Decomposing a particular fraction

Question

I am working on Problem I.4.7 in Lang's Complex Analysis. It asks to find the convergence of \begin{equation}\sum_\limits{n=1}^\infty \frac{z^{n-1}}{(1-z^n)(1-z^{n+1})}.\end{equation} The hint says to multiply and divide each term by $1-z$ and do a partial fraction decomposition to get a telescoping sum. I tried this and got \begin{equation}\frac{A}{(1-z^n)(1-z)} + \frac{B}{(1-z^{n+1})(1-z)} + \frac{C}{(1-z^n)(1-z^{n+1})} = \frac{z^{n-1}-z^n}{(1-z^n)(1-z^{n+1})(1-z)}.\end{equation}

By inspection, I found that

\begin{equation} \begin{split} A&= 0\\ B&=-\frac{1}{z}+1 \\ C&= \frac{1}{z} \end{split} \end{equation}

is a solution. This however, does not give me a telescoping sum. The resources I have found only talk about partial fraction decomposition after reducing to irreducible quadratics, which does not seem tenable here. Are there any more general techniques to partial fraction decomposition might help here?

Edit: Here is the decomposition the above obtains, in case it is useful:

\begin{equation} \frac{z^{n-1}}{(1-z^n)(1-z^{n+1})}=\frac{1}{(1-z^{n+1})(1-z)} - \frac{1}{z(1-z^{n+1})(1-z)} + \frac{1}{z(1-z^{n+1})(1-z^n)} \end{equation}

Answer 1

The hint says to multiply and divide each term by $1-z$ and do a partial fraction decomposition

Not entirely sure what the hint meant to say, but the following telescopes nicely regardless:

$$ \begin{align} \frac{z^{n-1}}{(1-z^n)(1-z^{n+1})} &= \frac{z^{n-1}}{(1-z)^2(1+z+\ldots+z^{n-1})(1+z+\ldots+z^{n-1}+z^n)} \\[5px] &= \frac{z^{n-1}}{(1-z)^2} \cdot \frac{1}{z^n}\left(\frac{1}{1+z+\ldots+z^{n-1}} - \frac{1}{1+z+\ldots+z^{n-1}+z^n}\right) \end{align} $$

Then:

$$ \begin{align} z(1-z)^2\sum_\limits{n=1}^{N} \frac{z^{n-1}}{(1-z^n)(1-z^{n+1})} &= \sum_\limits{n=1}^{N} \left(\frac{1}{1+z+\ldots+z^{n-1}} - \frac{1}{1+z+\ldots+z^n}\right) \\[5px] &= 1 - \frac{1}{1+z+\ldots+z^N} \\[5px] &= 1 - \frac{1-z}{1-z^{N+1}} \\[5px] &= \frac{z(1-z^N)}{1-z^{N+1}} \end{align} $$

Answer 2

You may also ignore the hint and perform the following formal manipulations: $$ \frac{x^{n-1}}{(1-x^n)(1-x^{n+1})}=\sum_{a,b\geq 0}x^{(a+b+1)n+(b-1)} $$

$$ \sum_{n\geq 1}\sum_{a,b\geq 0}x^{(a+b+1)n+(b-1)}=\sum_{a,b\geq 0}\frac{x^{a+2b}}{1-x^{a+b+1}}=\sum_{m\geq 0}r(m)\,x^m $$ where $r(m)$ is the number of $(a,b,c)\in\mathbb{N}^3$ such that $(c+1)a+(c+2)b+c = m$.
$r(m)$ is clearly less than $(m+1)^3$, hence the radius of convergence of the last power series is at least one, and it cannot be more than one since for any prime $p$ the $p$-th partial sum of the original series has a singularity at $x=\exp\left(\frac{2\pi i}{p}\right)$.