I'm trying to prove that $f(z)=\displaystyle \sum_{n=1}^{\infty}{\displaystyle \frac{nz^n}{1-z^n}}$ is an holomorphic function over the open disc $D_1(0)$. I know that $f_n(z)=\displaystyle \sum_{i=1}^{n}{\displaystyle \frac{iz^i}{1-z^i}}$ is an holomorphic funtion in the disc $D_1(0)$, for all $n \in \mathbb{N}$, so if $f_n$ converges uniformly to $f$ then $f$ is an holomorphic funtion; I did the following trying to prove the uniform continuity, let $ \varepsilon >0$:
$|f_n(z)-f(z)| = \displaystyle \left|\displaystyle \sum_{i=1}^{n}{\displaystyle \frac{iz^i}{1-z^i}}-\sum_{i=1}^{\infty}{\displaystyle \frac{nz^n}{1-z^n}} \right| = \left| \sum_{i=n+1}^{\infty}{\displaystyle \frac{iz^i}{1-z^i}}\right| \leq\sum_{i=n+1}^{\infty}{\displaystyle \frac{|iz^i|}{|1-z^i|}} \leq \sum_{i=n+1}^{\infty}{\displaystyle \frac{i}{|1-|z|^i|}}$
Because $|z|<1$. but I don't know how to go from here, any hint?
First, a notational gripe: It's probably best to not use $i$ as an indexing variable when you're doing complex analysis. The work you've shown was really confusing for a moment.
As for the question itself, you're trying to prove something too strong. In complex analysis we almost always work with "uniform convergence on compact sets". This is weaker than full uniform convergence, but is still enough to conclude that a limit of analytic functions is analytic.
With that in mind, what do we want to show?
Why does it matter that we're restricting attention to compact subsets of the disk? Because every such $K$ will be bounded away from $1$. That is, every such $K$ will satisfy $\sup_{z \in K} |z| < 1 - \delta$ for some positive $\delta$.
But now we know how to finish off your work!
$$ \sum_{k = n}^\infty \left | \frac{k z^k}{1 - z^k} \right | \leq \sum_{k=n}^\infty \frac{ |k z^k| }{\delta} = \frac{1}{\delta} \sum_{k=n}^\infty k |z|^k $$
But now bounding this tail is a matter of real analysis. Do you see how to finish from here?
I hope this helps ^_^