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 $|z|\lt1$.