6.2. ABSOLUTE CONVERGENCE 157

Thus the even partial sums are decreasing and the odd partial sums are increasing. Theeven partial sums are bounded below also. (Why?) Therefore, the limit of the even partialsums exists. However, it must be the same as the limit of the odd partial sums becauseof the last equality above. Thus limn→∞ Sn exists and so the series converges. Now I willshow later below that ∑k

12k and ∑k

12k−1 both diverge. Include enough even terms for the

sum to exceed 7. Next add in enough odd terms so that the result will be less than 7. Nextadd enough even terms to exceed 7 and continue doing this. Since 1/k converges to 0, thisrearrangement of the series must converge to 7. Of course you could also have picked 5 or−8 just as well. In fact, given any number, there is a rearrangement of this series whichconverges to this number. Calculus is not algebra! No such thing happens with finite sums!

Theorem 6.2.5 (comparison test) Suppose {an} and {bn} are sequences of non neg-ative real numbers and suppose for all n sufficiently large, an ≤ bn. Then

1. If ∑∞n=k bn converges, then ∑

∞n=m an converges.

2. If ∑∞n=k an diverges, then ∑

∞n=m bn diverges.

Proof: Consider the first claim. From the assumption, there exists n∗ such that n∗ >max(k,m) and for all n ≥ n∗ bn ≥ an. Then if p ≥ n∗,

p

∑n=m

an ≤n∗

∑n=m

an +k

∑n=n∗+1

bn ≤n∗

∑n=m

an +∞

∑n=k

bn.

Thus the sequence,{∑pn=m an}∞

p=m is bounded above and increasing. Therefore, it convergesby completeness. The second claim is left as an exercise.

Example 6.2.6 Determine the convergence of ∑∞n=1

1n2 .

For n > 1, 1n2 ≤ 1

n(n−1) .Now

p

∑n=2

1n(n−1)

=p

∑n=2

[1

n−1− 1

n

]= 1− 1

p→ 1 as p → ∞

Therefore, letting an =1n2 and bn =

1n(n−1) the conclusion follows from Theorem 6.2.5.

A convenient way to implement the comparison test is to use the limit comparison test.This is considered next.

Theorem 6.2.7 Let an,bn > 0 and suppose for all n large enough,

0 < a <an

bn≤ an

bn< b < ∞.

Then ∑an and ∑bn converge or diverge together.

Proof: Let n∗ be such that n ≥ n∗, then anbn

> a and anbn

< b and so for all such n,abn <an < bbn and so the conclusion follows from the comparison test.

The following corollary follows right away from the definition of the limit.

Corollary 6.2.8 Let an,bn > 0 and suppose limn→∞anbn

= λ ∈ (0,∞) . Then ∑an and∑bn converge or diverge together.