evaluating $\sum_{n=1}^\infty \frac{n^2}{3^n}$ I am trying to compute the sum $$\sum_{n=1}^\infty \frac{n^2}{3^n}.$$
I would prefer a nice method without differentiation, but if differentiation makes it easier, then that's fine.
Can anyone help me?
Thanks.

We have 
\begin{gather*}
S=\sum_{n=1}^{\infty}\frac{n^2}{3^n}=\frac{1}{3}\sum_{n=0}^{\infty}\frac{(n+1)^2}{3^{n-1}}=\frac{1}{3}\left(\sum_{n=0}^{\infty}\frac{n^2}{3^{n-1}}+2\sum_{n=0}^{\infty}\frac{n}{3^{n-1}}+\sum_{n=0}^{\infty}\frac{1}{3^{n-1}}\right)=\frac{1}{3}\left(S+2 S_1+S_2\right).
\end{gather*}
Then 
$$
S_2=\sum_{n=0}^{\infty}\frac{1}{3^{n-1}}=\frac{3}{2},
$$
$$S_1=\sum_{n=0}^\infty \frac{n}{3^n}=\sum_{n=1}^\infty  \frac{n}{3^n}= \frac{1}{3}\sum_{n=1}^\infty  \frac{n}{3^{n-1}}=  \frac{1}{3}\sum_{n=0}^\infty  \frac{n+1}{3^{n}}= \frac{1}{3}\left[\sum_{n=0}^\infty  \frac{n}{3^{n}}+\sum_{n=0}^\infty  \frac{1}{3^{n}}\right] =\frac{1}{3}\left[S_1+ S_2\right].$$
By solving this equation you get $S_1=\dfrac{3}{4}.$
Thus we have the equation
$$
S=\frac{1}{3}(S+2 \cdot \frac{3}{4}+\frac{3}{2}).
$$
By solving it you will get $S=\dfrac{3}{2}.$
