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 $m\to\infty$?
 A: 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,
$$f(z) = \frac{z(1+z)}{(1-z)^3}$$
A: There is another similar way to do it using the small trick $n^2=n(n-1)+n$
$$f(z)=\sum_{n= 0}^\infty n^2 x^n=\sum_{n= 0}^\infty [n(n-1)+n] x^n$$
$$f(z)=\sum_{n= 0}^\infty n(n-1) x^n+\sum_{n= 0}^\infty n x^n=x^2\sum_{n= 0}^\infty n(n-1) x^{n-2}+x\sum_{n= 0}^\infty n x^{n-1}$$
$$f(z)=x^2 \left(\sum_{n= 0}^\infty  x^{n} \right)''+x \left(\sum_{n= 0}^\infty  x^{n} \right)'$$
The smae would have been applied if you had $n^3$ or $n^4$ instead of $n^2$ since
$$n^3=n(n-1)(n-2)+3n(n-1)+n$$
$$n^4=n(n-1)(n-2)(n-3)+6n(n-1)(n-2)+7n(n-1)+n$$
