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

where r < R < 1. Then

for all such n. Therefore,
 (5.7) 
and so if m > n, then
To verify the divergence part, note that if r > 1, then 5.7 can be turned around for some R > 1. Showing lim_{n→∞}
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. Before presenting this test, it is necessary to first prove the existence of the p^{th} root of any positive number. This was shown earlier in Theorem 2.11.2 but the following lemma gives an easier treatment of this issue based on theorems about sequences.
Lemma 5.3.8 Let α > 0 be any nonnegative number and let p ∈ ℕ. Then α^{1∕p} exists. This is the unique positive number which when raised to the p^{th} power gives α.
Proof: Consider the function f

By the nested interval theorem, there exists a unique point x in all these intervals. Generalizing Theorem 4.4.6 slightly to include the product of of the terms of finitely many sequences, it follows from Theorem 4.4.11 that

Hence f

If there are infinitely many values of n such that

Proof: Suppose first that

Therefore, for such n,

and so the comparison test with a geometric series applies and gives absolute convergence as claimed.
Next suppose
Stated more succinctly the condition for the root test is this: Let

then

To see the test fails when r = 1, consider the same example given above, ∑ _{n}
A special case occurs when the limit exists.
Proof: The first and last alternatives follow from Theorem 5.3.9. To see the test fails if r = 1, consider the two series ∑ _{n=1}^{∞}