Stuck on finding the value of $\sum_{n=0}^\infty {n(n+1) \over 3^n}$ I am trying to find the value of the series 
$$
\sum_{n=0}^\infty {n(n+1) \over 3^n}
$$
Here's what I have done so far:
$$
\sum_{n=0}^\infty {n(n+1) \over 3^n}=\sum_{n=0}^\infty {n^2 \over 3^n}+\sum_{n=0}^\infty {n \over 3^n}
$$
Determining the values of the both terms separately:
\begin{eqnarray}
S_1 = \sum_{n=0}^\infty {n \over 3^n} = {1 \over 3^1}+{2 \over 3^2}+{3 \over 3^3}+ \dots \\
= {1 \over 3^1} + {1 \over 3^2} + {1 \over 3^3} + \dots \\ +{1 \over 3^2} + {1 \over 3^3} + \dots \\ +{1 \over 3^3} + \dots \\ 
= {1 \over 3^1} + {1 \over 3^2} + {1 \over 3^3} + \dots \\
+{1 \over 3^1}\left( {1 \over 3^1}+{1 \over 3^2}+\dots\right) \\
+ {1 \over 3^2}\left( {1 \over 3^1}+\dots\right)\\
+ \dots\\
= \sum_{n=1}^\infty{1 \over 3^n} \cdot \sum_{n=0}^\infty {1 \over 3^n}
\end{eqnarray}
Next evaluating $\sum_{n=1}^\infty{1 \over 3^n}$:
\begin{align*}
3S_1 - S_1 = \sum_{n=1}^\infty{1 \over 3^{n-1}} - \sum_{n=1}^\infty{1 \over 3^n} = \\ 
1 - \require{enclose}\enclose{updiagonalstrike}{1 \over 3^1} + \enclose{updiagonalstrike}{1 \over 3^1} - \enclose{updiagonalstrike}{1 \over 3^2} + \enclose{updiagonalstrike}{1 \over 3^2} - \enclose{updiagonalstrike}{1 \over 3^3} + \enclose{updiagonalstrike}{1 \over 3^3} - \dots - {1 \over 3^n} \\
= 1-{1 \over 3^n} = 2S_1;
\end{align*}
$$
S_1=\sum_{n=1}^\infty{1 \over 3^n} = \frac 12 \lim_{n \to \infty} \left( 1-{1 \over 3^n} \right) = \frac 12
$$
And thus we receive 
$$
\sum_{n=1}^\infty{n \over 3^n} = \frac 14;
$$
I was trying to evaluate the second term $\sum_{n=0}^\infty {n^2 \over 3^n}$, however it wasn't a success. Could you give me a hint, how I can proceed with this series?
 A: $|x|<1:\\\sum_{n=0}^{\infty }x^{n}=\frac{1}{1-x}\Rightarrow 1+\sum_{n=1}^{\infty }x^{n}=\frac{1}{1-x}\\\Rightarrow \sum_{n=1}^{\infty }x^{n}=\frac{x}{1-x}\\\frac{d}{dx}\sum_{n=0}^{\infty }x^{n}=\frac{d}{dx}\frac{1}{1-x}\Rightarrow \sum_{n=1}^{\infty }nx^{n-1}=\frac{1}{(1-x)^2}\Rightarrow \sum_{n=1}^{\infty }nx^{n}=\frac{x}{(1-x)^2}\\\sum_{n=1}^{\infty }(n+1)x^n=\frac{x}{(1-x)}+\frac{x}{(1-x)^2}\\$
$
\begin{array}{l}
 \left( {\sum\limits_{n = 0}^{ + \infty } {\left( {n + 1} \right)x^n } } \right)^\prime   =\sum\limits_{n = 1}^{ + \infty } {n\left( {n + 1} \right)x^{n - 1} }  = \frac{1}{{\left( {1 - x} \right)^2 }} + \frac{{1 + x}}{{\left( {1 - x} \right)^3 }} \\ 
  \Rightarrow \sum\limits_{n = 0}^{ + \infty } {n\left( {n + 1} \right)x^n }  = \frac{x}{{\left( {1 - x} \right)^2 }} + \frac{{x^2  + x}}{{\left( {1 - x} \right)^3 }} \\ 
 x = \frac{1}{3} \Rightarrow \sum\limits_{n = 0}^{ + \infty } {n\left( {n + 1} \right)\left( {\frac{1}{3}} \right)^n }  = \frac{{\left( {\frac{1}{3}} \right)}}{{\left( {1 - \left( {\frac{1}{3}} \right)} \right)^2 }} + \frac{{\left( {\frac{1}{3}} \right)^2  + \left( {\frac{1}{3}} \right)}}{{\left( {1 - \left( {\frac{1}{3}} \right)} \right)^3 }} \\ 
  \Rightarrow \sum\limits_{n = 0}^{ + \infty } {n\left( {n + 1} \right)\left( {\frac{1}{3}} \right)^n }  = \frac{9}{4} \\ 
 \end{array}$
A: Write it as:
$$
\frac{1}{3} \sum_{n \ge 0} (n + 2) (n + 1) \left( \frac{1}{3} \right)^n
$$
This looks suspicious...
$$
\frac{\mathrm{d}^2}{\mathrm{d} z^2} \frac{1}{1 - z}
  = \sum_{n \ge 0} (n + 2) (n + 1) z^n
$$
and you are all set.
