Question

test the series for converge or divergence.
n^2e^(-n).
I'm not sure what method to use here. I tried integral test...but that didnt seem to work.

Answer 1

use the ratio test:$$\lim_{n\to\infty}\left|\frac{(n+1)^2e^n}{n^2e^{n+1}}\right|=\lim_{n\to\infty}\left(\frac{n+1}{n}\right)^2\frac{e^n}{e^{n+1}}=\frac1e\lim_{n\to\infty}\left(1+\frac1n\right)^2=\frac1e$$

Answer 2

\(1/e<1\) hence the series converges

Answer 3

thanks oldrin! Youre the man. How did you know to use the ratio test?

Answer 4

oh ok...I just learned this stuff today, so Im still working on it, but thanks for your help!

Answer 5

oops I did the root test incorrectly:$$\lim_{n\to\infty}\sqrt[n]{n^2e^{-n}}=\frac1e\lim_{n\to\infty}n^{2/n}=\frac1e<1$$

Answer 6

@Requiem no problem; once you get used to them it's a lot easier to know which test seems adequate

Answer 7

I'd also suggest you try proving that the tests work :-p

Answer 8

:P one step at a time oldrin!