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.
 A: Here is a non-calculus method which works in these cases:
Put
$$S = \frac{1}{3} + \frac{4}{9} + \frac{9}{27} + \frac{16}{81} + \dots$$
then $$\frac{1}{3}S = \frac{1}{9} + \frac{4}{27} + \frac{9}{81} + \dots$$
subtract:
$$\frac{2}{3}S = \frac{1}{3} + \frac{3}{9} + \frac{5}{27} + \dots$$
This has simplified the problem (since the numerators are now linear instead of quadratic), and if you repeat the process, you get the answer, using the sum of a geometric progression.
A: Hints: 
For $\;|x|<1\;$ :
$$f(x):=\frac1{1-x}=\sum_{n=0}^\infty x^n\implies f'(x)=\frac1{(1-x)^2}=\sum_{n=1}^\infty nx^{n-1}\implies$$
$$f''(x)=\frac2{(1-x)^3}=\sum_{n=2}^\infty n(n-1)x^{n-2}=\sum_{n=2}^\infty n^2x^{n-2}-\sum_{n=2}^\infty nx^{n-2}\;\;\ldots$$
A: Here's a bit of a slick trick. Let's put $$S=\sum_{n=1}^\infty\frac{n^2}{3^n}$$ and $$L=\sum_{n=1}^\infty\frac{n}{3^n}.$$ You should be able to determine (via ratio test or some such) that both series are convergent. (You'll see why I brought up the other series in a moment.)
Note that $$\begin{align}S &= \frac13+\sum_{n=2}^\infty\frac{n^2}{3^n}\\ &= \frac13+\sum_{n=1}^\infty\frac{(n+1)^2}{3^{n+1}}\\ &= \frac13+\sum_{n=1}^\infty\frac{n^2}{3^{n+1}}+2\sum_{n=1}^\infty\frac{n}{3^{n+1}}+\sum_{n=1}^\infty\frac1{3^{n+1}}\\ &= \frac13+\frac13 S+\frac13 L+\frac13\sum_{n=1}^\infty\frac1{3^n}\\ &= \frac13 S+\frac13 L+\frac13\sum_{n=0}^\infty\frac1{3^n},\end{align}$$ so $$\frac23 S=\frac13 L+\frac13\sum_{n=0}^\infty\frac1{3^n},$$ so $$S=\frac12 L+\frac12\sum_{n=0}^\infty\frac1{3^n}.\tag{$\star$}$$ Now a similar trick shows us that $$\begin{align}L &= \frac13+\sum_{n=2}^\infty\frac{n}{3^n}\\ &= \frac13+\sum_{n=1}^\infty\frac{n+1}{3^{n+1}}\\ &= \frac13L+\frac13\sum_{n=0}^\infty\frac1{3^n},\end{align}$$ so $$\frac23 L=\frac13\sum_{n=0}^\infty\frac1{3^n},$$ so $$L=\frac12\sum_{n=0}^\infty\frac1{3^n},$$ and so by $(\star),$ we have $$S=\frac34\sum_{n=0}^\infty\frac1{3^n}.$$ All that's left is to evaluate the geometric series.
A: 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 $S=\dfrac{3}{2}.$
