Dose the series diverge on the boundary when $\theta $ is irrational?

Question

series: $$\sum_{n=1}^{\infty}\left(\ln n\right)^2z^n$$ where $z\in \Bbb{C}$ , by Hadamard's formula, the radius of convergence is $1$, and I try to discover the status of convergence when $|z|=1$.
writer $z=e^{i\theta}$,the series became: $$\sum_{n=1}^{\infty}\left(\ln n\right)^2e^{in\theta}$$ where $\theta \in [0,2\pi)$,when $\theta \in \Bbb{Q}$, $\left(\ln n\right)^2e^{in\theta} \not\rightarrow 0$,so it is divergent. but what about $\theta \in [0,2\pi)\backslash\Bbb{Q}$?

thanks very much

Answer 1

The sequence of general term $(\ln n)^2\mathrm e^{\mathrm in\theta}$ does not converge to zero for every real number $\theta$, whether rational or irrational.

Hence the series $\sum\limits_{n\geqslant1}(\ln n)^2z^n$ diverges for every $z$ such that $|z|=1$.