evaluating $\sum_{n=1}^\infty \frac{n^2}{3^n}$ I am trying to compute the sum $$\sum_{n=1}^\infty \frac{n^2}{3^n}.$$
I would prefer a nice method without differentiation, but if differentiation makes it easier, then that's fine.
Can anyone help me?
Thanks.

Here is a non-calculus method which works in these cases:
Put
$$S = \frac{1}{3} + \frac{4}{9} + \frac{9}{27} + \frac{16}{81} + \dots$$
then $$\frac{1}{3}S = \frac{1}{9} + \frac{4}{27} + \frac{9}{81} + \dots$$
subtract:
$$\frac{2}{3}S = \frac{1}{3} + \frac{3}{9} + \frac{5}{27} + \dots$$
This has simplified the problem (since the numerators are now linear instead of quadratic), and if you repeat the process, you get the answer, using the sum of a geometric progression.
