Compute $\sum_{n=1}^\infty\frac{1}{(1-z^n)(1-z^{n+1})}z^{n-1}$ and show its uniform convergence Given the power series $$P:=\displaystyle\sum_{n=1}^\infty\frac{1}{(1-z^n)(1-z^{n+1})}z^{n-1}$$ I want to show that $P$ converges uniformly in $\mathbb{C}$ and compute its limit.
I've tried to multiply $P$ with $(1-z)z$ which yields $$\tilde{P}=\displaystyle\sum_{n=1}^\infty\left(\frac{1}{z^{n+1}-1}+\frac{1}{z^n-1}\right)$$ However, I'm unable to continue from here. What's the trick now?

Hint: Note that
$$
\frac{z^{n-1}}{(1-z^n)(1-z^{n+1})}
=\frac1{1-z}\left(\frac{z^{n-1}}{1-z^n}-\frac{z^n}{1-z^{n+1}}\right)
$$
Therefore,
$$
\begin{align}
\sum_{n=1}^N\frac{z^{n-1}}{(1-z^n)(1-z^{n+1})}
&=\frac1{1-z}\left(\frac1{1-z}-\frac{z^N}{1-z^{N+1}}\right)\\
&=1-\frac1{1-z}\frac{z^N}{1-z^{N+1}}
\end{align}
$$
You should be able to use this on compact subsets of $|z|\lt1$.
