Computing $\sum_{n=1}^ \infty n^2 \cdot \left(\frac{2}{3}\right)^n$ I've been dealing with the following series for a while now, without real progress.
$$\sum_{n=1}^ \infty n^2 \cdot \left(\frac{2}{3}\right)^n$$
After using WolframAlpha, I know it converges to $30$, but I can't see how to calculate it by myself.
Any leads would be greatly appreciated!

For $-1<x<1$, $\displaystyle \sum_{n=1}^{k}x^n=\frac{x(1-x^{k})}{1-x}$.
Differentiating,
\begin{align*}
\sum_{n=1}^knx^{n-1}&=\frac{(1-x)[1-(k+1)x^k]-(x-x^{k+1})(-1)}{(1-x)^2}\\
&=\frac{1-(k+1)x^k+kx^{k+1}}{(1-x)^2}\\
\end{align*}
So.
\begin{align*}
\sum_{n=1}^knx^{n}&=\frac{(1-x)[1-(k+1)x^k]-(x-x^{k+1})(-1)}{(1-x)^2}\\
&=\frac{x-(k+1)x^{k+1}+kx^{k+2}}{(1-x)^2}\\
\end{align*}
Then differentiate again and multiply by $x$. Put $x=\frac{2}{3}$. Take $k\to\infty$.
