1550251515-Classical_Complex_Analysis__Gonzalez_

Sequences and Series 187

Proof Since

Sm - Sn = an+1 + · · · +am

( 4. 7-1) follows at once from the Cauchy condition for sequences (Theo-

rem 4.4).

Letting m = n + p, with p an arbitrary positive integer, the inequality

above can also be written as

lan+l + an+2 + · · · + an+pl < €

where n > Ne and p > 0 is arbitrary.

Corollary 4.1 If I:~ an converges, then limn-+oo an = 0.

Proof By taking p = 1 in (4.7-2), we obtain

\an+ 1 \ < € for n > N or \an\ < € for n > N + 1

as a necessary condition for convergence.

(4.7-2)

That this condition is not sufficient for convergence is seen with the

example l:(l/n) (the so-called harmonic series). Here lim(l/n) = 0, yet

the series diverges since for e :::;^1 / 2 and p = n, we have

1 1 1 1 n 1

--+ .. ·+->-+ .. ·+-=-=-

n + 1 2n. 2n 2n 2n 2

so (4.7-2) cannot be satisfied no matter how large n is taken.
As a consequence of Corollary 4.1, we have:

Corollary 4.2 If liman f 0, then I:~ an diverges (test for divergence).

Theorem 4.6 If a series converges absolutely, then it converges.

Proof If I:~ \an\ converges, then by the Cauchy condition, corresponding

to any given e > 0 there is N such that for n > N and p arbitrary, we have

\an+1\ + · · · + \an+p\ < €

But this implies that

lan+l + · · · + an+p\ :::; \an+i \ + · · · + \an+pl < €

for n > N. Hence again by the Cauchy condition (sufficiency part), I:~ an
converges.

Example Consider the geometric series I: qn, where q is a complex num-

ber. This series converges absolutely (so it converges) if \q\ < 1, and the

series diverges if \q\ ;::: 1.

In fact, the real. geometric series I:~ lq\n is .known to be convergent if

\qi < 1. On the other hand, if \q\ ;::: 1 we have \q\n ;::: 1, so qn does not tend

to zero as n --+ oo, which shows that I:~ qn diverges in this case.