Evaluate a power series by relating it to a geometric series (where coefficients depend on the index of summation). I am trying to evaluate the power series $f\left(z\right)=\sum_{n\geq0}n^{2}z^{n}$ by relating it to a geometric series. By the ratio test, the series converges (absolutely) for $\left|z\right|<1$.
I assume it's not as simple as writing $f\left(z\right)=\sum_{n\geq0}\left(n^{2/n}z\right)^{n}$? 
Should I proceed as in the case of the standard geometric series, i.e., by writing down an expression for the $m$th partial sum and then taking the limit as $m\to\infty$?

Start with the fact that
$$\frac{1}{1-x} = \sum_{n\geq 0} x^n$$
(This is the simplest formula for an infinite geometric series)
Differentiate to obtain
$$\frac{1}{(1-x)^2} = \sum_{n\geq 0} nx^{n-1}$$
and multiply by $x$:
$$\frac{x}{(1-x)^2} = \sum_{n\geq 0} nx^{n}$$
Repeat the top two steps and you get a closed formula
$$\frac{1+x}{(1-x)^3} = \sum_{n\geq 0} n^2 x^{n-1}$$
$$\frac{x(1+x)}{(1-x)^3} = \sum_{n\geq 0} n^2 x^{n}$$
Therefore,
$$f(z) = \frac{z(1+z)}{(1-z)^3}$$
