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!

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Hint : $n^2=n(n-1)+n$ – Philippe Malot Apr 25 '13 at 17:14
Thanks! It made it clear :) – darenn Apr 25 '13 at 17:17
For reference purposes, this is essentially the polylogarithm $\mathrm{Li}_{-2}(1/7)$. – deoxygerbe Apr 25 '13 at 18:19

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.

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Personally, I'd rather differentiate first, then multiply by $x$. – Harald Hanche-Olsen Apr 25 '13 at 17:11
@HaraldHanche-Olsen True. =) You're absolutely right. – Pedro Tamaroff Apr 25 '13 at 17:15
Thanks for this addition! =) – darenn Apr 25 '13 at 17:19