158 CHAPTER 6. INFINITE SERIES

Example 6.2.9 Determine the convergence of ∑∞k=1

1√n4+2n+7

.

This series converges by the limit comparison test above. Compare with the series ofExample 6.2.6.

limn→∞

(1n2

)(

1√n4+2n+7

) = limn→∞

√n4 +2n+7

n2 = limn→∞

√1+

2n3 +

7n4 = 1.

Therefore, the series converges with the series of Example 6.2.6. How did I know what tocompare with? I noticed that

√n4 +2n+7 is essentially like

√n4 = n2 for large enough

n. You see, the higher order term n4 dominates the other terms in n4 + 2n+ 7. Therefore,reasoning that 1/

√n4 +2n+7 is a lot like 1/n2 for large n, it was easy to see what to

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 6.2.10 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: ∑

nk=1 2ka2k−1 ≥ ∑

2n

k=1 ak ≥ ∑nk=0 2k−1a2k .

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 .

In case it is not clear why the claim implies the assertion, first suppose ∑∞n=0 2na2n con-

verges. Then 2∑∞n=0 2na2n = ∑

∞n=1 2na2n−1 is finite. Then from the claim, ∑

2n

k=1 ak ≤∑

n+1k=1 2ka2k−1 ≤ ∑

∞n=1 2na2n−1 < ∞ and so the partial sums are bounded. Since the terms of

the series are nonnegative, the infinite series converges as shown earlier. In case ∑∞n=0 2na2n

diverges, a similar argument shows the partial sums of the original series are unbounded.

Example 6.2.11 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 in