I want to find the largest domain where $f(z)=\sum_{n=1}^\infty \frac{nz^n}{1-z^n}$ is holomorphic. I know $f(z)$ is holomorphic in the unitary disk, $\mathbb{D}$, and I believe that it is the largest domain.
I am trying to prove it by contradiction. Suppose $f$ is holomorphic in $U$ open, then $\mathbb{D}\subset U$. Take $z\in U\backslash \mathbb{D}$, then $|z|\geq 1$. So I am trying to prove that the series diverges in that case, but I have been able only to prove that when $|z|>1$ then the series doesn't converge absolutely.
So my questions is:
How can I prove it doesn't converge when $|z|\geq 1$?
Best Answer
Let |z|=1. Thus $z=e^{i\theta}$. The set of all such $z$ (for $0 \leq \theta < 2\pi$) forms a multiplicative group, using the multiplication of $\mathbb{C}.$
If $z$ has finite order in the group, then infinitely often the general term $f_n(z)$ is undefined. So for such $z$ there is no convergence.
Otherwise $z$ has infinite order in the circle group, so the general term $f_n(z)$ is always defined. We have: $$|f_n(z)|\,=\,\frac{n\,|z|^n}{|1-z^n|}\,\geq\,\frac{n\,|z|^n}{1+|z|^n}\,=\,\frac{n}{1+1},$$
so there is no convergence of the series for any $z$ on the unit circumference. If $|z|>1$, we estimate:
$$|f_n(z)|\,=\,\frac{n\,|z|^n}{|1-z^n|}\,\geq\,\frac{n\,|z|^n}{1+|z|^n}\,>\,\frac{n|z|^n}{2|z|^n}\,=\,\frac{n}{2}.$$