Convergence of $\sum_{n=1}^\infty \frac{z^n}{2^n(1-z^n)}$ I'm solving the following problem:

Find the maximal open set, $\Omega,$ where the following series converges:
$$\sum_{n=1}^\infty \dfrac{z^n}{2^n(1-z^n)}.$$
Extra: Prove that the series define a holomorphic function in $\Omega.$

I conclude that a possible election is $\Omega = \mathbb{C} \setminus \partial \mathbb{D}$ doing the following:

*

*Let $z \in \bar{B}(0,r)$ with $r < 1.$ We know that $|z|\leq r.$ Furthermore we have that

\begin{align*}
|f_n(z)|=\bigg| \frac{z^n}{2^n(1-z^n)} \bigg| &= \frac{|z^n|}{2^n|1-z^n|} \leq \frac{|z^n|}{2^n|1-|z|^n|}\\ &= \frac{|z|^n}{2^n(1-|z|^n)} \leq \frac{r^n}{2^n(1-r^n)} \leq \frac{1}{2^n(1-r^n)} = a_n.
\end{align*}
The series $\sum_{n=1}^\infty a_n < +\infty$ since $$\lim_{n \rightarrow \infty} 
 \frac{a_n}{1/2^n} = \lim_{n \rightarrow \infty} \frac{1}{1-r^n} = 1.$$
We conclude that the original series is absolutely convergent.

*

*Let $z \in \mathbb{C} \setminus {B}(0,r)$ with $r > 1.$ Hence $r \leq |z|$ and we have that

\begin{align*}
|f_n(z)|=\bigg| \frac{z^n}{2^n(1-z^n)} \bigg| &= \frac{|z^n|}{2^n|1-z^n|} = \frac{1}{2^n|1/|z^n|-z^n/|z^n||}\\ &\leq \frac{1}{2^n|1-1/|z^n||} = \frac{1}{2^n(1-1/|z^n|)} \leq \frac{1}{2^n(1-1/r^n)} = b_n.
\end{align*}
As above, comparing with $\sum_{n = 1}^{\infty} 1/2^n$ we conclude the absolute convergence of the original series.
The Weierstrass M-criterion gives us the uniform convergence of the series and we conclude that the series define a holomorphic function in $\Omega = \mathbb{C} \setminus \partial\mathbb{D}.$
I can't decide if the series converges for some $z \in \partial \mathbb{D}$ and I hope someone could help me.
Thanks everyone!
 A: The commentary of Lord Shark the Unknown was adecuated since I think that the maximal open set I'm looking for is $\mathbb{C} \setminus \partial \mathbb{D}.$ And the commentary of The Count gives the last piece of the problem.
I will prove it as follows:
Let $\Omega$ be the maximal open set. Clearly, for every root of unit, the series is not defined. Suppose that exists some point $p$ of $\partial \mathbb{D}$ such that the series is convergent at $p$. Suppose that $p \in \Omega.$ Since $\Omega$ is open, there exists $r>0$ such that $B(p,r) \subset \Omega.$ Hence, there is a neighbourhood of $p$ where the series is convergent at every point of it. But $p$ is in $\partial \mathbb{D}$ and the set of roots of unity is dense in $\partial \mathbb{D}$ hence there are roots of unity in $B(p,r)$ where the series converges, which is absurd since the series is not defined there.

Note that I'm not proving the divergence of the series on $\partial \mathbb{D},$ I'm proving that the points of convergence of the unit circle aren't in the maximal open set where the series converges. Furthermore, this fact is true for all Lambert series $\sum_{n=1}^{\infty} a_n\frac{z^n}{1-z^n}.$

