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Prove that the series $$f(z)=\sum_{n>1}\frac{z^n}{1-z^n}$$ converges in the unit disc $D=\{z:|z|<1\}$ and defines there a holomorphic function.

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Do you need any help to translate? Puedo ayudar. Podés escribir todo en español y lo traducimos después. – Pedro Tamaroff Jan 9 '13 at 23:03
Pues si, es fácil traducir pero los términos en matemáticas... – Rafael Jiménez Guerra Jan 9 '13 at 23:09
El ejercicio dice: Prueba que esa seri converge a D(0,1) y que su suma es una función holomorfa. – Rafael Jiménez Guerra Jan 9 '13 at 23:10
Que denota $D(0,1)$? – Pedro Tamaroff Jan 9 '13 at 23:11
Es el disco de centro 0 y radio 1, pensaba que esa nomenclatura era internacional... – Rafael Jiménez Guerra Jan 9 '13 at 23:12
up vote 1 down vote accepted

Fix a disc $D_r=\{|z|\leq r\}$ for $r\in(0,1)$. Then, if $z\in D_r$, $$|z|^n\leq r^n$$ and $$1-|z|^n\geq 1-r^n\geq 1-1/2=1/2\;.$$ if $n$ is big enough, depending on $r$, let us say, for $n>N(r)$.Therefore $$\sum \frac{|z|^n}{1-|z|^n}\leq\frac{N(r)}{1-r}+2\sum r^n=\frac{2+N(r)}{1-r}$$ that is, the series converges absolutely, hence uniformly, on $D_r$, for $r\in (1/2, 1)$. Every compact $K\Subset D(0,1)$ is contained in some $D_r$ with $r\in (1/2,1)$, so we have that the series converges uniformly on every compact set of $D(0,1)$. Then the limit is a holomorphic function on $D(0,1)$.

EDIT: As noted in the comments, $1-|z|^n\geq 1-|z|$, therefore $$\sum\frac{|z|^n}{1-|z|^n}\leq\sum\frac{r^n}{1-r}=\frac{1}{(1-r)^2}\;.$$

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Hm yes, sorry, I was too fast in writing.. – wisefool Jan 10 '13 at 0:40

The brute force method works: assume $|z|\lt1$, then $\sum\limits_n\frac{z^n}{1-z^n}=\sum\limits_nz^n\sum\limits_kz^{kn}=\sum\limits_n\sigma_1(n)z^n$, where $\sigma_1(n)$ is the number of divisors of $n$, converges absolutely since $\sigma_1(n)\leqslant n$ and $\sum\limits_nn|z|^n$ converges.

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Hint: the series converges uniformly on compact sets. You can use a ratio test.

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thank you but i don't know how... – Rafael Jiménez Guerra Jan 9 '13 at 23:29

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