Series with radius of convergence 1 that diverges on roots of unity, converges elsewhere on the circle. I recall from complex analysis a while ago that the series
$$\sum_{n=1}^\infty \frac{z^{n!}}{n}$$
has radius of convergence 1, diverges on roots of unity, and converges elsewhere on the circle. It's simple enough to see that it has to diverge on roots of unity: consider $z$ such that $z^k = 1$ for some $k \in \mathbb{N}.$ Plug this into the series and you see that past the $k$'th term, $z^{n!} = (z^k)^m = 1$ for some $m \in \mathbb{N}$; hence, the series becomes harmonic and must diverge.
But I don't recall how to show that the series converges for other values on the circle. If this is simple I'd like just a hint, if it is involved I would prefer a sparse outline of the proof. Thank you!
 A: Convention
On the unit circle, we write $z=e^{2\pi i \theta}$ with $\theta\in\mathbb{R}$. Then the roots of unity are exactly when $\theta\in\mathbb{Q}$. Let $\{\alpha\}$ be the fractional part of $\alpha$. Since $z^{n!} = e^{2\pi i \theta n!}$, the distribution of $\{\theta n!\}$  becomes important to consider. 
There is an irrational $\theta$ that the series diverges
If we take $\theta = e$, then the series diverges. This is because $\{en!\}$ converges to $0$. 
The set of all $\theta$ that the series diverges is a $G_{\delta\sigma}$ set
This result is intensively discussed in this post in MO. This is because the set of all $\theta$ that the series converges is a $F_{\sigma\delta}$ set. 
For almost all $\theta$ the series converges
This is due to a result by Lebeque. The function $g(\theta, n) = \theta n!$ satisfies the hypotheses of Theorem 1 in the paper. Therefore, for every $\epsilon>0$, the series
$$
\sum_{n=1}^{\infty} \frac{ e^{2\pi i \theta n!} } {n^{1/2+ \epsilon}}
$$
converges for almost all $\theta$. This shows that the series with $\epsilon=\frac12$ converges for almost all $\theta$. Hence the set of all $\theta$ that the series diverges is a $G_{\delta\sigma}$ set of Lebesgue measure zero which contains all rational numbers and $e$, and it does not contain $e/2$ by @user254665's answer. 
A: A specific example.
For $n\geq 1$, the integer part of $n!e$ is odd when $n$ is even, and even when $n$ is odd. 
So for $z=\exp (e\pi i )$ we have  $z^{n!}/n=(-1)^{n+1}/n +O(1/n^2)$ as $n\to \infty$, and the series converges.
