Let $f(z)=\sum\limits_{k=1}^\infty\frac{z^k}{1-z^k}$. I want to show that this series represents a holomorphic function in the unit disk. I'm, however, quite confused. For example, is $f(z)$ even a power series? It doesn't look as such. Here's what I have so far come up with.
Proof:
$$ \sum\limits_{k=1}^\infty\frac{z^k}{1-z^k}=-\sum\limits_{k=1}^\infty\frac{1}{1-z^{-k}}=-\sum\limits_{k=-\infty}^{-1}\sum\limits_{n=0}^\infty z^{kn}$$
[If the above expression is correct then we have a Laurent series of a power series].
Now, $|c_n|=1$ for all $n$. By Parseval's formula,
$$2\pi\sum\limits_{k=-\infty}^{-1}\sum\limits_{n=0}^\infty \rho^{2kn}=\sum\limits_{k=-\infty}^{-1} \int\limits_0^{2\pi}\left|g(\rho^ke^{it})\right|^2dt=^? \int\limits_0^{2\pi} \sum\limits_{k=1}^{\infty} \left|g(\rho^{-k}e^{it})\right|^2dt$$
where $0\le\rho\le 1$, and $g$ is some function (holomorphic) on the unit disk. So we know that this integral (above in the middle) exists.
Now, I believe, what remains to be proved is that the infinite series of this integral also exists. Do you think this approach is OK, or did I make some mistakes in it? How can we prove that the series $\sum\limits_{k=-\infty}^{-1}\sum\limits_{n=0}^\infty \rho^{2kn}$ converges? And then, since it converges, does this imply that $G$ is holomorphic?
Thanks a lot.
Best Answer
Suppose $0<r<1.$ Then $r^k \to 0.$ Thus $r^k\le 1/2$ for large $k.$ For these $k$ and $|z|\le r$ we have
$$\left| \frac{z^k}{1-z^k}\right | \le \frac{r^k}{1-r^k} \le 2r^k.$$
Since $\sum 2r^k<\infty,$ Weierstrass M implies our series converges uniformly on $\{|z|\le r\}.$ This is true for every $r\in (0,1),$ hence the series converges uniformly on compact subsets of the open unit disc. This proves the series converges to a homolorphic function in the open unit disc.