Show uniform convergence and pointwise convergence for $\sum_{n=1}^ \infty \frac{z^ {n-1}}{(1-z^n)(1-z^ {n+1})}$

Question

Consider the series: $$\sum_{n=1}^ \infty \frac{z^ {n-1}}{(1-z^n)(1-z^ {n+1})}$$ show this converges to:

(a) $\frac{1}{(1-z)^2}$ for $|z|<1$

(b) $\frac{1}{z(1-z)^2}$ for $|z|>1$

Finally, show that this convergence is uniform for $|z| \leq c <1$ in (a) and for $|z| \geq c >1$ in (b).

This is my first course in complex analysis and I'm struggling a bit with seeing why this is true. Question (a) suggests recognising the product of two geometric series $$\frac{1}{1-z} \cdot \frac{1}{1-z}= \sum z^ k \cdot \sum z^ k $$ But I do not see how to rewrite the expression. I have no idea how one would derive the expression in question $(b)$ any hints on that would be greatly appreciated. Once I know how to derive these expressions I will maybe also know a way to apply the Weierstrass M-test for uniform convergence. Since the uniform limit must equal the pointwise limit all we need to show that the series converges uniformly on some domain of definition. Can someone provide some guidance on how to tackle this question, drop some small hints so I can maybe proceed?

Answer 1

Use partial fraction decomposition and exploit a telescoping sum to obtain the pointwise convergence results in (a) and (b). For example, we have

$$\frac{z^{n-1}}{(1-z^n)(1-z^{n+1})} = \frac{z^{n-1}}{z^n - z^{n+1}}\left(\frac{1}{1-z^n} - \frac{1}{1-z^{n+1}} \right) \\ = \frac{1}{z(1-z)}\left(\frac{1}{1-z^n} - \frac{1}{1-z^{n+1}} \right)$$

I'll leave the remainder of the details to you. The technique is illustrated in more detail below in the argument for uniform convergence.

Regarding uniform convergence, note that

$$\begin{align}\left|\frac{z^{n-1}}{(1-z^n)(1-z^{n+1})} \right| = \frac{|z|^{n-1}}{|1-z^n||1-z^{n+1}|} \leqslant \frac{|z|^{n-1}}{(1-|z|^n)(1-|z|^{n+1})} \end{align}$$

where the last inequality follows from application of the the reverse triangle inequality, $|1 - z^m| \geqslant 1 - |z|^m$, to the expressions in the denominator.

If $|z| \leqslant c < 1$, then we have

$$\tag{*}\left|\frac{z^{n-1}}{(1-z^n)(1-z^{n+1})} \right| \leqslant \frac{c^{n-1}}{(1- c^n)(1- c^{n+1})} = \frac{c^{n-1}}{c^n -c^{n+1}} \left(\frac{1}{1-c^n}-\frac{1}{1 - c^{n+1}}\right) \\ = \frac{1}{c(1-c)} \left(\frac{1}{1-c^n}-\frac{1}{1 - c^{n+1}}\right)$$

The series with terms appearing on the RHS of (*) is telescoping and convergent

$$\sum_{n=1}^\infty \left(\frac{1}{1-c^n}-\frac{1}{1 - c^{n+1}}\right)= \lim_{m \to \infty}\sum_{n=1}^m \left(\frac{1}{1-c^n}-\frac{1}{1 - c^{n+1}}\right) = \frac{1}{1-c} - \lim_{m\to \infty}\frac{1}{1 - c^{m+1}} \\ = \frac{1}{1-c}-1$$

Therefore, we have uniform convergence for $|z| \leqslant c < 1$, by the Weierstrass test.

For the case where $|z| \geqslant c > 1$, rearrange as

$$\left|\frac{z^{n-1}}{(1-z^n)(1-z^{n+1})} \right| = \frac{|z|^{-(n+2)}}{|1 - z ^{-n}||1- z^{-(n+1)}|} \leqslant \frac{|z|^{-(n+2)}}{(1 - |z |^{-n})(1- |z|^{-(n+1)})}, $$

and proceed in a similar way as before to apply the Weierstrass test.