6.3. RATIO AND ROOT TESTS 159

this 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.

Example 6.2.12 Determine the convergence of ∑∞k=1

1√n2+100n

.

Use the limit comparison test. limn→∞

( 1n )(1√

n2+100n

) = 1 and so this series diverges along

with ∑∞k=1

1k .

Sometimes it is good to be able to say a series does not converge. The nth term testgives such a condition which is sufficient for this. It is really a corollary of Theorem 6.1.8.Here is the nth term test.

Theorem 6.2.13 If ∑∞n=m an converges, then limn→∞ an = 0.

Proof:Apply Theorem 6.1.8 to conclude that limn→∞ an = limn→∞ ∑nk=n ak = 0.

It is very important to observe that this theorem goes only in one direction. That is,you cannot conclude the series converges if limn→∞ an = 0. If this happens, you don’tknow anything from this information. Recall limn→∞ n−1 = 0 but ∑

∞n=1 n−1 diverges. The

following picture is descriptive of the situation.

∑an converges

liman = 0

an = n−1

6.3 Ratio and Root TestsA favorite test for convergence is the ratio test. This is discussed next. There are exactlythree possible outcomes for this test: failure, spectacular divergence, and absolute conver-gence.

Theorem 6.3.1 Suppose |an|> 0 for all n and suppose limn→∞|an+1||an| = r. Then

∞

∑n=1

an

diverges if r > 1

converges absolutely if r < 1

test fails if r = 1

.

Proof: Suppose r < 1. Then there exists n1 such that if n ≥ n1, then 0 <∣∣∣ an+1

an

∣∣∣ < R

where r < R < 1. Then |an+1|< R |an| for all such n. Therefore,∣∣an1+p∣∣< R

∣∣an1+p−1∣∣< R2 ∣∣an1+p−2

∣∣< · · ·< Rp |an1 | (6.6)

and so if m > n, then |am| < Rm−n1 |an1 | . By the comparison test and the theorem on geo-metric series, ∑ |an| converges. This proves the convergence part of the theorem.