divergent and conditionally convergent series whose product is absolutely convergent

Divergent and Conditionally Convergent Series Whose Product is Absolutely ConvergentAuthor(s): Florian CajoriSource: Transactions of the American Mathematical Society, Vol. 2, No. 1 (Jan., 1901), pp. 25-36Published by: American Mathematical SocietyStable URL: http://www.jstor.org/stable/1986304 .

Accessed: 20/05/2014 22:42

Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at .http://www.jstor.org/page/info/about/policies/terms.jsp

.JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range ofcontent in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new formsof scholarship. For more information about JSTOR, please contact [email protected].

.

American Mathematical Society is collaborating with JSTOR to digitize, preserve and extend access toTransactions of the American Mathematical Society.

http://www.jstor.org

This content downloaded from 194.29.185.56 on Tue, 20 May 2014 22:42:15 PMAll use subject to JSTOR Terms and Conditions

http://www.jstor.org/action/showPublisher?publisherCode=ams

http://www.jstor.org/stable/1986304?origin=JSTOR-pdf

http://www.jstor.org/page/info/about/policies/terms.jsp


DIVERGENT AND CONDITIONALLY CONVERGENT SERIES WHOSE PRODUCT IS ABSOLUTELY CONVERGENT*

BY

FLORIAN CAJORI

? 1. Introduction.

It has been shown by ABEL that, if the product: n= a

Z (U0vn + u 11-v + + un vO)

of two conditionally convergent series:

n= m n=M

EU, ZVn~ n=O n=O

is convergent, it converges to the product of their sums. Tests of the conver- gence of the product of conditionally convergent series have been worked out by A. PRINGSHEIM,t A. Voss,4 and myself.? There exist certain conditionally- convergent series which yield a convergent result when they are raised to a certain positive integral power, but which yield a divergent result when they are raised to a higher power. Thus,

n=1 n

where r - 7/9, is a conditionially convergent series whose fourth power is convergent, but whose fifth power is divergent.11 These instances of conditionally

* Presented to the Society April 28, 1900. Received for publication April 28, 1900. t Mathematische Annalen, vol. 21 (1883), p. 327 ; vol. 26 (1886), p. 157. t Mathematische Annalen, vol. 24 (1884), p. 42. ? Anmerican Journal of Mathematics, vol. 15 (1893), p. 339 ; vol. 18 (1896), p. 195;

Bulletin of the American Mathematical Society, (2) vol. 1 (1895), p. 180. 11 This mnay be a convenient place to point out a slight and obvious extension of the results

which I have published in the American Journal of Mathematics, vol. 18, p. 201. It was proved there that the conditionally convergent series

n=w ~~1 E (_1)n(+l_<rc1), n=1 n

when raised by Cauchy's multiplication rule to a positive integral power q, is convergent whenever (q - 1)/q < r; but the power of the series is divergent, if (q-1 )/q > r. The case for which (q - )/q = r was left untouched, but an examination of the formulae appearing in that article readily yields the result that in this case the series is divergent. Hence the condition for diver- geucy should be written (q -1 )/q : r.

25



26 F. CAJORI: DIVERGENT AND CONDITIONALLY CONVERGENT [January

convergent series, yielding powers of the series which converge less rapidly the higher the power and which for a sufficiently high power yield a divergent result, suggest the following questions: Does the product qf two conditionally convergent series never converge mnore r apidly than do one or both of the factor- series ? Can the product of two conditionally convergent series or of a conditionally convergent and a divergent series in no case be absolutely convergent ?

The first doubt of the correctness of a negative reply arose in connection with the conditionally convergent series:

n=oo (_ 1 )n+l

n=l1}$

This yields conivergent results, no nmatter to how high a positive integral power it may be raised.* If the series is squared, the constituents, 1,/(n - r) (r + 1), which go to make up the nth term of the product, namely,

r=n-11 1

r=o n-r r +?i

are all positive or all negative for that term. Hence the numerical value of the nth term is the sum of the numerical values of all the products 1/(n - r)(r + 1). If the square of the series is indicated by

=a, n=1

then all the constituents, an-r * 1/(r + 1) , which enter into the composition of the nth term in the cube of the series have all like signs in that term. Simi- larly for higher powers. Now it has been shown that all the positive integral powers of the conditionally convergent series:

n=-o (_ l)n+l

n=1 n continue to converge, eveui though each term of a given power of the series is numerically the sum of the numerical values of its constituents, and it is evi- dent that, if the constituents of each term were partly positive and partly negative, each term would be numerically smaller. Can the signs of the terms of two conditionally convergent series and the numerical values of its terms be so chosen that the product of the two series is absolutely convergent?

In ani important article on the multiplication of conditionally convergent seriest A. PRINGSHEIM states that he knows no a priori reason why there might not be special cases in which stuch a product is absolutely convergent. i; Aller-

* American Journal of Mathematics, vol. 18, p. 204. t Matheinatische Annalen, vol. 21, p. 332.



1901] SERIES WHOSE PRODUCT IS ABSOLUTELY CONVERGENT 27

dings," he continues, " war es mir bisher nicht m6glich an irgend einem bestimm- ten derartigen Falle die unbedingte Convergeinz zu constatiren, und es diirfte dies -wenn es iiberhaupt solche Fille giebt, was ja immerhin als fraglich erscheinen muss - wegen der zusammeng,esetzten Beschaffelnheit der Reibenglieder w, eben nur sehr schwer gelingen."

? 2. Two typieal examples. We have succeeded in devising special cases in which an absolutely convergent

series is obtained as the result of multiplying two conditionally convergent series together, or one conditionally convergent series bv a divergenlt series. The series obtained by removing, the parentheses from the series:

s1 i(4p + 4P+4?4p+1 4p+4)'

s PEX 1 1 1 A S2 - O =4p + 1 +4p + 4 4p + 1 4p + 4)'

are both conditionally convergent. Let the four terms obtained in either series, or the terms in a similar series, for any one positive integral value of p, be called a " set " of terms. The 4n-th term of the product of S1 and 82 is

p=o 4n - 4p 4p +1 4n- 4p-3 4p+4 1 1 1 1 \

+4n- 4p 4p+1 4n 4p? ?3 4p+4p

We see that the sum of each set of constituents for the 4n-th terni is identically zero; therefore the entire 4n-th term is identically zero.

The (4n + 1)-th term of the product series is as follows:

.p=n'? 1 1 1 1

= n-4p +1 4p + 1 4n-4p 4p+4 1 1 1 1

4n-4p -3 4p+1 +4n-4p 4p+4J p=n'_ 1 1 1 1

+=O p0 n-4p + l 4]p+ 1 ? 4n - 4 4p + 4 1 1 1 ___

4n-4p-34p+1 4n- 4p 4pP?4) a

where n' = -1 or l(n -1), accordincg as n is even orodd. When n is odd,

1 1 1 1 1 1 1 1 2n+2 2n+2 2n- 21 2n+2 2n+2 2n-1 2n-1l




and when n is even, 1 1

2nu+1 2n+1'

In the first summation we are adding 4n' + 4 constituents froin one end of the column or series yielding the (4n + 1)-th term of the product; in the second summation we are adding 4n' + 4 constituients from the other end. In the middle of the series the sets of constituents will overlap, if n is odd, and will not touch if n is even. When n is odd, then a represents the sum of the constituents which have been counted twice; when n is even, then a stands for the negative value of the constituent niot counted at all. Each of these constituents is of the degree - 2 with respect to n and as their number is finite, their sum cannot be of a higher degree with respect to n than the degree - 2.

The (4n + 1)-th term, simplified, is

2=n 1 --4 o94P + 1 (4n - 4p + 1)(4n- 4p - 3)1 -a.

For any positive integral value of n the fraction 4/(4n - 4p + 1)(4n - 4p- 3) has its maximum value when p = n'. Writing n' forp, we obtain a fraction which is less than 1 /n2. Hence for any value of p,

1 4 1 1. 4p + 1 (4n - 44 J 1)(4n- 4p-3) < 4p + 1

It is easy to see that 2n 1

-< a log (2n),

where a is a constant greater than 3/2 log 2. Hence

p=n' 1

4p 1 <alog (2n), and

p=n 1 4 p=n/ 1 1 a log (2n) p=4p + 1 (4n-4p + 1)(4n 4p-3) <pfO4p + 1 n2 n 2

But a number m can be found, such that, when n> m, we have

a log'(2n) 1 n2 n (log n)0'

where I > 1. Therefore, for n > m, the (4n + 1)-th term is numerically less than the nth term of a series known to be absolutely convergent.




The (4n + 2)-th term of the product-series is

p=nf 1 1 1 1

p=Ot 4n 4p+4 4p+1 ? 4n-4p+1 4p 4

_ _ _ _ 1 1 _\ +4n-,4p 4p + 1 4n- 4p -3 4P +4)

+ E + ~ ~ ~ ~~+-

PoI 4n-4p+4 4p+1 4n-4p+- 4p+4 ~~~~1 1 1 1\

4n -4p 4p?+ 1+ 4n- 4p- 3 4p + 4

where n' is defined as before and where

1 1 1 1 2n?222n -1 2n-I 2n+2

when n is odd, and

c +2n+42n+1 2n+1 2n+4 '

when n is even. We see that the entire (4n + 2)-th term is identically equal to zero.

The (4n + 3)-th term of the product-series is

P=o 4n-4p+? 4p+1 4n-4p+4 4p+4

1 1 1 1 N 4n-4p+1 4p+1 + 4n-4p 4p+4J

'=n 71 1 1 1 ?=d 4n-4p+1 4p+1 4n-4p+4 4p+4

1 1 1 +4n-4p+1 4p+14n- 4p 4p+4

p-nf ~~~4 -=2( 4p+4 (4n- 4p+4)(4n -4p) a,

where 1 1

2n+2 2n+2' when n is odd, and

1 1 2 1 1 1 -2n+1 2n?1 + 2n?4 2n?4 + 2n?1 2n?1'




when n is even. The reasoning which we applied to the (4n + 1)-th term will show that this term, too, when n> m, is lnumerically less than 1/n(log n)0, 0> 1.

We have now shown that the 4n-th termi anid the three immnediately succeeding terms are either identically zero, or, for n> m, numnerically less than 1/n(logn)0. Henee the product of the series 8 and 82 converges absolutely.

Again, take the conditionally conivergent series, obtained by dropping the parentheses in the series

P=00 1 11 1 X T=E v + o 1- <67 + 2 6p + 3 6p + 1 6p + 2 6p + 3

and the divergent series, obtained by dropping the parentheses in the series:

P= o(6p + 1 6p + 2 6p + 3 6p + 1 6p + 2 6p+3)

The 6n-tll terml of their product, viz.,

p=1n-1 . 1 1 1 1 1

PO 6n -6p + 3 6p + 1 + 6n-6p?+2 6p+2 6n-6p+1 6p+3

1 1 ___ __ ) +6n-6p+3 6p+1 6n-6pJ2 6p+2 6n-6p+l 6p+3

is seen to be identica]ly zero. The (6n + 1)-th term of the product of T, and T, is

v 1 _ 1 _ 1 1 11_ x-o <6n-6p+5 6p+1 6n-6p-3 6p+2 6n-6p-4 6p+3

1 1 1 1 1 1 = 6n-6p=5l 6p+1+ 6n-6p)-3 6p+2 J 6n-6p=4 6p+3}

6n-6p-5 6p+1- 6n-6p-3 6p+2 6n-6p-4 6p+3, a

p=o(6p + 1 (6n 6p + 1)(6n 6p 5))

where n' is defined as before and where




1 1 1 1 1 1 1 1 1 1 axx + _ 3n 3n-1 3-1 3n 3n-2 3n-2 ' 3n3n-1 3n- 1 3n

1 1 3n-2 3n-2'

when n is odd, and a 1/(3n + 1)(3n + 1), whenl n is even. By repeating the reasoning used in previous cases we find that this term, forn>m, is numerically less than 1/n(log n)0, 0s> 1 .

The (6n + 3)-th, (6n + 4)-th, and (6n + 5)-th terms in the product of T, and T12, when subjected to this course of reasoning, all yield the same conclu- sion. The (6n + 2)-th term is identically zero. Hence the product-series is absolutely convergent.

? 3. The general method.

The method by which the above factor-series were constructed is simple. The harmonic series was modified by adopting some simple law of succession of signs. To render the series conditionally convergent, make the number of + terms in a set equal to the number of - terms. If these n-umbers are unequal, then the series is divergent. Then add suitable finite numbers to the denominators of some of the terms in the set, so that, when the product of the two series is formed, the constituents of any set of the nth term in the product, when added, yield a fractional sum having denomiinators whose degree in n is higher by two than the degree of the numerators. We were able to prove the absolute con- vergence from the fact that in each complete set of constitutents of the nth term in the product-series there are as mnany positive constituents as there are negative ones. To secure this even distribution of signs among the constituents for the sets in every term of the product, it is necessary that in one of the factor- series the nuniber of + terms in any set be equal to the number of - terms in that set. This condition is not sufficient, because among the terins in the product some might have a preponderance of + constituLents and others a preponderance of - constituents. This happens, for instance, when we multiply the series + + - - + + - - etc. by itself.

In the proper selection of signs the followinig rule, which we give without proof, may be .of service. Let a, denoting the niumber of terms in each set of the series Zm, be an even nlumber and equal to 2c; also let the first c terms in each set be +, the remaining terms -. Let the nuimber of terms in each set of the second series, EVm, be b. Then, in order to secure as many + constituents in each set of a term of the product of the two series Eum and 2vm, as there are negative constituents, it is sufficient that d should be a factor of c, where d is the smallest value, numerically, satisfying the congruence b d (mod a).




In explaining the further process of constructing our series we shall start out with a particular selection of signs, but the process is the same as for any other selection, similarly made. Let a 4, b=3, then c 2 and d= -1, a factor of c. Accordingly, the signs of terms in each set of the first series are + + -- . Let the signs of the b termiis in each set of EVm be + + -. We have now the two factor-series obtained by dropping the parentheses from the series on the right hand sides of the formulas:

p1 4p+ a, 4p + 1 + a2 4p + 2 + a3 4p + 3 + a1 4 and

P=.M 1 1 1 +

, pvn=, 3P + 'a 3P + 1 + 12 3 + 2 + 13,

in which the Greek letters are quantities to be chosen, if possible, so that the product of the two series shall be absolutely convergent.

For brevity, we shall, in the series Zum9 represent by {u ap + a } a fraction 11(ap + a), and, similarly, in the series Ev., by {vbp + ,B} a fraction 1/(bp + 13). Accordinigly we have, dropping the parentheses as before,

P=00

EZU m ({u4++at} + {u4,+1+a2}- {u4u +2+a3}- {u4p+3+aa,}) p=1

and

EZ V ({V3 + /1} + JV3p + 1 + 82} {v3? + 2 +3}). p=l

In the product of Eum and Evm, a set will consist of 12 constituents. The

first set in the (12n)-th term will be (since {v12-9 + 134{v3? 134, and similarly in other cases)

I{u12 + 3 + 4} {v12 9 ? 13 - {unu+ 2 + aI} {v12 8 + 12}

- {Ul2l + 1 + a2} {V1- 7 + 13} + {U12 + a} {t12 -6 + 13} - {ul.- 1 + a4} {V-2-5 + /34 + {u12 -2 + a3} {V- 4 + /34

(I) ? {tu12:3?a4} {vl2-3+32'i} + {JU12i 4+ 3} {V12-2+13}

+ { u12n -5 + a} I {IVl2 - + 183' } {12.- 6 + a 3} { fV12 + 31 }

+ { -12, -7 + a2} {t12 + 1 + i } - -8 + a,} {v12 + 2 + /3 }

Here each of the quantities 81, 82, 83 occurs four times. We have broad- ened our assumptions by attaching a single accent to each letter the first time it is repeated and a double and triple accent the secoud and third time it is repeated. Accordingly /3k, 1, 1Bi' 11, 1% /32 132' 13 13 3 3 may in general represent each a different number. Similarly, the letters al, a2 , a3 , a4 receive a single accent the first time they are repeated in a set of constituents and a double accent the second time they are repeated.



1901] SERIES WHOSE PRODUCT IS ABSOLUTELY CONVERGENT g

We shall limit ourselves to the case where, in any terin of the product series, each set, when multiplied by n, yields a product that converges toward zero when n = oo. Accordingly, if the set I be multiplied by n, we get, for n oo (supposing the Greek letters to be positive constants, or variables which approach positive finite limits), the first of the two following equations:

-{V12 -9 + 1} - {12-8 + 1821 - V12 + ?83} + {V126 + I} - {V12 5+fl2} + {V12-4+134 + {v12 3?+31} + {12 2+ f2}

I + {12 1 + 3} - {V12 + 31} + {v12+1+/2 }-{v12 22?13 }=O,

(II) l+ {~1 29?8} - {128+/2}?{v12 T+3} + {v126+1`1} + {V12-5 + 12} + {V12 +33} {V12 3?+1}+ {V122 + 2}

- {V12 1 + V3 {v12 + 1 V -{v12 + 1 + i }- {12 + 2 + 83`} 0.

The second equation in II is got from the (12n + 1)-th term by the same procedure. All the even terms between the (12n)-th and the (12n + 12)-th yield the same equation as the (12n)-th term; all the odd terms yield the same equation as the (12n + 1)-th term.

The last set in the (12n)-th term of the product is

-{V12. + 2 + )33} {Ul2 -8 + a, + {V12,n + 1 + %2} {U12 - 7 + a2}

12n 1 + 1i3} {U12 6 ? 3} + {V12n 1 + 33} {tU12 - + a4}

III) ? {v12n 2 +l2 }{u12-4+ a } + {a123+l1 }{u123+a2 } + {V12 4 + fl3} {U12 2 + a3} - {V12 -5 + 0f} {U12 1 + a3 }

+ {V12n 6+ 31}{u2 a} {V12T +7 + 3} {u12 +J1 + a2 } - -812 8 + /82} {U12 + 2 + V13} -{12 9 + 331} {U12 + 3 + a2}.

Assuming set III, when multiplied by n, to yield a product which converges toward zero, when n = o, we get the first of the following three equations:

- {U12-8 + a,j + {U12-7 +2} - {U12 6 + a3} + {u12-5? + a}

+ {u12-4 + a} + {af12 3+ a} + { -12 2 +a3} {12 1 + a'} ? {u12?al}-{u 1 + a12} -{u ?2 ? +a3} - {u12?3 ?K} =0,

? {u12-8?+a1}-{ub12-7?+cz2}-{u12-6?cz3}-{ub12-5?+z4}

(IV) -{u12-4?+ al} + {u12-3?}a1 + a2 - 2 + ac } + { 312-? + a + {U12+ctl} ? {U12?1+a2} ? {u12+2?+t3} - {u12+3?+ a} 0 ,

? {u28 + al} ? {u72 7+ a2} + au12 6?+ a}} {u125+ct4J + {U12 4Ia)} - {u12 3+ a2} {U12 2?+3} {U12 1+ct4}

- { {U + al} + {U12+1 -+6 -{u12?2+a'3} + {bU12 +3+ 4} = 0

The second and third equations in IV are derived from the last set in the (12n + 1)-th and (I 2n + 2)-th term, respectively. The (12n + 3)-th,

Trans. Am. Math. Soc. 3




(12n + 6)-th, and (12n + 9)-th terms yield the same equation in IV as the (12n)-th term; the (12n + 4)-tb, (12n + 7)-th, (12n + 10)-th terms the same as the (12n + 1)-th term; the (12n + 5)-th, (12n + 8)-th, (12n + 11)-tb the same as the (12n + 2)-th term.

The equations II, when simplified and b the substitution of v for v12 made to apply to otber sets in the terms of tb )roduct, are

{v2- 8 + I82} + {Vl2P- 6 +?/1} + {v12p 4 + 13} --

{v,12p,-2 +? } - -{v12p+ 2 +?3` 0,

l2p - 9 +)31Y + {v ? f} - {v83 P + {V12p?- 5 + {12 += -{v 2 3 +)B'1'1} {V12p-1 +)3"I- {V12p,+ + 182" j =?

The equations IV, when simplified, u 12p being written for u121 become

{Ut1 -7+ a2+ {ul2p - 4 + al - {ub12p-l + a4}- {u12P+ 2 + a" = 0

(VI) { u12- 8 +al} - {u12 -5 + a} {-uP2P 2 +a-}+{ul2p +1 + a"} = 0 {u12 -6+a3} -j{u12P-3+aj- {u12p +a,"I}+{u12P+3+a4}- 0.

If we aim to assign to the Greek letters constant values, then, in every equation in V each fraction must admit of being associated singly with another fraction in that equation which is numlierically equal to it, but of opposite sign. This relation is brought about by the following six equations:

[{12p 9 +1}- = {vl2p 3+i3}, {1 1 ++2p i+j- fv12 +i31},

(VII) {vl2p-8+/2} VL2 " f2+I2J { V2p 5?+1'- {vl2 +1+I32},

IV - 7 +,83 = {v12r- 1 + 3} {v12p- 4?+3 1 {vl2p+2=+ }

The same considerations applied to equations VI yield the following six rela- tionS:

I{U12 - 8 + a1} = ul- 2 + aO}, {u12 -5 + a,} =u12p +1 + a},

(VIII) {u12- 7.+ a2} = {u12p -1 + a}' {1u12p 4 + a1} ={U12, + 2 + a"} {UQ2 _6 +a3j = {um12p + a"}, {gM - 3 + a'l = {u,b2, + 3 + a'"} 12p ~~~~~12p, +2} {1++4}

Applying relations VIII, one may write the series Eu2m thus, the parentheses being suppressed:

EU 3E({ u12p- 8 + all + {u12p - 7 + a2} - {2 6 + a3} (IX) p _ {ul2 5 + a4} + {ul2p -4 + a} + {u12p- 3 + a'}

{~~Um~~Z({~u,2-8+a - {u12

4a, {u12 -+6 })a

-{u,2, 8 + a1} - {u12p - 7 + a2} - {fl2p - 6 + a?3}

I lrq, fill Sq,41 rr_q 1 2'l




Applying relations VII, one may write the series Evm thus, the parentheses being suppressed:

P=00

EVM = - ({Vl2p 9 + i31} + {v12, -8 + 2} - -2p 7 + l83Y p=l

(X) -+ {V 6 + 1 } + {vl2p -5 + i3} - {v12p -4 +I3,}

jZl m V+ IV12p 9 + 81t + {V12P-5+I32}- jv12-+I33})

l?+ {V fl 6 + ? + {vl2p -5 +2} - {V12p 4 + ?3}).

Series IX is conditionally convergent; series X divergent. In both series any constant positive values may be assigned to the Greek letters. The product of the two series is absolutely convergent. This may be shown by examining the 12n-th term and the eleven immediately succeeding terms of the product-series in the same way as we did the terms in the product of S1 and S2. We shall here write down the expression for the (12n + 1)-th term only, viz.,

p=n,

Z(+ {u127-12P 16 + cl} {v2p 9 + i31} {-l2n-12p + 9 2 p=1

+ {u2~ -12 +?8 +ctJ{v1} l2p7 +f3} + {u12n1 + 7 + a4} {v12p 1-6+ }

+ {U2 _12p + 6 ? a3} {vl2p -5 + %3 ? {U122n-12p + 5 + 21 {v12- 3}

- {Ul2n-12 + 4 + a} {v12p 9 + ,I%} + {U12n-12p + 9 ? 2} {V12p 8 + 32}

~{U1 12p + 8 + all} {vl2p -7 +,i3}{u12n12p+ +a 4}{v12p 1-6I3 -

2t12n-19p + 6 + a3} {vl2p 5 ? 7} - 12p + a ? 2} {vl2p -4 ? i3})

+ (+ {v12n-12p + + + I} {u+2- 8 + al {-V12p12p + 8 + 3} {u12- 7 + a2} - 12-12jp + 6 + a3} {U12p

- 6 + M} - {v12n12p + 6 + a} {U12p - + a,=}

-n 1212+ 5 +33}{12p 4+a} + I 12 212p 2 3

- {v12n-12p+3 +,1}{u12p- 8 + al} + {V12 {12p+8+3} l82p -7 + a2

+ {v22-12p + 7 + l} {U12P 6 + a3} + {2''l2n-12p + 6 + 31} {ul2p - 5 + a4}

+V 12n-12p + 5 + 13} {U12P 4 + al 31 - 4

p=n, - 2l {v2 9 + 1j {u12n-12p + + + V12 , u12p 12p + + 2 1}

p=n'

127 { ,8'1l2p -8 + al {+2n 62 + 5 } 8'n12 +u12 - + + a,j ? V2-12p{u 2~- 31 {-2a1,

p=nl

where n' = in when n is even, and n' 2(n + 1) when n is odd, and where a is used in the same way as before. In this case a consists of one constituent when n is even, and of eleven constituenits when n is odd. By the very same reasoning as that applied in the series S, and S2 it follows that the (12n + 1)-th



36 F. CAJORI: ON THE MULTIPLICATION OF SERIES

term of the product of IX and X, for n > m, has a numerical value less than 1/n(log n)0, where 0> 1. The two factor-series IX and X differ from the pairs S1 S2 and T,, TJ2 in this-that (suppressing in each series each parenthesis which encloses a set) corresponding terms in IX and X need not have the same numerical values. However, the three pairs of factor-series have in common one property, to which allusion has not yet been made; viz., in one of the series of each pair, the sum of the terms in each set is identically zero, so that the sum of the series is itself zero.

COLORADO COLLEGE, COLORADO SPRINGS.



divergent and conditionally convergent series whose product is absolutely convergent

Documents