- $0<x<1$
- $u_0 \in \mathbb{N}$ and $u_0 \geq 2$
- $u_{n+1}= u_n ^{ u_n}$
- $\alpha= \sum_{n=0}^{+ \infty} \dfrac{1}{u_n}$
- We want to prove that the series $\sum \dfrac{x^n}{ \sin ( \pi \alpha n) }$ diverges.
My attempt :
The sequence $(u_n)_n$ has been studied here
$
\begin{align*}
\forall j, u_{j+1}&=u_j^{u_j} \in \mathbb{N} \\
\forall j, \forall k<j, u_k &| u_j \\
\forall k, k <n, \dfrac{u_n}{u_k} &\in \mathbb{N} \\
\forall n, u_n \sum_{k=0}^{n} \dfrac{1}{u_k} &\in \mathbb{N}
\end{align*}
$
$
\begin{align*}
\sin( u_n \pi \alpha) &= \sin( u_n \pi \sum_{k=0}^{n} \dfrac{1}{u_n} + u_n \pi \sum_{k=n+1}^{\infty} \dfrac{1}{u_n} ) \\
\sin( u_n \pi \alpha)&= \sin( u_n \pi \sum_{k=n+1}^{\infty} \dfrac{1}{u_n} ) \\
|\sin( u_n \pi \alpha)| & \leq | u_n \pi \sum_{k=n+1}^{\infty} \dfrac{1}{u_n} | \\
|\sin( u_n \pi \alpha)| & \leq u_n \pi \sum_{k=n+1}^{\infty} \dfrac{1}{u_n} \\
& \leq\dfrac{ \pi C }{ u_n^{u_n -1} } \\
\end{align*}
$
$C$ is an intercept, that does not depend on $n$.
$$\dfrac{x^{u_n} }{ |\sin( u_n \pi \alpha)| } \geq \dfrac{1}{ \pi C } x^{u_n} u_n^{u_n -1} (\star) $$
We expect to get the divergence from this last inequality called ($\star$). $u_n \to \infty$
We can prove that $\alpha \notin \mathbb{Q}$
Assume that $\alpha= \dfrac{p}{q}$ where $p,q \in \mathbb{Q}$
then
$
\begin{align*}
u_n \sum_{k=0}^{n} \dfrac{1}{u_k} &\in \mathbb{N} \\
\alpha=\sum_{k=0}^{\infty} \dfrac{1}{u_k} &\in \mathbb{N}\\
q u_n \sum_{k=n1}^{\infty} \dfrac{1}{u_k} &\in \mathbb{N} \\
q u_n \sum_{k=n1}^{\infty} \dfrac{1}{u_k} &\to 0
\end{align*}
$
It is a contradiction so $\alpha \not \in \mathbb{Q}$
We get the divergence from $\star$ because $u_n > n+1$
Best Answer
Provided that $\alpha$ is irrational, then the sequence given by $z_n=e^{i\pi\alpha n}$ is dense in $\mathbb{S}^1$ (irrational rotations are transitive). So, there is a subsequence $n'$ along which $z_n\rightarrow e^{ -i\pi\alpha}=(\cos\pi\alpha,-\sin\pi\alpha)$. From $$ \frac{|\sin(\pi\alpha n)|}{|\sin(\pi\alpha (n+1))|}=\frac{|\sin\pi\alpha n|}{|\sin(\pi\alpha n)\cos(\pi \alpha)+\sin(\pi\alpha)\cos(\pi\alpha n)|}=\frac{1}{|\cos(\pi\alpha)+\sin(\pi\alpha)\cot(\pi\alpha n)\big|} $$ we have that along the subsequence $n'$, $\frac{|\sin\pi\alpha n|}{|\sin(\pi\alpha(n+1))|}\rightarrow\infty$. Thus, $$ \infty =\limsup_n\frac{|\sin(\pi\alpha n)|}{|\sin(\pi\alpha (n+1))|}\leq \limsup_n \sqrt[n]{\frac{1}{|\sin(\pi\alpha n)|}}$$ This means that the radius of convergence of the power series $\sum_{n\geq1}\frac{z^n}{\sin\pi\alpha n}$ is $0$.