I tried to solve this exercise (Remmert Theory of Complex Functions, p. 107, exercise 1 ), but I didn't get very far:
Proposition: $\sum \frac{z^{2n}}{1-z^{n}}$ is normally convergent in $\mathbb{E}$
What does $\mathbb{E}=\{z\in \mathbb{C} | |z|<1 \}$ stand for? For normal convergence, it suffices if one finds a majorant series whose absolute value is less than infinity?
so (most likely it is $|z|<1$ for all $z\in \mathbb{C})$ : $\left|\sum \frac{z^{2n}}{1-z^{n}} \right| \le \sum \left| \frac{r^{2n}}{1-r^{n}}\right| \le \sum |r^{n}| < \infty$
Does anybody see if this is right?