Calculus- Early Transcendentals, 2021a

9.7. The Ratio and Root Tests 361
Proof. The example above essentially proves the first part of this, if we simply replace 1/5 byL and 1/2
by r. Suppose that L > 1, and pick r so that 1 < r < L. Then for n ≥ N, forsomeN,
|a n+1 |
|a n |
> r and |a n+1 | > r|a n |.
This implies that |a N+k | > r k |a N |, but since r > 1 this means that lim |a N+k | ≠ 0, which means also
k→∞
that lim a n ≠ 0. By the divergence test, the series diverges.
n→∞
To see that we get no information when L = 1, we need to exhibit two series with L = 1, one that
converges and one that diverges. The series ∑1/n 2 and ∑1/n provide a simple example.
♣
The ratio test is particularly useful for series involving the factorial function.
Example 9.43
Analyze
∞
∑
n=0
5 n
n! .
Solution.
5 n+1 n!
lim
n→∞ (n + 1)!
Since 0 < 1, the series converges.
5 n = lim
n→∞
5 n+1 n!
5 n (n + 1)! = lim 5 1
n→∞ (n + 1) = 0.
♣
A similar argument justifies a similar test that is occasionally easier to apply.
Theorem 9.44: The Root Test
Suppose that lim
n→∞
|a n | 1/n = L. If L < 1 the series ∑a n converges absolutely, if L > 1 the series
diverges, and if L = 1 this test gives no information.
The proof of the root test is actually easier than thatoftheratiotest,andisleftasanexercise.
Example 9.45
Analyze
∞
∑
n=0
5 n
n n .
Solution. The ratio test turns out to be a bit difficult on this series (try it). Using the root test:
( 5
n
) 1/n
(5 n ) 1/n
lim
n→∞ n n = lim
n→∞ (n n )
1/n
= lim
n→∞
5
n = 0.
Since 0 < 1, the series converges.
♣
The root test is frequently useful when n appears as an exponent in the general term of the series.