<s>[INST] Calculate $\lim \sin(2\pi n!e)$. [SOLVED] 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 $0$. 
 [/INST] 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 $0$. 
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