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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!

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  • $\begingroup$ 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 $\partial\mathbb{D}$. $\endgroup$ – The Count May 7 at 3:20
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    $\begingroup$ 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)? $\endgroup$ – Lord Shark the Unknown May 7 at 4:15
  • $\begingroup$ @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 $x.$ $\endgroup$ – DrinkingDonuts May 7 at 9:52
  • $\begingroup$ Dear @LordSharktheUnknown, my last comment was absolutely wrong. I think my problem is concluded with the answer I posted. $\endgroup$ – DrinkingDonuts May 8 at 16:41
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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 $\sum_{n=1}^{\infty} a_n\frac{z^n}{1-z^n}.$

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