A favorite test for convergence is the ratio test. This is discussed next. It is at
the other extreme from the alternating series test, being completely oblivious
to any sort of cancellation. It only gives absolute convergence or spectacular
divergence.

Theorem 10.4.7Suppose

|an|

> 0 for all n and suppose

|an+1|
nl→im∞ -|an|--= r.

Then

(
∑∞ { diverges if r > 1
an( converges absolutely if r < 1 .
n=1 test fails if r = 1

Proof: Suppose r < 1. Then there exists n_{1} such that if n ≥ n_{1}, then

||an+1||
0 < ||-an-|| < R

where r < R < 1. Then

|an+1| < R |an|

for all such n. Therefore,

2 p
|an1+p| < R |an1+p−1| < R |an1+p−2| < ⋅⋅⋅ < R |an1| (10.7)

(10.7)

and so if m > n, then

|am|

< R^{m−n1}

|an1|

. By the comparison test and the theorem on
geometric series, ∑

|an|

converges. This proves the convergence part of the
theorem.

To verify the divergence part, note that if r > 1, then 10.7 can be turned around for
some R > 1. Showing lim_{n→∞}

|an|

= ∞. Since the n^{th} term fails to converge to 0, it
follows the series diverges.

To see the test fails if r = 1, consider ∑n^{−1} and ∑n^{−2}. The first series diverges
while the second one converges but in both cases, r = 1. (Be sure to check this last
claim.) ■

The ratio test is very useful for many different examples but it is somewhat
unsatisfactory mathematically. One reason for this is the assumption that a_{n}≠0,
necessitated by the need to divide by a_{n}, and the other reason is the possibility that the
limit might not exist. The next test, called the root test removes both of these
objections.

Theorem 10.4.8Suppose

|an|

^{1∕n}< R < 1 for all n sufficiently large.Then

∞∑
an converges absolutely.
n=1

If there are infinitely many values of n such that

|an|

^{1∕n}≥ 1, then

∞∑
an diverges.
n=1

Proof: Suppose first that

|a |
n

^{1∕n}< R < 1 for all n sufficiently large. Say this holds
for all n ≥ n_{R}. Then for such n,

n∘|an| < R.

Therefore, for such n,

n
|an| ≤ R

and so the comparison test with a geometric series applies and gives absolute convergence
as claimed.

Next suppose

|an|

^{1∕n}≥ 1 for infinitely many values of n. Then for those
values of n,

|an|

≥ 1 and so the series fails to converge by the n^{th} term test.
■

Stated more succinctly, using Definition 3.2.12 the condition for the root test is this:
Let

r = lim sup |an|1∕n
n→ ∞

then

(
∞∑ { converges absolutely if r < 1
ak test fails if r = 1
k=m ( diverges if r > 1

To see the test fails when r = 1, consider the same example given above, ∑_{n}

1n

and∑_{n}

1n2

.

A special case occurs when the limit exists.

Corollary 10.4.9Suppose lim_{n→∞}

|an|

^{1∕n}exists and equals r. Then

∞ ( converges absolutely if r < 1
∑ a { test fails if r = 1
k=m k( diverges if r > 1

Proof: The first and last alternatives follow from Theorem 10.4.8. To see the test
fails if r = 1, consider the two series ∑_{n=1}^{∞}

1n

and ∑_{n=1}^{∞}

n12

both of which have r = 1
but having different convergence properties. The first diverges and the second converges.
■