# $\sum \frac{z^{2n}}{1-z^{n}}$ 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?

-
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
Wouldn't writing it as $$\sum \frac{1}{z^{-n}-z^n}$$ help? @DavideGiraudo The exponent in the numerator is $2n$. –  Pedro Tamaroff Feb 12 '12 at 22:14
add comment

## 1 Answer

I'm not really sure how far this will be correct, but

\eqalign{ & \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{z^{2n + 2}}}}{{1 - {z^{n + 1}}}}\frac{{1 - {z^n}}}{{{z^{2n}}}}} \right| = \cr & \mathop {\lim }\limits_{n \to \infty } \left| {{z^2}\frac{{1 - {z^n}}}{{1 - {z^{n + 1}}}}} \right| = \cr}

Now I use

$$\frac{{{a_n}}}{{{b_n}}} \sim \frac{{{a_{n + 1}} - {a_n}}}{{{b_{n + 1}} - {b_n}}}$$

So $$\mathop {\lim }\limits_{n \to \infty } \left| {{z^2}\frac{{1 - {z^n}}}{{1 - {z^{n + 1}}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {{z^2}\frac{{{z^n} - {z^{n + 1}}}}{{{z^{n + 1}} - {z^{n + 2}}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {z\frac{{{z^{n + 1}} - {z^{n + 2}}}}{{{z^{n + 1}} - {z^{n + 2}}}}} \right| = \left| z \right|$$

Thus we need $$\left| z \right| < 1$$ to ensure convergence (D'Alambert).

-
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
@VVV You're welcome. –  Pedro Tamaroff Feb 13 '12 at 11:58
add comment