Sum of a simple infinite series 
Evaluate:
  $$\sum\limits_{n=1}^\infty \frac{n^2}{3^n}.$$

By the ratio test, $\displaystyle\lim_{n\to\infty} \frac{(n+1)^2}{3^{n+1}}\cdot\frac{3^n}{n^2}=1/3,$ which is less than 1, therefore the series is convergent.
Now I am stuck on how to evaluate this series, without the $n^2$ on top, it can be easily calculated by the geometric series formula. Any help would be appreciated.

Just to expand on David's comment:
\begin{align*}
\frac{1}{1-x} &= 1 + x + x^2 + \cdots + x^n + \cdots \\
\frac{d}{dx} \left( \frac{1}{1-x} \right) =
\frac{1}{(1-x)^2} &= 1 + 2x + 3x^2 + \cdots + nx^{n-1} + \cdots \\
\frac{x}{(1-x)^2} &= x + 2x^2 + 3x^3 + \cdots + nx^{n} + \cdots \\
\frac{d}{dx} \left( \frac{x}{(1-x)^2} \right) = 
\frac{1 + x}{(1 - x)^3} &= 1 + 2^2x + 3^2x^2 + \cdots + n^2x^{n-1} + \cdots \\
\frac{x(1 + x)}{(1 - x)^3} &= x + 2^2x^2 + 3^2x^3 + \cdots + n^2x^{n} + \cdots \\
\end{align*}
Now plug in $x = \frac13$.
