Show That This Complex Sum Converges For complex $z$, show that the sum
$$\sum_{n = 1}^{\infty} \frac{z^{n - 1}}{(1 - z^n)(1 - z^{n + 1})}$$
converges to $\frac{1}{(1 - z)^2}$ for $|z| < 1$ and $\frac{1}{z(1 - z)^2}$ for $|z| > 1$.
Hint: Multiply and divide each term by $1 - z$, and do a partial fraction decomposition, getting a telescoping effect.
I tried following the hint, but got stuck on performing a partial fraction decomposition.  After all, since all polynomials can be factored in $\mathbb{C}$, how do I know what the factors of an arbitrary term are?  I tried writing
$$\frac{z^{n - 1}(1 - z)}{(1 - z^n)(1 - z^{n + 1})(1 - z)} = \frac{z^{n - 1}}{(1 - z)^3(1 + z + \dotsb + z^{n - 1})(1 + z + \dotsb + z^n)} - \frac{z^n}{(1 - z)^3(1 + z + \dotsb + z^{n - 1})(1 + z + \dotsb + z^n)},$$
but didn't see how this is helpful.

HINT: Use
$$
   \frac{z^{n}-z^{n+1}}{(1-z^n)(1-z^{n+1})} = \frac{1}{1-z^n} - \frac{1}{1-z^{n+1}}
$$
