Prove that $\sum z^n/(1-z^n)$ is holomorphic on the unit disc. 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.
 A: 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.
A: 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
$$\sum\frac{|z|^n}{1-|z|^n}\leq\sum\frac{r^n}{1-r}=\frac{1}{(1-r)^2}\;.$$
A: Hint: the series converges uniformly on compact sets.  You can use a ratio test.
