1549901369-Elements_of_Real_Analysis__Denlinger_

8.4 The Cauchy Product of Series 489

Example 8.4.2 (Two Convergent Series Whose Cauchy Product Series Di-

verges) Let I: ak and I: bk both be the same series, with ak = bk = ( -J(.

Then both I: ak and I: bk converge, but their Cauchy product series diverges.

Proof. The two series converge by the alternating series test. On the other
hand, the kth term of the Cauchy product series is
k (-l)i+l (-l)k+l-i k (-l)k+2 k k 1
Ck.= L.. = L.. = (-l) L. ..
i=l A vk + 1 - i i=l v'i-Jk + 1 - i i=l v'i-Jk + 1 - i
k 1 k 1 k
Thus , lckl 2: L h h = 2:-k = -k = 1.
i=l v kv k i=l

So, the Cauchy product series diverges, since it fails the general term test. D

Now we show the connection between absolute convergence and the behav-
ior of Cauchy product series.

Theorem 8.4.3 The Cauchy product of two absolutely convergent series is
absolutely convergent, and its sum is the product of their sums.

Proof. Suppose I: ak and I: bk converge absolutely, and denote their sums
by A and B , respectively. Let I: Ck denote their Cauchy product. That is ,

k
ck = L aibk+l-i·
i=l
Then the terms ck represent the sums along the indicated diagon als of the
infinite matrix

fafb~ _q,.1-b; _l}i:b'; ~1.ff,( ~1.ffs' · · ·

>' l}·fb{' ,-' !115;' ,-' <t<zlf( ,/ <t<zlf4,, a2 bs

.fa:1b1 _q,.3152 _q,.3153 a3b4 a3bs

>' 1}4V;'l}4V;' / a4b3 a4b4 a4bs

jl5b~ asb2 asb3 asb4 asbs

(14)

Define the series I: dk as t he series obtained by removing the parentheses
from the series I: ck. That is,

dk = a 1 b1 +a1b2+a2b1 +a1b3+a2b2+a3bi +a1b4+a2b3+a3b2+a4bi + a1bs +
a2b4 + +a3b3 + a4b2 + asb1 + · · ·.

Denote the partial sums of these four series by

n n n n

An= Laki Bn = Lbk , Cn = L Ck, Dn = Ldk.

k=l k=l k=l k=l