Using the identity $\frac{1}{1-z} = 1 + z + z^2 + \ldots$ for $|z| < 1$, find closed forms for the sums $\sum n z^n$ and $\sum n^2 z^n$.
My solution: Because $\displaystyle1 + z + z^{2} + \ldots = \frac{1}{1-z}$, $\displaystyle1 + 2z + 3z^2 + \ldots = \sum_{n=1}^\infty n z^{n-1} = \frac{1}{(1-z)^2}$ and $\displaystyle\sum_{n=1}^\infty n z^n = \frac{z}{(1-z)^2}.$ Similarly, $\displaystyle\sum_{n=1}^\infty n^2 z^{n-1}=\frac{1+z}{(1-z)^3}$ and $\displaystyle\sum_{n=1}^\infty n^2 z^n=\frac{z(1+z)}{(1-z)^3}$.
Could you please improve this exercise?