# Behavior of $\sum_{n=1}^\infty \frac{1}{n} z^{n!}$ on the unit circle [duplicate]

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?

## marked as duplicate by kingW3, Paul Frost, Lord Shark the Unknown, Arnaud D., Namaste calculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Nov 16 '18 at 20:16

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 $$0$$.