Sum of power series $\sum_{n\geq0}\frac{n^2}{7^n}$ I'm having trouble with this power series : $$
\sum_{n\geq0}\frac{n^2}{7^n}$$
I have to solve it using differentiation/integration. I guess I have to approach it as $\sum_{n\geq0}n^2x^n$, where $x = \frac{1}{7}$ but I'm kinda lost as to what to do next.
Thanks in advance!
 A: Take $$\frac 1{1-x}=1+x+x^2+\cdots$$
Differentiate and multiply by $x$. Repeat once more. Profit.
ADD In general, let $D=x\frac{d}{dx}$ mean "differentiate then multiply by $x$", and let $D^k$ mean "do it $k$ times". Then $$D^k\frac{1}{1-x}=\sum_{n=1}^\infty n^kx^n$$
We have the following first cases:
$$\eqalign{
  & \frac{1}{{1 - x}} = {D^0}\frac{1}{{1 - x}}  \cr 
  & \frac{x}{{{{\left( {1 - x} \right)}^2}}} = {D^1}\frac{1}{{1 - x}}  \cr 
  & \frac{{{x^2} + x}}{{{{\left( {1 - x} \right)}^3}}} = {D^2}\frac{1}{{1 - x}}  \cr 
  & \frac{{{x^3} + 4x + x}}{{{{\left( {1 - x} \right)}^4}}} = {D^3}\frac{1}{{1 - x}} \cr} $$
The numbers that pop up on the left hand side, $1;1,1;1,4,1;\dots$ are the Eulerian numbers
ADD${}^*$ There is an alternative way of obtaining the expansion of  $$\sum_{n=1}^\infty n^kx^n$$ directly, and is by means of the so called falling factorial:
$$x^{\underline{n}}=x(x-1)\dots (x-n+1)$$
We have that $$x^k=\sum_{i=0}^k \left\{\begin{matrix} k \\ i\end{matrix}\right\}x^{\underline{i}}$$
where the "curly" binomials are the Stirling numbers of the second kind, which count the number of ways of partitioning a $k$ object set into $i$ nonempty sets. If you let $x=n$, we get that $$\begin{align}
  {n^k}{x^n} = \sum\limits_{i = 0}^k \left\{\begin{matrix} k \\ i\end{matrix}\right\}n\left( {n - 1} \right) \cdots \left( {n - i + 1} \right){x^n} \\
  \sum\limits_{n > 1} {{n^k}{x^n}}  = \sum\limits_{i = 0}^k\left\{\begin{matrix} k \\ i\end{matrix}\right\}\sum\limits_{n > 1} n\left( {n - 1} \right) \cdots \left( {n - i + 1} \right){x^n}  \\ 
   = \sum\limits_{i = 0}^k \left\{\begin{matrix} k \\ i\end{matrix}\right\}\sum\limits_{n \geq  i} n\left( {n - 1} \right) \cdots \left( {n - i + 1} \right){x^n}  \\
 = \sum\limits_{i = 0}^k \left\{\begin{matrix} k \\ i\end{matrix}\right\}x^i\sum\limits_{n \geq  i} n\left( {n - 1} \right) \cdots \left( {n - i + 1} \right){x^{n-i}}  \\
   = \sum\limits_{i = 0}^k \left\{\begin{matrix} k \\ i\end{matrix}\right\}{x^i}\frac{d^i}{dx^i}\frac{1}{1 - x}  \end{align} $$
so you can express your sum in terms of the "plain" derivatives.
