I need to calculate $\sum_{n=0}^\infty (r e^{2 \pi i \alpha})^{n!}$ for $\alpha \in \mathbb Q$ and $r \in \mathbb R$.
My Attempt:
$\sum_{n=0}^\infty (r e^{2 \pi i \alpha})^{n!}=\sum_{n=0}^\infty r e^{2 \pi i \alpha n!}=\sum_{n=0}^\infty r (\cos (2 \pi \alpha n!) + i \sin (2 \pi \alpha n!)),$ and I'm stuck here, I think I should use a trigonometric identity to deal with the $\cos$ and $\sin$ of $2 \pi \alpha n!$, but I don't know which identity I could use.
If $\alpha=p/q$ and $n\ge q$ then $\alpha n!$ is an integer and so $$(re^{2\pi i\alpha})^{n!}=r^{n!}\ .$$ So the tail of the sum is $$r^{q!}+r^{(q+1)!}+r^{(q+2)!}+\cdots\ ;$$ this diverges if $|r|\ge1$. It converges if $|r|<1$, but I don't know of any simple expression for the sum.