1549901369-Elements_of_Real_Analysis__Denlinger_

8.4 The Cauchy Product of Series 491

is a partial sum of a rearrangement of L dk. By Theorem 8.3.13, every rear-

rangement of an absolutely convergent series converges to the same sum. Thus,

lim A nBn = lim Dn

n-too n-.oo

i.e., :L:akLbk = Ldk.

Therefore, L ak L bk = L Ck· •

Actually, for the Cauchy product of two convergent series to converge, it is
not necessary that both of the series converge a bsolutely. As the next theorem
shows, it is sufficient that one of them converges absolutely.

* Theorem 8.4.4 (Mertens' Theorem) The Cauchy product of an abso-

lutely convergent series and a convergent series converges (but not necessarily

absolutely), and its sum is the product of their sums.

Proof. Suppose L ak co nverges absolutely to A and L bk converges to B.
Using the same notation used in the proof of Theorem 8.4.3, let L ck denote
their Cauchy product, and let An, Bn, and Cn denote the partial sums of L ak,
L bk, and L Ck, respectively. We define a new sequence {Bn} by

B n = Bn -B.

Then, Cn =~Ck =~ (t,aibk+I-i )

n

= LakBn+I-k

k=l

(show, in Exercise 13)

n

= L ak(B + Bn+I-k)

k=l n

= AnB + L akBn+I-k·

k=l

(17)

Now, lim AnB = AB. Thus, our proof will be complete if we can prove
n-->oo
n
the following claim: n-.oo lim '""""'akBL-t n+I-k = 0.
k=l
To prove this claim, let c: > 0. Since L ak converges absolutely, we may let
A' = L lakl· Since B n --+ 0, 3M > 0 3 Vn EN, IBnl :=:; Mand 3 no E N 3

k ~no::::? IBkl :=:;
2

~, · (We can assume A' > 0, since if A'= 0 there is nothing

to prove.)