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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
@DavideGiraudo Apparently the definition Remmert uses is different from what we were thinking. – Pedro Tamaroff Sep 26 '14 at 1:52
up vote 3 down vote accepted

The definition of normal convergence according to Remmert is the following:

A series of functions $\sum f_n$, $f_n:X\to\Bbb C$, is normally convergent in $X$ if for each $x\in X$ there is a nbhd $U$ of $x$ such that $\sum |f_n|_U<+\infty$ where $|f_n|_U=\sup_U |f_n|$

Moreover, $B(0,1)$ is locally compact, so it suffices you show that for each ball $B(0,r)$ with $0<r<1$, we have $$\sum |f_n|_{B(0,r)}<+\infty$$

for in a locally compact space $X$, normal convergence is equivalent to $\sum |f_n|_K<+\infty$ for each $K$ compact in $X$.

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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
Thank you, Peter. – VVV Feb 13 '12 at 10:20

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