192 Chapter4Theorem 4.14 The series ( 4.8-5) obtained by Cauchy's rule is convergentwhenever one of the series A= I:~ an, B =I:~ bn is convergent and the
other is absolutely convergent. Furthermore, the sum of ( 4.8-5) is AB
(Mertens [8]).Proof Suppose that the series I:~ bn converges absolutely. We haveCn =Ci+ Cz + · · · + Cn = Aibn + Azbn-i + · · · + Anbi
Letting An = A - en, ( 4.8-6) becomesCn = ABn - (6bn + 6bn-i + · · · + enbi)
and since Bn --; B, it suffices to show that(J'n = 6bn + 6bn-i + ... + enbi --; 0
( 4.8-6)as n __, oo. To do this, let B' = lbi I+ lb2I + · · · + lbnl + · · ·. We have B' > 0
if we rule out the trivial case bn = 0 for every n. Clearly, en __, O, so thatfor any given E > 0 there is N such that n > N implies that lenl < c/2B'.
With N fixed, let K = max(l61, ... , leNI). Then we have, for n > N,
l(J'nl::; (16llbnl + · · · + leNllbn-N+iD + (leN+illbn-NI + · · · + lenllbil)::; K(lbnl + ... + lbn--N+il) + 2~' (lbn-NI + ... + lbil) (4.8-7)
But( 4.8-8)and by the Cauchy condition for convergence, there exists Ni > N such
that n > Ni implies that
Elbn-N+1 I + · · · + lbnl < 2 1{
By using ( 4.8-8) and ( 4.8-9) in ( 4.8-7), we obtainl(J'nl < E
Hence limn->oo Cn = AB.( 4.8-9)Remark If neither I:~ an nor I:~ bn is absolutely convergent, the prod-
uct series may not be convergent (see example in Exercises 4.2, problem 5).
However, if the factors are convergent and have sums A and B, and the
product series, as obtained by Cauchy's rule, is also convergent with sumC, then C =AB. This theorem is due to Abel, and it will be proved later
as consequence of Abel's limit theorem (Corollary 8.7).
Cesaro [3] has shown that if I: an is summable ( C, r) and I: bn is
summable (C,s) (r > -l,s > -1), then L:cn is summable (C,r + s + 1),