I'm being tasked with showing that:
$\sum_{n=0}^\infty n^2 z^n= \frac{z(1+z)}{(1-z)^3}$ for $|z|<1$
I know that the series converges and the completeness of $\mathbb{C}$ ensures that: $\sum_{n=0}^\infty n^2 z^n=\lim_{n\rightarrow \infty}\sum_{n=0}^k n^2 z^n$.
However calculating $\sum_{n=0}^k n^2 z^n$ and taking the limit is a rather ugly endeavor (picture related).
Does anybody have a more elegant solution/hints?
$$\sum_{n=0}^\infty z^n=\dfrac{1}{1-z}$$ for $|z|<1$, then differentiation gives $$\sum_{n=1}^\infty nz^{n-1}=\dfrac{1}{(1-z)^2}$$ or $$\sum_{n=1}^\infty nz^{n}=\dfrac{z}{(1-z)^2}$$ another differentiation shows $$\sum_{n=1}^\infty n^2z^{n}=\dfrac{z(1+z)}{(1-z)^3}$$