5.5. DOUBLE SERIES 93

Then from Theorem 4.4.13,

f (x) f (x) = limk→∞

f (ak) f (bk)≤ 0

Hence f (x) = 0.

Theorem 5.4.9 Suppose |an|1/n < R < 1 for all n sufficiently large. Then

∞

∑n=1

an converges absolutely.

If there are infinitely many values of n such that |an|1/n ≥ 1, then

∞

∑n=1

an diverges.

Proof: Suppose first that |an|1/n < R < 1 for all n sufficiently large. Say this holdsfor all n ≥ nR. Then for such n, n

√|an| < R. Therefore, for such n, |an| ≤ Rn and so the

comparison test with a geometric series applies and gives absolute convergence as claimed.Next suppose |an|1/n ≥ 1 for infinitely many values of n. Then for those values of n,

|an| ≥ 1 and so the series fails to converge by the nth term test.Stated more succinctly the condition for the root test is this: Let r = limsupn→∞ |an|1/n

then∞

∑k=m

ak

 converges absolutely if r < 1test fails if r = 1diverges if r > 1

To see the test fails when r = 1, consider the same example given above, ∑n1n and ∑n

1n2 .

Indeed, limn→∞ n1/n = 1. To see this, let en = n1/n− 1 so (1+ en)n = n By the binomial

theorem, 1+nen +n(n−1)

2 e2n ≤ n and so e2

n ≤ 2nn(n−1) showing that en→ 0.

A special case occurs when the limit exists.

Corollary 5.4.10 Suppose limn→∞ |an|1/n exists and equals r. Then

∞

∑k=m

ak

 converges absolutely if r < 1test fails if r = 1diverges if r > 1

Proof: The first and last alternatives follow from Theorem 5.4.9. To see the test failsif r = 1, consider the two series ∑

∞n=1

1n and ∑

∞n=1

1n2 both of which have r = 1 but having

different convergence properties. The first diverges and the second converges.

5.5 Double SeriesSometimes it is required to consider double series which are of the form

∞

∑k=m

∞

∑j=m

a jk ≡∞

∑k=m

(∞

∑j=m

a jk

).