# $\sum \frac{z^{2n}}{1-z^{n}}$ normally convergent in $\mathbb{E}$

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?

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I don't know it depend what $\mathcal E$ is? Isn't it written earlier in the book? –  Davide Giraudo Feb 12 '12 at 21:58
Excuse this attitude if it pleases you, it is same as: $E=\{z\in \mathbb{C} | |z|<1 \}$ –  VVV Feb 12 '12 at 22:05
$E$ is in fact the open unit disc. But I'm not sure the convergence is normal on this disk because for a fixed $n$ the supremum $\sup_{x\in E}\left|\frac{z^n}{1-z^n}\right|$ is infinite. (but the convergence is normal on any disc of the form $\{|z|<r\}$ where $0<r<1$. –  Davide Giraudo Feb 12 '12 at 22:09

$E$ is in fact the open unit disc. But I'm not sure the convergence is normal on this disk because for a fixed$n$ the supremum $\sup\limits_{|z|<1}\left|\dfrac{z^{2n}}{1-z^n}\right|$ is infinite. (but the convergence is normal on any disc of the form $\{|z|<r\}$ where $0<r<1$)
For a proof of his claim that the supremum is infinite, note that over the line $(x,0),0<x<1$ we have $$\frac{x^n}{1-x}$$
Taking $x=1- \varepsilon$ gives $\varepsilon^{-1}(1-\varepsilon)^n$. As we approach $\varepsilon =0$, this blows up.
In the german edition there is a discussion in this chapter (before the exercise) which shows that if $\sum \vert f_\nu \vert_{B_r(c)} < \infty$ for all $0<r<s$, then $\sum f_\nu$ is normal convergent on the open Disc $B_s(c)$. So you are done. –  Blah Feb 12 '12 at 22:52