# 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!

• Not to be a drag, but unless I'm greatly mistaken, the series certainly does not converge for $z=1$... but I'm sure you meant the rest of $\partial\mathbb{D}$. – The Count May 7 at 3:20
• Are you happy for a solution for the given problem (easy to finish), or now do you want to determine all $z$ for which the series converges (much harder)? – Lord Shark the Unknown May 7 at 4:15
• @TheCount Yes, you are correct. I forget this case since I was inspired by a plot and the representation missed some points and details. Perhaps I must exclude more values of $x.$ – DrinkingDonuts May 7 at 9:52
• Dear @LordSharktheUnknown, my last comment was absolutely wrong. I think my problem is concluded with the answer I posted. – DrinkingDonuts May 8 at 16:41

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.
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 $$\sum_{n=1}^{\infty} a_n\frac{z^n}{1-z^n}.$$