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!