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.