TITLE: Behavior of $\sum_{n=1}^\infty \frac{1}{n} z^{n!}$ on the unit circle
QUESTION [5 upvotes]: I'm trying to understand the behavior of $\sum_{n=1}^\infty \frac{1}{n} z^{n!}$ on the unit circle.
Since for each $m$th root of unity $\zeta_m$
$$\sum_{n=1}^\infty \frac{1}{n} \zeta_m^{n!} = C + \sum_{n=m}^\infty \frac{1}{n} = \infty$$
holds for some $C \in \mathbb{C}$, the series diverges for all $e^{\varphi \pi i}$ with $\varphi \in \mathbb{Q}$.
But what happens for $\varphi \in \mathbb{R} \setminus \mathbb{Q}$? 
Does the series diverge everywhere, or are there points where it is convergent?

REPLY [5 votes]: The sum converges for $z = e^{2\pi i\varphi}$ for $\varphi = \frac{1}{2}\frac{1}{1!}+\frac{1}{2}\frac{1}{3!}+\frac{1}{2}\frac{1}{5!}+\frac{1}{2}\frac{1}{7!}+...$. The reason is that for $n$ odd, $n!\varphi \pmod{1}$ is basically $\frac{1}{2}$, while for $n$ even, $n!\varphi \pmod{1}$ is basically $0$.