n

12. Examples

The following examples illustrate the technique of using series to make numerical computations: Example 1. Find a rapidly convergent series suitable for calculating 7c.

Solution. Setting x = 1/ .3 in the expansion

x' (-1 x 1) arctanx=x--+---+ x3

x5

3

5

7

(Ruds., p. 115), we obtain

-a = arc tan 6

1( N/3

(1

I

3 1

1

1

1

3 3 + 5 32

1

1

7 33

+ ...

,

Infinite Series: Ramifications

64

which is much more suitable for calculating a than the slowly convergent series

n= 1- 1+ 1- 1+ 4 3 7

...

5

To get a series which is even more suitable for calculating a, let 1

a = arc tan - . 5

Then 2 1

1 -

5

10

tz

120

1 - 44

119

tan 4a =

5

tan 2a =

tan a

12

25

Since tan 4a is close to 1, it is clear that the angle 4x is close to a/4. Setting

we have 120

_ 1

tanY = 119

=

1

239

1 + 119

so that

= arc tan 239

It follows that

7r =16a-4(7 1

C1

16

3 53

5 1

11

1

1+ 511

1

1

1

1

7 57

5 55 4

1

1

1

1

9 59 1

(1)

239

3 2393

This series is very rapidly convergent (see Problem 2).

Computations Involving Series

65

Example 2. Find a rapidly convergent series suitable for calculating In 2. Solution. Setting

x=

1

2n+1

in the expansion In

1 + x = 2x I+

1-x

x2

+

x4

3

+

5

1)

)

(Buds., p. 117), we get

1+1

2

Inn+1-

n

1

3 (2n + 1)2

2n + 1

+1

1

+

5 (2n + If (2)

(cf. Ruds., p. 124). For n = 1, this gives the rapidly convergent series

In 2 = 2 (1 +

3

3 9 + 5 92

+

(3)

(see Problem 3).

Example 3. One way of calculating roots is to use a table of logarithms. Another way is to use the binomial series

(1 +x)'"= 1

+mx+m(m- 1)Y2

2!

m (m - 1) (m - 2) 3!

x3 + ... (IxI < 1)

(4)

(Buds., p. 120). Let a be an approximate value of the root VIA, where a may be an overestimate or an underestimate. To improve this approximation, suppose, say, that A ak 5

Fichtenholz (2094)

=1+x,

Infinite Series: Ramifications

66

where Jxl is a small proper fraction. Then

VA=ak/Q =a(l+x)llk,

(5)

and we can use (4) with m = Ilk. Alternatively, it is often more convenient to start from the formula ak --1+x', A

where Ix'I is again a small proper fraction. This time

a = a (1 + x')-llk,

VA=

(5')

ak

and we can use (4) with m = -1/k. Thus, for example, to sharpen the approximation

J2 : 1.4, we write 2 J2= 1.4J196

1.4

.04

1

1.4(1

+01.96

ill + 49) 1

or 1.4

J2 =

1.96 NI

=1.4 11-50

1.4

0.04

1

2

2

To keep the calculations simple, we naturally prefer the second formula, which leads to the rapidly convergent expansion

J2 = 1.4 1 + 35

1

3

1

2 50 +

8

502

1

1

63

1

+ 16 503

1

+ 128 504 + 256 505 (see Problem 8).

5

F

J

(6)

Computations Involving Series

67

PROBLEMS

1. Prove that az

1

1

4

2

3 23

1

1

1

1

1

5 25 1

1

+ I3 - 3 33 + 5 35 1

Hint. Choose x = -, y = 3 in the formula

arc tan x + arc tan y = arc tan

x +y

1 - xy

valid for

Iarc tan x + arc tan yj < a-` 2

(explain this condition). 2. Use formula (1) to calculate a to seven decimal places, mak-

ing sure to take round-off error into account. Hint. It is enough to compute the terms already written out in (1).

Ans. r = 3.1415926... 3. Use formula (3) to calculate In 2 to nine decimal places. Hint. Nine terms of the series suffice. In fact, if A is the error due to dropping all terms from the tenth on, then

A =-23
0)

(1)

be a convergent series, where for convenience we write, the kth term in the form (- 1)k akxk without assuming that all the a, > 0. For the sequence {ak} we introduce the successive differences

dak = ak+1 - ak, 42ak = dak+1 - dak = ak+2 - 2ak+1 + ak, and, in general, d Dak = AD-1a,+1

-

A--lak

+ (-1)'ak,

= ak+D - C'ak+p_1 + Czak+y-2

(2)

where C; is the binomial coefficient o _

C'

P!

.i! (p - i)!

We then write the series (1) in the form as

ajx - aox

S(x) =

+

a2x2 - a1x2 _ a3x3 - a2x3 + .. 1 + x

1 + x

(3)

Infinite Series: Ramifications

72

This is permissible since the (k + 1)th partial sum of the new series differs from the kth sum of (1) only by the term k+1

ak+lx

k+1

1 + x

which approaches zero as k -+ oc because of the convergence of the original series (Ruds., Theorem 5, p. 8). We now use the differences to simplify (3), obtaining

S(x)=

1

x

(ao - dao

Retaining the first term, we write the remaining series

-

x

1+x

(dao - dal x + dal x2 - )

like S(x) itself in the form x

1

1+x 1+x

(dap - d'a, . x + d2a1 . x2 - ...),

so that, splitting off the first term again, we get S(x) =

dao x (1 +x)2

ao

1 +x +

xz

(1 + x)2

(d2ao

Continuing in this way, we obtain

S(x) _

ao

+x

+ (-

-

dao x + d2ao (1 + x)2 (1 + x)3 dp-la0

1)p-1

(1 + x)p

xp-1 + RP(x)

(4)

Computations Involving Series

73

after p steps, where XP

RP(x)

(d°ao - deal x + LPa2 x2 - ...)

(1 + x)" = (-1)P

XP

+ x)" ko

(1

`-1)kAPakXk.

To prove that RP(x) - 0 asp -> oo, we replace the pth difference iPak by its expression (2) and reverse the order of summation, obtaining TP

k1( (- 1)k+P xk+P `

RP(x)

(1 + X)P 1

P

CPiXi

(1 + x)" i=0

r k=0

L (- 1)i Ciak+P-i

1) k+p-i

ak+P-IX

k+P

-i

Denoting the remainder of the original series (1) by rn(x) _

(- 1)k +n ak+nxk+n

(n = 0, 1, 2, ...),

k=0

we can finally write RP(x) in the form

Y C'xirP-i (x) RP(x) =

Y Cix"-iri (x)

i=0

(1 + x)P

(1 + x)P

and hence RP(x) -> 0 as p -> oc, by Problem 3, p. 25, since rn(x) -+ 0 as n - oc. Taking the limit asp -+ cc in (4), we get S(x) =

1

l+ x

jao - 4ao

+ (-1)PAPa0 (

X

l+ x

1 + X )P

+ 12ao

...

( X )2 \ l+ x (5)

Infinite Series: Ramifications

74

Comparing (1) and (5), we finally obtain Euler's transformation

Y(-1)kakxk =

k=0

>

1

1+ X D=0

(- 1)Dd"ao

x

1+ x

J

(6)

Equation (6) is most frequently used with x = 1, and it then transforms one numerical series into another: (- 1)D ADa0

Y (- 1)k ak

(7)

2P+ 1

D=0

k=0

Example 1. Let 1

ak =

z+

'

k

where z is any constant different from 0, -1, -2, ... Then the series

1)k

k=o z + k is a series of the Leibniz type if we drop a sufficiently large number of initial terms, and hence converges. The successive differences Aak, A2ak, ... are easily calculated. In fact, using mathematical induction, we find that

dal, = (-1)D

pi

(z+k)(z+k+ 1)...(z+k+p)

and, in particular, ADa0 = (-1)D

z(z+

p!

(8)

It follows from (7) and (8) that I)k

k0

z+k

p

1

D=O 2D+1 z(z+1)...(z+p)

(9)

Setting z = 1 in (9) leads to the following transformation of the familiar series for In 2:

In 21)m m=1

m

n=1 2 nn

Computations Involving Series

75

The second series is obviously more suitable than the first for numerical calculations. In fact, to get an accuracy of 0.01 we need 99 terms of the first sequence and only 5 terms of the second! Example 2. Let z + 2k

ak

where z is a constant different from 0, -2, -4, ... Representing ak in the form ak

we can use formula (8) to write d°ao =

1)°

1

2 2

(2

lp + 1/ ...

( \2

+

2°+lpi

1

_ (-1)OP 2 z (z + 2)

(z + 2p)

P (10)

Then Euler's transformation (7) takes the form 1k k=o

1

1

( - ) z+2k

°°

Y

pi 11

2P=

z = I in (11) leads to the following transformation of the familiar series for r/4 (Ruds., p. 115): 1

4

k=O

k

1

2k + 1

_

1

p!

Y 2 ,,==o (2p + 1) (2p - 1) ... 3- 1

Example 3. In calculations using a transformation of a series, it is often convenient to first calculate some terms of the series directly and afterwards subject only the remainder of the series

Infinite Series: Ramifications

76

to the transformation. As an illustration, suppose we calculate n by using the series 2

(1 +

\

1

1

+

2

2

1

3

+...

3.5.7

3.5 1.2...p 3

(12)

1) derived in the preceding example. Since the ratio of the last

written term to the preceding term is p 2p + 1

1

2

the discarded remainder of the series is always less than the pre-

ceding calculated term. For example, we get n to six decimal places by calculating 21 terms of the series (12), since 2

1

2

20

3

= 0.00000037... < 0.0000005.

However, if we calculate the first 7 terms, say, of the original series directly and transform only the remainder after 7 terms, we get

n=4 1-+++ 1

3

2

1

1

1

1

1

5

7

9

11

13

1 + 15.17 1

+

.2 1

15.17.19

15

1 2. ..p

+...

..\

+

The eighth term of the transformed series is now within the required limits, since 2

1

45 6 15.17..29

2

3

7

= 0.0000002...

Computations Involving Series

77

Hence to get the same accuracy as before, we must now only calculate 8 more terms besides the 7 retained terms, i.e. only 15 terms in all as opposed to the previous 21 terms! PROBLEMS

1. Prove that

arc tan x =

(- 1)k k=O

x2k+1

1

2k + 1

2p (2p - 2)...4.2

X

x2 I

P

(cf. Ruds., p. 115).

Hint. Choose ak = 11(2k + 1) in (6), using (10) with z = 1 to calculate A°ak. 2. Give examples showing that subjecting a convergent series

to Euler's transformation does not always improve its rate of convergence. Comment. In comparing the rate of convergence of two series X

I Ck,

k=O

CL

k=O

with terms of arbitrary sign, we start from the behavior of the ratio of the corresponding remainders y and i.e., if 0 as n -> oo, we say that the first series converges more rapidly than the second and the second more slowly than the first (cf. Ruds., Definition 1, p. 53). Ans. The series ly,&n'l

(-1)

k=0

1

k

2k

78

Infinite Series: Ramifications

is transformed into the more rapidly convergent series 1

1

D=o 2 4P

while the series

is transformed into the more slowly convergent series °p

1

3

D=o 2

4

D

14. The Transformations of Kummer and Markov

We have just seen how Euler's transformation, based on a well-defined rule, leads to a unique transformed series, which, however, is not always suitable for computational purposes (see Problem2 above). We now describe a method of transforming series due to Kummer which has a high degree of built-in arbitrariness (requiring much ingenuity on the part of the calculator), but which in return is more specifically directed to the goal of simplifying the given calculation. Thus let

Acs)+Acz)+... +AM+...

(1)

be a convergent series, and suppose we want to calculate the sum

of (1) to a given accuracy. Clearly A(k) - 0 as k -' oc. Suppose we find another series ali) F alz) + ... + alk + ... (2)

converging to an easily calculable finite sum Al such that the general terms A(k) and alk) of the series (1) and (2) are equivalent, indicated by writing A(k) - a(l') (as k --+ oo), in the sense that (k)

lim = 0, k-. A(k)

(3)

Computations Involving Series where

79

a1 (k) = A(k) - a1

(4)

1

Formula (3) is usually written more concisely as a(k) = o(A(k)).

(5)

1

Note that (5) implies a(lk) = o(a(,k))

as well, since a (k)

lim

lim

(k)

a1

k

m

A(k)

a (k)

lim A(k) 1- a1(k) - k-.ao

1 - al(k)

- 0.

A(k)

It follows from (4) that 00

A(k) k=1

=

M

X

.0

kL=J1

k=1

k=1

r aik) + I ai) = Al + I aik)

(6)

Thus we have reduced the calculation of the sum of the original series (1) to that of the sum of a transformed series whose terms certainly approach zero more rapidly than A(k), because of (5). Example. To calculate the sum of the series

we recall that 1

k1 k (k + 1)

=1

(Ruds., Prob.2, p. 4). Since 1 limk2

k-. ao

k(k+ 1) =limk2(k+ 1) _0 1

k2

k-. co

1

k2

Infinite Series: Ramifications

80

it follows from (6) that

Y1=1+

k=1 k2

1

k=1

k2 (k + 1)'

(7)

where the transformed series is clearly more suitable for calculation than the original series. Repeating the same process, suppose we can find still another series a2

+aZ2) +... +aZ) +...

converging to an easily calculable finite sum A2 such that aZ) is equivalent to a(k) in the indicated sense. Then aD

A(k)=Al+A2+Ia2), k=1 k=1 so that the calculation of the sum of the original series (1) reduces

to that of the sum of a series whose terms IX(k) = IX (k) 2 1

-

a(k) 2

= O ( IX(k) 1 )

converge to zero more rapidly than a1 . Similarly, repeating the process p times, we arrive at the formula YAtk)=Al+A2+...+Ap+yIXDk),

k=1

k=1

(8)

where X

A; =

k)

1,a'

k=1

(1= 1, 2,...,p)

are known sums of series which are successively "split off" from the original series, so that the problem reduces to calculating the sum of the series in the right-hand side of (8).

Next we discuss another method, due to Markov, for transforming a given convergent series Y A(k) = A. k=1

(9)

Computations Involving Series

81

Like Kummer's method, this method also has much built-in arbitrariness. Suppose every term of (9) can be represented as the sum of a convergent series cc A(k) _ > a(Jk) (k = 1, 2, ...), (10) J=1

and consider the infinite rectangular matrix A(1)

al(1)

A(2)

a (2) a (2) a (2) ... 3 1 =

A(3)

al(3)

A(k)

a2(1) a3(1) ... aJ(1)

...

a (2) J

...

a2(3) a3(3) ... aJ(3)

...

a (k) a (k) a (k) ... (k) aJ 2 3 1

...

made up of the terms of the series (10). Then the number A is just the sum of the iterated series

A=

a k) k=1J=1

corresponding to this matrix. Assuming that every column of the matrix gives a convergent series (k)

k=1

aj

= Aj,

Markov established a necessary and sufficient condition for the series M

Aj J=1

to converge to the same sum A. Correspondingly, Markov's transformation consists in replacing the original iterated series 6

Fich(enholz (2094)

Infinite Series: Ramification

82

(summed over rows) by the other iterated series (summed over columns): M

A=YA(k'=YAK. k=1

j=1

A sufficient condition for applying Markov's transformation is given by Theorem 3, p. 34, say. However, the theorem proved by Markov is much more general, and does not even assume the absolute convergence of the series involved. PROBLEMS

1. Prove that ak -

F'k

(as k -- oo) if and only if

lim ak = 1.

k-.ao fl k

2. Show that 1+ 1

k2

1 + 2!k=1 1 k2 (k + 1) (k + 2) 1

22

k2

1

+

1 1 + 3! > 22

+

1

k=1 k2 (k + 1) (k + 2) (k +

32

3).

and more generally, that x

11= 1+ 1 +...+ 1 +pt Y 00

k=1 k2

22

1

k=1 k2 (k + 1)

p2

(k + p) (12)

Hint. Starting from (7), use (8) and the formula

k1k(k+ 1)...(k+p) (Ruds., Prob. 2, p. 4).

Comment. Thus the calculation of the sum of the slowly converging series

1

1

k=1 k2

Computations Involving Series

83

is reduced to the calculation of the sum of p of its terms and the sum of a transformed series which converges rapidly (even for moderate values of p).

3. Show that 1

k

1

k2

+

1

=3

1

1

+

2.22 3

2

.+

2!

1

3 .23 5-3-1

(P - 1 t p 2°(2p-1)(2p-3)...3 )!

1

+

1

(P!)3

2" (2p - 1) (2p - 3) ... 3 x

°°

x 1

1

(13)

k=1 k2 (k + 1)2 ... (k + p)2

Hint. Since k + y

1

2p- 1 k2(k+ 1)2...(k+p- 1)2

k+1+y (k + 1)2 (k + 2)2 ... (k + p)2

r (2p - 1) k2 + p (p + 2y) k + yp2 2p-1 k 2 (k + 1)2 ... (k + p - 1)2 (k + p)2 1

I

k2(k+ 1)2 ... (k + p - 1)2 as k -> oo, where y is temporarily undetermined, it follows from (6) that Sp

°°

_

1

k=1 k2 (k + 1)2 ... (k + p - 1)2

r2 P -

+

L

k=1

P In - 1

1

l+y

2p - 1 (p!)2

(P + 2y)l k +

P2 L

k2(k+ 1)2...(k+p)2

YP

z

- In 1

-j

Infinite Series: Ramifications

84

Now choose

thereby causing the term containing k to vanish in the numerator of the series on the right. With this choice of p, S°

_

3p

P3

S

2(2p- 1) (p!)' + 2(2p-1)

D+

Therefore

1=S)=3+1 S2, 2 2

&=i k2 1

3

SZ =

2

2

(2!)3

1

22 3

+

S3, 22

[(p - 1)!]3

2°-' (2p

- 3) (2p - 5) ... 3 .

3

(p-1)!

3

1

3° 1

1) (2p _ 3)...3. 1

2° (2p - 1) (2p - 3) ... 3 .

1

S°+1,

which together imply (13).

4. Suppose we choose p = 5 in (13) and also retain 5 terms of the transformed series. What is the accuracy of the resulting calculation of the sum of the original series?

5. Interpret Example 3, p. 44 as an instance of Markov's transformation, discussing the results of using the two representations a2)+...+a(k)1+rk, a, +

Computations Involving Series

85

where a(k)

(k - 1)!

=

.1(.1+1)...(j+k).

Comment. As already noted on p. 45, Markov's transformation gives nothing new if we use the second representation.

6. Show the connection between Kummer's transformation and Markov's transformation. Hint. Take the limit asp - oc in formula (8), assuming that lim

p-.ao k=1

(14)

a(pk) = 0.

7. Verify that (14) holds in the case of the expansion (13), thereby proving that

1= k=1k2

3

1+

+

1

3.235.3.1

2.22

2

2!

1

+ ...

(p - 1)!

1

p . 2p (2p - 1) (2p - 3) ... 3- 1

3>

(p

-1)!

1)(2p-3)...3.1

(15)

Hint. The sum in the last term in the right-hand side of (13) does not exceed 1

°°

1

(p!)2 k l k2 '

and hence the whole term does not exceed the quantity p!

2p (2p -

°°

1

k2

which obviously approaches zero asp -p oo. Comment. Note that (15) is equivalent to formula (5), p. 46.

86

Infinite Series: Ramifications

8. Give an example showing that taking the limit (14) does not always lead to a useful result. Hint. Taking the limit asp - co in (12) merely gives the identity °°

1

k=lk2

P=Ip2

Comment. Thus Markov's method is a very general technique

offering the calculator many possibilities but requiring much ingenuity on his part.

CHAPTER 4

Summation of Divergent Series 15. Introduction

So far, we have defined the sum A of a given numerical series

I

e0

a°=ao+a1+a2+... +a.+...

(1)

as the limit

A = lim A.

(2)

n- 00

of its partial sums +a2+...+an,

assuming that this limit exists and is finite or equal to ± oo. On the other hand, the sum of an "oscillating" divergent series, for which the limit (2) fails to exist, has not been defined, and we have systematically avoided such series. In the second half of the nineteenth century, however, consideration of various situations encountered in mathematical analysis, such as the fact that the product of two convergent series can diverge (see Example 1, p. 27), naturally brought to the fore the problem of trying to sum divergent series in some appropriate new sense, perforce

different from the usual sense associated with formula (2). Certain of these "summation" methods (to be studied in detail below) turned out to be particularly fruitful. It should be pointed out that divergent series were quite often encountered in mathematical practice prior to the creation by Cauchy of a rigorous theory of infinite series, based in turn on a 87

Infinite Series: Ramifications

88

rigorous theory of limits. Although the use of divergent series in proofs was disputed, nevertheless attempts were sometimes made to assign numerical values to such series. For example, the number z was assigned to the oscillating series

1-1+1-1+1-1+... even in Leibniz's time, a choice which Euler justified by observing that the expansion 1

1+x

= 1 -x+x2-x3+x4-x5+...

(which is actually valid only for lxi < 1) becomes 1

=1-1+1-1+1-1+

2

if x is formally set equal to 1. There was a kernel of truth in this observation, but the whole approach lacked clarity. For example one could equally well deduce the formula

-=1-1+1-1+1-1+ 3

2

by setting x = 1 in the expansion

1 +x

1+x+x2

1 -x2 = 1 -x2+x3-x5+x6-xs+...,

1-x3

The whole problem is posed differently in modern analysis. The starting point of all discussion is some precisely stated definition of the "generalized sum" of a series, which is applicable to a whole class of numerical series rather than being contrived for a particular series of interest at the moment. The legitimacy of this approach cannot be contested. The reader need only re-

call that even the ordinary concept of the "sum of a series," as simple and natural as it may seem, was introduced on the basis of a tentative definition, subsequently justified only by its expediency! However, regardless of how such a "generalized sum" is defined, it is usually required to satisfy the following two conditions:

Summation of Divergent Series

89

1) If the series Y, an

n=1

is assigned the generalized sum A and the series M

Y bn n= 1

is assigned the generalized sum B, then the series W

Y pan + qb.), n=1 where p and q are arbitrary constants, must be assigned the generalized sum pA + qB. A method of summation satisfying this condition is said to be linear. 2) The new definition of summation must include the ordinary definition as a special case. More exactly, a series which converges to the sum A in the ordinary sense must have a generalized sum, and this sum must also equal A. A method of summation satisfying this condition is said to be regular. Naturally, we are only interested in regular methods of summation which allow us to sum a larger class of series than can be

summed by the ordinary method of summation, since only in this case does it actually make sense to talk about "generalized summation." The next few sections are devoted to two methods of generalized summation of particular importance in the applications. PROBLEM

Justify writing m n

where m and n are arbitrary positive integers.

Infinite Series: Ramifications

90

Hint. Note that

1 -x"

I

16. The Method of Power Series First we consider the method of power series, primarily due to Poisson who applied it to the study of trigonometric series. DEFINITION. Given a numerical series

+a2 +... n=0

suppose the associated power series W

Y ax" = ao + a1x + a2x2 +

+ a"x" +

n=0

converges for all 0 < x < 1 to a sum function f(x), and suppose that

Um f(x) = A.

X-1-

Then the number A is called the generalized sum (in the sense of Poisson) of the given series. Example 1. As already noted on p. 88, the series

1-1+1-1+1-1+ has the generalized sum I in the sense of Poisson, since 1

lim

1

x-.1-1+x

1

2

Summation of Divergent Series

91

Example 2. Use the method of power series to sum the divergent series 2

+ Y 1cosnO

(-r < 0 < n)

(1)

(see Problem 1). Solution. Form the power series 1

- + I x" cos n0 2

(0 < x < 1),

n=1

whose sum is easily seen to be'

1 -x2 1 -2xcos0+x2

1

2

But lira

(2)

1 -x2 - 1 - 2xcos =0 0 + x2 1

2

if 0 0

0. Hence the generalized sum of the series (1) equals 0 if 0. If 0 = 0, (2) reduces to 1

2

1 -x2 _ (1

x)2

1

1 Y-x

2 1+x

which approaches + oc as x - 1-. Hence (1) has the generalized sum + co if 0 = 0. Note that (1) obviously has the ordinary sum

+ooif0=0. It is immediately clear that the present method of generalized summation is linear. Its regularity is a consequence of the follow1. In fact, multiplying

1+22x"cosnO by 1 - 2x cos 0 + x2 and noting that 2 cos nx cos x = cos (n + 1) x + cos (n - 1) x, we get 1 - x2 after cancelling terms.

Infinite Series: Ramifications

92

ing theorem, proved by Abel in his investigations on the theory of the binomial series. Accordingly, the method of power series will henceforth be called the Poisson-Abel method (of summa-

tion). THEOREM I (Abel). If the series x Y_ a" n=0

converges to A (in the ordinary sense), then the power series x a"x"

(3)

n=0

converges for 0 < x < 1 and its sum approaches A as x - 1-. Proof. First we note that the radius of convergence of the series (3) is no less than 1 (Ruds., p. 68), and hence (1) actually

converges for 0 < x < 1. Consider the identity

I anx" = (1 - x) > Anx",

n=0

n=0

(4)

where

An = ao + a, + ... + an (see Prob. 3, p. 83 or Ruds., Prob. 1, p. 85). Subtracting (4) term by term from the obvious identity

A=(1-x)YAx", n=0

we get

A - Y anx" = (1 - x) Y anx", n=0

(5)

n=0

2. There is no doubt that the general formulation of the method of power series, used by Poisson only in a special case, stems from Abel's theorem. The Poisson-Abel method is often simply called Abel's method, although Abel himself regarded the idea of "summing" a divergent series as very much of an oddity.

Summation of Divergent Series

93

where a" = A - A. Since an -> 0 as n -> co, given any e > 0, we can find an integer N > 0 such that Ia"I < E for all n > N. The right-hand side of (5) is the sum of the two expressions

(1 - x) > oC"X",

(1 - X) I %"X", n=0

(6)

n=N+1

the second of which can be estimated at once (independently of x): M a"Xn

(1 - x)

n=N+1

1< (1 - X) n=N+1

< E (l - x)

lanI X"

n=N+1

X" < E.

(7)

As for the first of the expressions (6), it approaches zero as

x -- 1-, and hence N

(1 - x) > a"X"

<E

(8)

n=0

for x sufficiently near 1. Combining (7) and (8), we finally get A -

a"x" < 2E. n=1

We have already seen that even if the series an

(9)

n=0

is Poisson-Abel summable with sum A, it need not have a sum in the ordinary sense. In other words, the existence of the limit X

lim I a"x" = A

X-1- n=0

(10)

does not in general imply the convergence of (9). The question naturally arises of what extra conditions must be imposed on the terms of (9) to make (10) imply the convergence of (9), so that

94

Infinite Series: Ramifications

(9) has the sum A in the ordinary sense. The first result along these lines is the following theorem due to Tauber: THEOREM 2 (Tauber). Suppose the series I anx" n=0

converges for 0 < x < 1 and satisfies (10). Moreover, suppose the terms of the series (9) are such that lim

a1 + 2a2 +

n-m

+ nan

n

Then

M

Y an=A.

n=0

Proof. The proof will be carried out in two steps. First we assume that3

lim nan = 0,

(12)

n-.a,

or equivalently

an=o(n). Thus if 8n = max Ikakl, k>_ n

then 8" decreases monotonically to zero as n -+ co. Given any integer N > 0, we have M=O

a,, - A= n=0 an(1

- x")anx"+ n=N+1

(I anx"-A n=0

and hence N

I an - A n=0

N

n=0

(1 - x)Nao +

n=N+1

D

n

00

i Inan1 (1 - x) + Y

Ina"1 x

n

6N+1

(N+1)(1-x)

+

n=0

+ n=0

a,,x" - A

anx" - A

3. Note that (12) implies (11), by Corollary 2, p. 23, but not conversely, so that we are first proving the theorem under a stronger hypothesis than (11).

Summation of Divergent Series

95

where we use the inequalities

1

1 -X

1

valid for 0 < x < 1. Choosing e > 0 arbitrarily small, we set

(1 -x)N=e, or equivalently

x=1--, N e

so that x - 1 as N - oc. Now let N be so large that SN+1 < e2 and the corresponding value of x is close enough to 1 to make anx" - AI < e. R=0

Then

N

1)

and hence -0

n=0

00

Vn

QnXn = a0 +

n=1 n

X"

ao+(1 -x)

vn-1

X" n=1

n=1 n

n

x"+

vn

n=1 n (n + 1)

x"+1.

(13)

Infinite Series: Ramifications

96

This time we have (11) or, equivalently, vnIn -+ 0 as n -> w. It follows that

lim (1 - x)

X-.1-

n=1 n

X. = 0.

(14)

To see this, we write

(1X) n=1 11"x"=(1-X) t-°x"+(1-x) n=N+1 LOX" n=1 n n n and then choose N so large that all the factors v"In in the second

expression on the right have absolute values less than some preassigned number e > 0. Then the second expression is itself less than E, regardless of x. Moreover, the first expression on the right, which contains only a finite number of terms, can also be made less than E, by simply choosing x close enough to 1. Thus, finally, combining (10), (13) and (14), we get Un

lim

x-.1-

n=1

n(n + 1)

X " +1

=A - ao.

But here we can apply the already proved special case of the theorem, obtaining cc

n=1

Vn

n (n + 1)

= A - a0.

On the other hand, we have

i

Vn

n=1 n (n + 1)

=

-

t nn

j

m+1

n=1 n + 1

n=1 n

-M+1 U,"

t'n

Vn

[

n =1

n

I

n=1

Vn-1

n

n'

+

an.

n=1

Since the first term on the right approaches zero as m -- oo, it follows that m

lim I an=A - ao.

m-.m n=1

Summation of Divergent Series

97

Remark. Subsequently, various authors have proved a whole series of delicate theorems of the same type (customarily called "Tauberian theorems"), modifying and extending Tauber's conditions. PROBLEMS

1. Show that the series

12 + n=1 cos n0 diverges for all 0 in the interval [-ar, ir]. Hint. If Olrr is a rational number, i.e., of the form pfq where p and q > 0 are integers, then cos n O = + I for values of n which are multiples of q. This violates the necessary condition for convergence of a series (Ruds., Theorem 5, p. 8). If the ratio O/n is irrational, then, expanding 01rr as an infinite continued fraction, we find that4 0

m

7r

n

< n2

for appropriate rational numbers min, and hence In0-m7zl oc. We can now give a definitive answer to the question of the connection between the Poisson-Abel method of summation and Cesaro's method: THEOREM 1 (Frobenius). If the series W

Y an n=0

is summable by the method of arithmetic means to a finite "sum" A, then it is also summable by the Poisson-Abel method to the same sum A. Proof. By hypothesis, an -> A as n - oo. Then the power series x

J(x)=Ia,,x" n=0 clearly converges for 0 < x < 1, by the lemma. Applying Abel's transformation twice (as in Prob.3, p. 17 or Ruds., Prob. 1, p. 85), we get W

f(x)=(1-x)> Anx"=(1-x)2 n=0

W

n=0

(n+1)anx".

(2)

But, as is easily verified, 1

(1 - X)2

n=0

(n + 1)x"

or

W

1=(1-x)Z>(n+1)x"

(3)

n=0

for 0 < x < 1. Multiplying both sides of (3) by A and then subtracting (2) term by term from the resulting identity, we get W

A-f(x)=(1 -x)ZY(n+ 1)(A-an)x". n=0

Infinite Series: Ramifications

102

The right-hand side is the sum of the two expressions N

co

n=0

n=N+1

(1-x)2I(n+l)(A-a")x", (1-x)2 Y (n+l)(A-an)x". (4)

Let N be so large that IA - a"j < E for all n > N, where E is any preassigned positive number. Then the second of the expressions (4) is itself less than E is absolute value (independently of x), while the first expression can also be made less than a is absolute value by simply choosing x close enough to 1.5

Thus we have shown that whenever Cesaro's method is applicable, the Poisson-Abel method is also applicable and leads to the same generalized sum. The converse is not true, i.e., there exist series summable by the Poisson-Abel method but not by Cesaro's method. For example, the series

1-2+3-4+".

(5)

cannot be summed by Cesaro's method, since the necessary con-

dition a" = o(n) proved in the lemma obviously fails. On the other hand, the power series

1-2x+3x2-4x3+

(0<x

in

that any Am (with n < in An- k C. n

Then, summing over in, we find that

S>kAn -

k -C. 2

n

7. Changing the sign of all the terms of the series, we see that it is also sufficient to assume the opposite inequality mam < C

(m = 1, 2, ...).

In particular, the theorem is obviously applicable to all series with terms of constant sign.

Summation of Divergent Series

105

Comparing this with (8), we find that Am < an+k +

n+1 k

/(

lan+k - an) +

k n

(9)

C.

Now let n approach infinity while at the same time k is subject to the requirement that the ratio k/n approach a preassigned posi-

tive number E. Then the right-hand side of (9) approaches the limit A + EC, so that An < A + 2EC

(10)

for all sufficiently large n. In just the same way, by considering the sum n

S' =

I Am = kan_k + (n + 1) (an - an-k)

m=n -k+1

and deducing the upper bound

Am=An-(am+1 + . . . +an) 0, let N be so large that Ian I < e for all n > N. Then the second of the expressions (3) is less than a in absolute value (independently of x), while the first expression, being the product of a-x and a polynomial, can be made less than a in absolute value by simply

choosing x large enough. 1 2) Euler's method. It will be recalled from formula (7), p. 74

that

a0

a0

i (-1)k ak =D=oI (-1)° k=0

D

ao

2D+1

(4)

(Euler's transformation). As shown in Sec. 13, if the series on the left converges, then the series on the right converges to the same

Summation of Divergent Series

123

sum. However, the series on the right may converge to a finite sum A even if the series on the left diverges. The number A is then called the generalized sum (in the sense of Euler) of the series

on the left. The regularity of this method of summation, known as Euler's method, is an immediate consequence of the italicized assertion.

Starting from the series (4) without the factors of +I and recalling formula (2), p. 71 for the pth differences, we see that the generalized sum in the sense of Euler of the series (2) is just he ordinary sum (provided it exists) of the series

ao + Cial + C_2a2 + ... + Crap 2p+1

D=o

Remark. This concludes our discussion of methods for summing divergent series. We have presented enough material to give the reader some idea of the great variety of approaches to the problem. Note that we have always insisted that our methods of

generalized summation be regular. Had space permitted, more could have been said about the relations between different methods of summation. It may happen that two methods of summation have overlapping "domains of applicability" (neither of which is contained in the other, and it may even turn out that two methods assign different "generalized sums" to the same divergent series. PROBLEMS

1. Sum the series

1-1+1-1+1-1+ by Borel's method. Ans. lim X-ap

e_X

e+

e-X

=

-. 1

Infinite Series: Ramifications

124

2. Sum the following series by Euler's method:

a) 1 - 1 + 1 - 1 + ;

b) 1 - 2 + 3 - 4 +

;

c) 1

Hint. In each case, use formula (4). Ans.

a) A = 1; b) A°a° = 1, A'a° = 1, ADa° = 0

A=#-I=#;

if p > 1,

c)A'a°= 1,A=I-I+I-... =1;

d) A°a° = 1, Ala° = 7, AZa° = 12, A3a° = 6, A°a° = 0

if p>3,A= -'4+'8 - 6=-*. 3. Let

9T°(x),9"(x),...,9)(x),... be a sequence of functions defined on a domain X, where X has a (finite or positively infinite) limit point w. Given a numerical se-

ries (2), with partial sums A = a° + al +

+ an, construct

the "functional series" W

n=0

A.9,. (x) = A ogpo (x) + Alps (x) + ... + Anrpn (x) + .. . (5)

Suppose the series (5) converges, at least for x sufficiently near CO, and suppose the sum of (5) approaches a finite limit A as x --+ w. Finally, let A be called the "generalized sum" of the series (2).

This gives a very general method of summation, which is obviously linear. Prove that the method is also regular, provided the functions satisfy the following three conditions: 1) For every fixed n, lim rpn(x) = 0; x-.m

Summation of Divergent Series

125

2) For x close enough tow (i.e., for Ix - wl < 6 if co is finite

or x>Aif to = +oo), cc

Y I(pn(x)I co, then, given any e > 0, there is an N so large that IAn -

Al
N. It follows from the boundedness of the An and the absolute convergence of the series W

I qn(X)

n=0

that the series CO

I An9pn(x)

n=0

converges (at least for x close enough to w). Moreover, N

M

m

Y An9pn (x) - A = Y (A,, - A) 9gn(x) + I n=0 n=0

n=N+1

+

A( n_0

(A,, - A) q'n(X)

99n(x)-1J, /

and hence N

AI n=0

_

(An - A) 99,,(x)

I

n=0

W

+n=N+1 Y

I

I

A

I

n=O

The second term on the right is less thane because of (6), while the first and third terms can each be made less than a by choosing x close enough to co.

126

Infinite Series: Ramifications

Comment. Clearly, there is another version of the foregoing which involves the numbers a (the terms of the series) rather than the partial sums An. 4. Particularize the preceding problem to the case where X is the set of all nonnegative integers m, with limit point co = + oo. Hint. Instead of the sequence of functions {ggn(x)}, we now have an infinite rectangular matrix too

tot

t10

ti l

t20

t21

t22 ... t2m

tn0

tn1

tn2

... (7)

"' tnm

(tn = gin(m)), and the "generalized sum" of the series (2) becomes

lim (Aotom + Altlm + A2t2m + ... + Antnm + ".).

The regularity conditions are now the following: 1) For every fixed n,

lim

m-m

0;

2) For sufficiently large m, ac

n=0

Itnml

K = const;

3)

lim I tnm = 1 .

m-.OD n=0

Summation of Divergent Series

127

Comment. These ideas are essentially due to Toeplitz, who, however, considered the case where the matrix (7) reduces to an infinite triangular matrix. This special case is sufficient for most purposes. 5. Show that the general method of summation of Problem 3

contains the Poisson-Abel method and the Borel method as special cases. Ans. The Poisson-Abel method corresponds to

99"(x)=(1 -x)x", X = (0, 1), w= 1, while the Borel method corresponds to X"

4"(X) = ex -, X = (0, +ce), co = +oo n!

(the regularity conditions are clearly satisfied in both cases).

Index Abel, N.H., 92 Abel's theorem: on multiplication of series, 29

Equivalent terms of series, 78 Error, 61 Euler, L., 71, 88 Euler's method of summation, 122 regularity of, 123 Euler's transformation, 74

on regularity of Poisson-Abel summation, 92 Associativity of convergent series, 1 Bessel function(s), 54 generating function of, Borel's method of summation, 121 regularity of, 122

Fichtenholz, G. M., 4, 46, 119 Frobenius, G., 98 Frobenius' theorem, 101 Generalized Cesaro method of summation, 114 vs. Poisson-Abel method, 116 Generalized sum of divergent series,

Cauchy, A.L., 15, 87 Cauchy's theorem, 14 Cesaro, E., 98, 114 Cesaro's method of summation, 98 linearity of, 100 regularity of, 100 vs. Poisson-Abel method, 101-102

88ff.

in the sense of Bore], 122 in the sense of Cesaro, 99 in the sense of Euler, 123 in the sense of Poisson, 90 in the sense of Voronoi, 113

Commutativity of absolutely convergent series, 5 Computations involving series, 6186

Double limit, 35 Double series, 35 absolutely convergent, 38 arbitrary, 38 conditionally convergent, 38 convergent, 36 divergent, 36 partial sum of, 37 positive, 37 sum of, 36

Hardy, G. H., 103 Hardy-Landau theorem, 103 Holder's method of summation, 120

Iterated limit, 36 Iterated series, 31 absolutely convergent, 33 convergent, 31 sum of, 31 Khinchin, A. Y., 97 Kummer, E.E., 78 129

Index

130

Kummer's transformation, 78-80 Lambert series, 48 Landau, E., 103 Leibniz, G. W., 88

Power series in several variables, 60 Power series in two variables, 54 region of convergence of, 54 Product of two series, 14 in Cauchy's form, 15

Riemann zeta function, 19 Riemann's theorem, 9 Round-off error, 62

Markov, A.A., 80, 81, 82 Markov's transformation, 81 Mertens' theorem, 26 M e th od of ar ithme ti c means (Cesaro's method), 98

Method of power series (PoissonAbel method), 90

Summation of divergent series, 87ff. linear method of, 89 regular method of, 89

linea rity of , 91

regularity of, 91-93 Multiple series, 60 Multiplication of series, 110-112

13-21,

Poisson, S.D., 90, 92 Poisson-Abel method of summation (Method of arithmetic means), 92 vs. Cesaro's method, 101-102

Tauber, A., 94, 97 Tauberian theorem, 97, 103, 106 Tauber's theorem, 94 Toeplitz, 0., 127 Toeplitz's theorem, 21 Truncation error, 62 Voronoi's method of summation, 113 linearity of, 113 regularity of, 113