ratio test on $\sum_{n=1}^{\infty} \frac{n}{5^n}$

Question

$\sum_{n=1}^{\infty} \frac{n}{5^n}$

doing the ratio test i get

$ \frac{n}{5^n} * \frac{5^{n+1}}{n+1} $

which results in

$ L = 5 \frac{n}{n+1} $

as limit reaches infinity on $L$. i get that $L$ is greater than 1 which is undefined behavior. but wolfram confirms it is 5/16. help

Note that any series of the form $\sum_{n=1}^\infty nr^n$ converges if r is less than one. http://math.stackexchange.com/questions/30732/how-can-i-evaluate-sum-n-0-infty-n1xn — GTX OC, Dec 09 '13 at 12:48
5/16 is the sum that this series converges to. The result of this test is the limit of the n'th element. — Billie, Dec 09 '13 at 12:51

score 2 · Accepted Answer · answered Dec 09 '13 at 12:44

2

You divided thew wrong way around: $$L=\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=\lim_{n\to\infty}\frac{n+1}{5^{n+1}}\frac{5^n}{n}=\lim_{n\to\infty}\frac15\frac{n+1}{n}=\frac15<1$$ So, by the ratio test, the series converges.

answered Dec 09 '13 at 12:44

Dennis Gulko

15,640

score 0 · Answer 2 · answered Dec 09 '13 at 12:44

0

Double check how to apply the ratio test. You took the limit of the reciprocal of what you should have.

$\frac{n+1}{5^{n+1}}\cdot\frac{5^n}{n}=\frac{n+1}{5n}\rightarrow\frac15<1$ as $n\rightarrow\infty$

answered Dec 09 '13 at 12:44

David H

29,921

Salech Alhasov · Answer 3 · 2013-12-09T13:23:18.573

-1

Using Cauchy root test, gives us the following:

$$\lim_{n\to\infty}\sqrt[n]{\frac{n}{5^n}}=\frac{1}{5}\lim_{n\to\infty}\sqrt[n]{n}=\frac{1}{5}<1$$

Thus, the given series is converges.

Edit: Yes, I know that the OP asked the ratio test.

edited Dec 09 '13 at 13:23

answered Dec 09 '13 at 12:49

Salech Alhasov

6,780
2
29
47

It should be a comment. This answer irrelevant to the question. – Billie Dec 09 '13 at 12:54
@SalechAlhasov Watch your language! – Michael Greinecker Dec 10 '13 at 00:04

ratio test on $\sum_{n=1}^{\infty} \frac{n}{5^n}$

3 Answers3