For what numbers does $\lim_{n\to\infty}\sin(2\pi xn!)$ converge

Question

For any real number $x\in\mathbb R$, when does the following limit converge? $$ \lim_{n\to\infty}\sin(2\pi xn!) $$ For $\frac{p}{q}=x\in\mathbb Q$ it converges to $0$ beacuse for any sufficiently large $n:xn!\in\mathbb N$ and then we get $\sin$ of a whole multiply of $2\pi$. (actually you can take $\pi$, not $2\pi$.)
My question is, are there any $x\in\mathbb{R-Q}$ such that the limit converges?
What about $x\in\mathbb C$?

Answer 1

My friend found a particle answer, it converges for $e$. for $x\in[0,1]$: $$ e^x=\sum_{i=0}^n\frac{x^i}{i!}+\frac{C^{n+1}}{(n+1)!},C\in[0,1]\\ e=\sum_{i=0}^n\frac{1}{i!}+\frac{C^{n+1}}{(n+1)!}\\ e-\sum_{i=0}^n\frac{1}{i!}=\frac{C^{n+1}}{(n+1)!}\\ 2\pi n!e-\sum_{i=0}^n\frac{2\pi n!}{i!}=\frac{C^{n+1}}{(n+1)!}\overset{n\to\infty}{\to}0 $$ Now $\frac{2\pi n!}{i!}\in\mathbb{N}$ and $\sin$ uniformly converges so $\forall\epsilon>0\exists\delta>0:|x-y|<\delta\Rightarrow|\sin(x)-\sin(y)<|\epsilon$. The limit above converges so $\forall\delta\exists n:|2\pi n!e-\sum_{i=0}^n\frac{2\pi n!}{i!}|<\delta$ and then $|f(2\pi n!e)-f(\sum_{i=0}^n\frac{2\pi n!}{i!})|<\epsilon$. But $f(\sum_{i=0}^n\frac{2\pi n!}{i!})=0$ so $|f(2\pi n!e)|<\epsilon$

Answer 2

If $x \in \mathbb{R}$ is irrational, we may choose a sequence $(p_n,q_n)$ with $|x-p_n/q_n| \leq 1/(n! q_n)$ and $q_n \leq n!$. Note that $q_n \rightarrow \infty$, since $x$ is irrational.

Write $\beta_n := x - p_n/q_n$. Using the trigonometric addition formula for sine, we get $$\sin(2\pi x n!) = \sin( 2 \pi \beta_n n!)\cos( 2 \pi \tfrac{p_n}{q_n} n!)+\sin(2\pi \tfrac{p_n}{q_n} n!) \cos(2\pi \beta_n n!). $$ Since $ |\beta_n n!| \leq q_n^{-1} \rightarrow 0$ and $p_n n! /q_n \in \mathbb{N}$, we see that $\sin(2\pi xn!) \rightarrow 0$.