Trying to prove $e$'s irrationality Knowing that $\lim\limits_{x\to\ 0}\ $$\frac{\sin(x)}{x}$$= 1$ , $\frac{1}{n+1}<n!r_n<\frac{1}{n}$, where $r_n=e- \sum _{ k=0 }^{ \ n}{ \frac { 1 }{k!}} $
By studying $\lim\limits_{n\to\infty}\ n\sin(2πn!r_n)$ I have to show that :
$$\lim\limits_{n\to\infty}\ n\sin(2πn!e)= 2π,$$ 
and then prove that $e$ is irrational ?

Let $I_n=\sum\limits_{k=0}^n\dfrac{n!}{k!}\in \Bbb Z$, then
$$n\sin(2\pi n! e)=n\sin(2\pi I_n+2\pi n!r_n)=n\sin(2\pi n!r_n).$$
Note
$$\frac 1{n+1}\le n!r_n=\frac 1{n+1}+\frac 1{(n+1)(n+2)}+\cdots\le\sum_{k=1}^\infty\frac1{(n+1)^k}=\frac 1n.$$
By sandwich theorem, $n!r_n\to 0$ and $n\cdot n!r_n\to 1$ as $n\to\infty$. Using $\sin\theta\sim\theta$ as $\theta\to 0$, we get
$$\lim_{n\to\infty}n\sin(2\pi n! e)=\lim_{n\to\infty}n\sin(2\pi n! r_n)=2\pi.$$

To prove $e$ irrational, assume $e=p/q$, then when $n>q$, $n!e\in\Bbb Z$ implies $\sin(2\pi n! e)\equiv 0$, and 
$$\lim_{n\to\infty}n\sin(2\pi n! e)=\lim_{n\to\infty}n\cdot 0=0\ne 2\pi,$$
a contradiction.
