6.6. DOUBLE SERIES 163

Theorem 6.5.4 Let bn > 0 for all n such that bn ≥ bn+1 for all n and limn→∞ bn = 0and consider either ∑

∞n=1 (−1)n bn or ∑

∞n=1 (−1)n−1 bn. Then∣∣∣∣∣ ∞

∑n=1

(−1)n bn −N

∑n=1

(−1)n bn

∣∣∣∣∣ ≤ |bN+1| ,∣∣∣∣∣ ∞

∑n=1

(−1)n−1 bn −N

∑n=1

(−1)n−1 bn

∣∣∣∣∣ ≤ |bN+1|

See Problem 8 on Page 167 for an outline of the proof of this theorem along withanother way to prove the alternating series test.

Example 6.5.5 How many terms must I take in the sum, ∑∞n=1 (−1)n 1

n2+1 to be closer than110 to ∑

∞n=1 (−1)n 1

n2+1 ?

From Theorem 6.5.4, I need to find n such that 1n2+1 ≤ 1

10 and then n−1 is the desiredvalue. Thus n = 3 and so∣∣∣∣∣ ∞

∑n=1

(−1)n 1n2 +1

−2

∑n=1

(−1)n 1n2 +1

∣∣∣∣∣≤ 110

Definition 6.5.6 A series ∑an is said to converge absolutely if ∑ |an| converges. Itis said to converge conditionally if ∑ |an| fails to converge but ∑an converges.

Thus the alternating series or more general Dirichlet test can determine convergence ofseries which converge conditionally.

6.6 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

).

In other words, first sum on j yielding something which depends on k and then sum these.The major consideration for these double series is the question of when

∞

∑k=m

∞

∑j=m

a jk =∞

∑j=m

∞

∑k=m

a jk.

In other words, when does it make no difference which subscript is summed over first? Inthe case of finite sums there is no issue here. You can always write

M

∑k=m

N

∑j=m

a jk =N

∑j=m

M

∑k=m

a jk

because addition is commutative. However, there are limits involved with infinite sums andthe interchange in order of summation involves taking limits in a different order. Therefore,it is not always true that it is permissible to interchange the two sums. Whenever you