Calculate sum of an infinite series I have been struggling with this functional series. $$\sum_{n=1}^{\infty}{(-1)^{n-1}n^2x^n}$$ I need to calulate the sum.Any tips would be appreciated.
 A: Try considering what happens when you differentiate the following with respect to $x$:
$1/(1+x) = 1-x+x^2-x^3+x^4...$  
That should get you thinking in the right direction.
A: Note that $n^2=n(n-1)+n$ and hence the given series $\displaystyle\sum\limits_{n\geq 1} n^2 x^n=: f(n;x)$ can be written as 
$f(n;x)=\displaystyle\sum_{n\geq 1} n(n-1)x^n+\displaystyle\sum_{n\geq 1} n x^n=f_1(n;x)+f_2(n;x)$ respectively. Let us determine these $f_1(n;x)$ and $f_2(n;x)$ separately.
Now $f_1(n;x)=x^2 \displaystyle\sum_{n\geq 1} n(n-1)x^{n-2}=x^2 \frac{2}{(1+x)^3}=\frac{2x^2}{(1+x)^3}$ by using the fact $(1+x)^{-1}=\displaystyle\sum_{n\geq 1} x^n$ and two times successive differentiation.
On the other hand $f_2(n;x)=-x \frac{1}{(1+x)^2}=\frac{-x}{(1+x)^2}$.
Therefore $f(n;x)=\frac{2x^2}{(1+x)^3}-\frac{x}{(1+x)^2}$
A: Hints:
If you divide your series by $x$ and then integrate it, it might be a little simpler.  
Then ask yourself what you can do to make it even simpler.  If you are lucky you will eventually end up with a geometric series where you know the answer.
Then undo all the steps you have taken (in reverse order), and you should have a solution to your original question.
A: Incorporate $-1$ into $x$. Now consider the derivatives of $\sum z^n$. How can you get $n^2$ to show up?
A: The ratio of convergence is
$$\limsup_{n\to\infty}\frac{1}{\sqrt[n]{n^2}}=1,$$
so the series in convergent on $(-1;1)$. Now if move one $x$ factor before the sum, so that the exponent of $x$ in the sum is $n-1$, what happens if you integrate elementwise? Repeat!
