If $a\in\mathbb{Q}$, prove that the sequence $\{\sin(n!a\pi)\}_{n=1}^\infty $ has a limit. This exercise is from Methods of Real Analysis by Richard Goldberg.

If $a\in\mathbb{Q}$, prove that the sequence $\{\sin(n!a\pi)\}_{n=1}^\infty$ has a limit.

I think this proof relies on the fact that if $a\in\mathbb{Q}$ then $a=\frac{p}{q}$ for some $p,q\in\mathbb{Z}$, and since $n$ goes to infinity, $q$ is a factor of $n!$. Thus, $n!a\in\mathbb{Z} \ (\forall n\gt q)$, and therefore $\sin(n!a\pi)=0 \ \forall n$ when $n \to\infty$.
Is this idea correct? If it is, how can I write this formally?
 A: You're pretty much done. Here's how to finish up.
Given $a=p/q$ where $q \in \mathbb{N}$, let $N=q$. Then if $n \geq N$, then $n! a \in \mathbb{Z}$ (why?), so if $n \geq N$ then $\sin(n! a \pi)=0$. Hence the limit is zero.
A: The idea is correct. You can write it as you said it, more or less. 
Suppose $a$ is rational. There exists an integer $p$ and a positive integer $q$ such that $a= p/q$. 
Clearly, for all integers $1 \le m \le n$ one has $m \mid n!$. Thus, for all $n \ge q$ we have $q \mid n!$ and thus $a \ n!$ is integral. 
Consequently for all $n \ge q$ we have $\sin ( a n! \pi) = 0$ as $\sin (t\pi)= 0$ for every integer $t$. As the sequence is eventually constant, it is of course convergent. 
A: You basically got it. To make it slightly more rigorous you may want to introduce an $\varepsilon$ and an $N$ such that $$\left|\sin(an!\pi)\right|<\varepsilon$$ for all $n\geq N$ and $\varepsilon>0$. As you say, once $n!$ gets large enough it'll cause $n!a$ to be an integer. We can guarantee this by choosing $N = q$. Then we can say without question that:
Given $\varepsilon>0$, we know $$\left|\sin(an!\pi)\right|=\left|\sin(an!\pi)-0\right|<\varepsilon$$ for all $n\geq q$. By definition, $\sin(n!\pi)$ converges to $0$.
