# Decomposing a particular fraction

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}

• $$\frac{1}{1-z^n} - \frac{1}{1 - z^{n+1}} = \frac{z^n-z^{n+1}}{(1-z^n)(1-z^{n+1})}$$ – Daniel Fischer Jan 5 '18 at 21:32
• $A, B, C$ should not be $z$-dependent – N74 Jan 5 '18 at 21:36
• @n74 then there is no solution the way I set it up. Is there a way to set it up to ensure there is a solution? – Daniel Mourad Jan 5 '18 at 21:37
• Using the hint by @DanielFischer $\frac{z^{n-1}}{(1-z^n)(1-z^{n+1})}= \frac{1}{z(1-z)(1-z^{n})} -\frac{1}{z(1-z)(1-z^{n+1})}$ – N74 Jan 5 '18 at 21:43
• Oh, I see. We can just treat the (1-z) as part of the other factors. Is there an algorithmic way of doing it? (Just curious, this is enough for me to solve the problem. Thanks! – Daniel Mourad Jan 5 '18 at 21:45

## 2 Answers

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}

• The hint just circumvents having to factor the factors in the denominator. – Daniel Mourad Jan 6 '18 at 21:42

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 $x=\exp\left(\frac{2\pi i}{p}\right)$.