88 CHAPTER 5. INFINITE SERIES OF NUMBERS
compare with. Of course this is not always easy and there is room for acquiring skillthrough practice.
To really exploit this limit comparison test, it is desirable to get lots of examples ofseries, some which converge and some which do not. The tool for obtaining these exampleshere will be the following wonderful theorem known as the Cauchy condensation test.
Theorem 5.2.10 Let an ≥ 0 and suppose the terms of the sequence {an} are de-creasing. Thus an ≥ an+1 for all n. Then ∑
∞n=1 an and ∑
∞n=0 2na2n converge or diverge
together.
Proof: This follows from the inequality of the following claim.Claim:
n
∑k=1
2ka2k−1 ≥2n
∑k=1
ak ≥n
∑k=0
2k−1a2k . (5.5)
Proof of the Claim: Note the claim is true for n = 1. Suppose the claim is true for n.Then, since 2n+1−2n = 2n, and the terms an, are decreasing,
n+1
∑k=1
2ka2k−1 = 2n+1a2n +n
∑k=1
2ka2k−1 ≥ 2n+1a2n +2n
∑k=1
ak
≥2n+1
∑k=1
ak ≥ 2na2n+1 +2n
∑k=1
ak ≥ 2na2n+1 +n
∑k=0
2k−1a2k =n+1
∑k=0
2k−1a2k .
By induction, the claim is valid. Then passing to a limit in 5.5
2∞
∑k=0
2ka2k =∞
∑k=1
2ka2k−1 ≥∞
∑k=1
ak ≥∞
∑k=0
2k−1a2k =12
∞
∑k=0
2ka2k
Thus, if ∑∞k=0 2ka2k < ∞ then the partial sums of ∑
∞k=1 ak are bounded above by ∑
∞k=0 2ka2k
so these partial sums converge. If ∑∞k=0 2ka2k diverges, then
∞ = limn→∞
12
n
∑k=0
2ka2k ≤ limn→∞
n
∑k=1
ak
and so ∑ak also diverges. Thus the two series converge or diverge together.
Example 5.2.11 Determine the convergence of ∑∞k=1
1kp where p is a positive number.
These are called the p series.
Let an =1
np . Then a2n =( 1
2p
)n. From the Cauchy condensation test the two series
∞
∑n=1
1np and
∞
∑n=0
2n(
12p
)n
=∞
∑n=0
(2(1−p)
)n
converge or diverge together. If p > 1, the last series above is a geometric series havingcommon ratio less than 1 and so it converges. If p ≤ 1, it is still a geometric series but inthis case the common ratio is either 1 or greater than 1 so the series diverges. It followsthat the p series converges if p > 1 and diverges if p≤ 1. In particular, ∑
∞n=1 n−1 diverges
while ∑∞n=1 n−2 converges.
∑∞n=1
1np p > 1 converges
∑∞n=1
1np p≤ 1 diverges