# How do I calculate values of series? [duplicate]

Possible Duplicate:
How to find the sum of the following series

Hey there,

How do I (generally) calculate values of series? What tricks and theorems are there to make my life easier? E.g.: $$\sum_{n=1}^\infty \frac{n^2}{2^n} = ?$$

I know perfectly well how to show whether a series converges or diverges, but (at least in this and a lot of other cases) I have no idea how to calculate the value to which it converges. I'm looking forward to your help.

-
exact duplicate : math.stackexchange.com/questions/20418/… –  mercio Feb 24 '11 at 16:42
@Qiaochu: This question is most certainly not an exact duplicate of the prior question since this one asks about general techniques for summing series, whereas the prior question asks only about one specific series. Now that the question is closed, the OP has no way to get answers to the general question. –  Gone Feb 24 '11 at 17:47
@Bill: In fairness to Qiaochu, I did flag this as a duplicate for moderator attention, so maybe you should be mad at me. (Of course a priori I don't think there is any way to see who flagged what.) Having said that, maybe you are right, but it is definitely a close call . Perhaps the question should be changed to "How do I evaluate $\sum_{n=1}^{\infty}\frac{P(n)}{r^n}$ where $P(n)$ is some polynomial and $0<r<1$." In its current form it really looks like a duplicate, and I don't feel the OP's intent was a discussion on general evaluation of series. (But maybe it was?) –  Eric Naslund Feb 24 '11 at 18:51
@Bill: the fact that the OP accepted the answer specific to his question indicates to me that he was not really that interested in the general case. If you want to argue to reopen the question feel free to start a meta thread. –  Qiaochu Yuan Feb 24 '11 at 18:53
@Qiaochu: Accepting an answer or not does not change what was asked, viz. "How do I (generally) calculate the value of series". Yet another abuse of binding close votes. When will it end? Do I have to start a meta thread to drum up community support to end such? –  Gone Feb 24 '11 at 19:37
Consider the more general problem $\sum {n^2}{z^n}$. To handle this, start with $\sum {z^n}$ and $\sum {n}{z^n}$. The first of these should be easy.