Close form of a power series starting at $n=2$

Question

This is the power series I am looking at $\sum_{n=2}^{\infty}{n(n-1)z^n}$. I want to find the closed form of this power series.

This is my approach, if I divide the power series by $z^2$, then I will have $\sum_{n=2}^{\infty}{n(n-1)z^{n-2}}$. But,

$\sum_{n=2}^{\infty}{n(n-1)z^{n-2}}= (\sum_{n=0}^{\infty}z^n)''$

And I know that $1\over 1-z$$=\sum_{n=0}^{\infty}z^n$.

So, I took the second derivative of $1\over 1-z$ and get $2\over {(1-z)^3}$.

Getting back to the original series, since I divided by $z^2$, then the closed form of the power series $\sum_{n=2}^{\infty}{n(n-1)z^n}$ is $2z^2\over {(1-z)^3}$.

This makes sense to me, but did I make any logical mistakes? Please help me out.

Answer 1

Your reasoning is correct, but note that the series is valid for $|z|<1$