13.8. CONVERGENCE OF SUMS 245

Theorem 13.7.1 Let ak ∈ Fp,F is either R or C and consider ∑∞k=1ak. Then this series

converges absolutely iflim sup

k→∞

|ak|1/k = r < 1.

The series diverges spectacularly if limsupk→∞ |ak|1/k > 1 and if

lim supk→∞

|ak|1/k = 1,

the test fails.

Proof: Suppose first that limsupk→∞ |ak|1/k = r < 1. Then letting R ∈ (r,1) , it followsfrom the definition of limsup that for all k large enough,

|ak|1/k ≤ R

Hence there exists N such that if k ≥ N, then |ak| ≤ Rk. Let Mk = |ak| for k < N and letMk = Rk for k ≥ N. Then

∞

∑k=1

Mk ≤N−1

∑k=1|ak|+

RN

1−R< ∞

and so, by the Weierstrass M test applied to the series of constants, the series converges andalso converges absolutely. If

lim supk→∞

|ak|1/k = r > 1,

then letting r > R > 1, it follows that for infinitely many k,

|ak|> Rk

and so there is a subsequence which is unbounded. In particular, the series cannot convergeand in fact diverges spectacularly. In case that the limsup = 1, you can consider ∑

∞n=1

1n

which diverges by calculus and ∑∞n=1

1n2 which converges, also from calculus. However, the

limsup equals 1 for both of these. ■This is a major theorem because the limsup always exists. As an important application,

here is a corollary.

Corollary 13.7.2 If ∑k ak converges, then limsupk→∞ |ak|1/k ≤ 1.

If the sequence has values in X a complete normed linear space, there is no change inthe conclusion or proof of the above theorem. You just replace |·| with ∥·∥ the symbol forthe norm.

13.8 Convergence of SumsOne can consider convergence of infinite series the same way as done in calculus.