series convergence lang page 26 exercise 7 Show that the series
$$\sum_n \frac{z^{n-1}}{(1-z^n)(1-z^{n+1})},\quad z\in\mathbb{C},$$
converges to $\frac{1}{(1-z)^2}$ for $|z|<1$ and to $\frac{1}{z(1-z)^2}$ for $|z|>1$...
I try prove it so;
I know that $\sum z^n = \frac{1}{1-z}$ when $|z|<1$ thus $\sum (n+1)z^n = \frac{1}{(1-z)^2}$  in this way I would to prove that $ \sum (n+1)z^n = \sum \frac{z^{n-1}}{(1-z^n)(1-z^{n+1})}$
Some recommendation for continue it way or some other idea?
This exercise is in  $\textit{Complex Analysis - Serge Lang}$ - Third Edition, page $26$
thank your.

Note $$\frac{z^{n-1}}{(1-z^n)(1-z^{n+1})} = \frac{1}{z(1-z)}\left[\frac{1}{1-z^n} - \frac{1}{1-z^{n+1}}\right]$$ and $$\lim_{n\to \infty} \frac{1}{1-z^{n+1}} = \begin{cases}1&\text{if $\lvert z\rvert < 1$}\\ 0 & \text{if $\lvert z\rvert > 1$}\end{cases}$$
So the series telescopes to $$\frac{1}{z(1-z)}\left[\frac{1}{1-z} - 1\right] = \frac{1}{(1-z)^2}$$when $\lvert z\rvert < 1$, and to $$\frac{1}{z(1-z)}\left[\frac{1}{(1-z)}\right] = \frac{1}{z(1-z)^2}$$when $\lvert z\rvert > 1$.
