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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?

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marked as duplicate by kingW3, Paul Frost, Lord Shark the Unknown, Arnaud D., Namaste calculus Nov 16 '18 at 20:16

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  • $\begingroup$ I have no idea. It's a cute problem, though. $\endgroup$ – davidlowryduda Nov 16 '18 at 12:16
  • $\begingroup$ For the purposes of googling, such a series is said to lacunary. In that context one usually talks about gap theorems of various strengths (eg the Fabry gap theorem) and one of those may be illuminating. $\endgroup$ – Semiclassical Nov 16 '18 at 14:59
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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 $0$.

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