Calculate $\lim \sin(2\pi n!e)$. [SOLVED]

Question

I need to calculate $\lim \sin(2\pi n!e)$.

I put it into Wolfram and saw that it is likely to converge to 0.

Of course this would mean that the fractional part of $n!e$ should be always very close to either 0 or 0.5.

It means that the fluctuation of the fractional part is gonna decrease. But how do I show that it will stop around 0 or 0.5?

Cheers.

EDIT: Solved.

$n!e = k + \frac{n!}{(n+1)!}\left( \frac{1}{1} + \frac{1}{n+2} + \frac{1}{(n+2)(n+3)} + \ldots \right) < k + \frac{n!}{(n+1)!}\left ( 1 + \frac{1}{n+1} + \frac{1}{(n+1)^2} + \frac{1}{(n+1)^3} + \ldots \right) = k + \frac{1}{n} $

where $k \in Z$

Therefore, the fractional part of $n!e$ approaches $0$.

Answer 1

Hint: $n!\; e = (integer) + \sum_{k=n+1}^\infty n!/k!$

Answer 2

SOLUTION:

$n!e = k + \frac{n!}{(n+1)!}\left( \frac{1}{1} + \frac{1}{n+2} + \frac{1}{(n+2)(n+3)} + \ldots \right) < k + \frac{n!}{(n+1)!}\left ( 1 + \frac{1}{n+1} + \frac{1}{(n+1)^2} + \frac{1}{(n+1)^3} + \ldots \right) = k + \frac{1}{n} $

where $k \in Z$

Therefore, the fractional part of $n!e$ approaches $0$.