Is this a convergent sequence : $ n \sin (a e n!)$ with $a$ is a real number?

Question

I would be surprised if the following was true: $\lim_n \sin (a e n!)$ exists where $a$ is a real number. Really, Wolfram Alpha assumed that it is divergent for some values of the real number $a$, but in my opinion, when we use Stirling's formula and substitute $n !$ by $\sqrt{2\pi n}\frac{n^n}{e^n }$, we probably get something convergent in the $\sin$ function using the continuity of the $\sin$ function. Therefore my question: Is this a convergent sequence: $n \sin (a e n!)$ where $a$ is a real number?

Answer 1

Stirling's formula has little to do with the convergence (or lack of) of such sequences, the relevant fact is that $n!e$ is always pretty close to an integer since $$ n!e = \underbrace{\sum_{k=0}^{n}\frac{n!}{k!}}_{\in\mathbb{Z}}+\underbrace{\sum_{k>n}\frac{n!}{k!}}_{\approx\frac{1}{n+1}} $$ hence $\sin(\pi n! e)\to 0$, for instance, and $n\sin(\pi n! e)$ has a subsequence converging to $\pi$ and a subsequence convergent to $-\pi$.