Help determining if the series $\sum_{n=0}^{\infty }\frac{n^2}{3^n}$ converges or diverges

Question

$\sum_{n=0}^{\infty }\frac{n^2}{3^n}$

I've tried using the ratio test but I can't seem to figure out what I'm doing wrong:

$$\lim_{n \rightarrow \infty}\left|\frac{(n+1)^2}{3^{n+1}}\cdot \frac{3^n}{n^2}\right|=\lim_{n \rightarrow \infty}\left|\frac{(n+1)^2}{3n^2}\right|=\frac{1}{3}\lim_{n \rightarrow \infty}\left|\frac{(n+1)^2}{n^2}\right|$$

Where do I go from here? It's been a few years since I've done any sequences and series.

Answer 1

$$\frac{1}{3}\lim_{n\to\infty}\left|\frac{(n+1)^2}{n^2}\right| = \frac{1}{3}\lim_{n\to\infty}\left|\frac{(n^2+2n+1)}{n^2}\right| = \frac{1}{3}\lim_{n\to\infty}\left|1 + \frac{2}{n} + \frac{1}{n^2}\right| = \frac{1}{3}(1 + 0 + 0) = \frac{1}{3}$$

Which is less than 1, so by the ratio test the series converges.

Answer 2

Root or ratio test and you are done. Both of the methods return a limit of 1/3.