What is the limit of $n \sin (2 \pi \cdot e \cdot n!)$ as $n$ goes to infinity? I tried and got this 
$$e=\sum_{k=0}^\infty\frac{1}{k!}=\lim_{n\to\infty}\sum_{k=0}^n\frac{1}{k!}$$
$$n!\sum_{k=0}^n\frac{1}{k!}=\frac{n!}{0!}+\frac{n!}{1!}+\cdots+\frac{n!}{n!}=m$$
where $m$ is an integer.
$$\lim_{n\to\infty}n\sin(2\pi en!)=\lim_{n\to\infty}n\sin\left(2\pi n!\sum_{k=0}^n\frac{1}{k!}\right)=\lim_{n\to\infty}n\sin(2\pi m)=\lim_{n\to\infty}n\cdot0=0$$
Is it correct?
 A: Let's add another solution, which never hurts.
$$\lim_{n\rightarrow \infty}n\cdot \sin\left(2\pi e\cdot n!\right)= \lim_{n\rightarrow \infty}n\cdot \sin \left(2\pi \left(\sum_{k=0}^{\infty}  \frac{1}{k!}\right)n!\right)$$
Let's study now that series.
$$\sum_{k=0}^{\infty}\left(\frac{1}{k!}\right) n!= \left(\sum_{k=0}^n\frac{1}{k!}+\sum_{k=n+1}^{\infty}\frac{1}{k!}\right)n!=A+b_n$$
Where $A\in \mathbb{Z}$ and:
$$b_n=\frac{1}{n+1}+ o \left( \frac{1}{n} \right)$$
From this, we can say that:
$$\lim_{n\rightarrow \infty}n\cdot \sin(2\pi A +2\pi b_n) = \lim_{n\rightarrow \infty}n\cdot \sin(2\pi b_n)=\lim_{n\rightarrow \infty}n\cdot \sin \left(\frac{2\pi}{n+1}+o\left(\frac{1}{n}\right)\right)$$
And, by expanding it using Taylor formulas, we obtain that the limit is equal to:
$$\lim_{n\rightarrow \infty}n\cdot\frac{2\pi}{n+1}+o\left(\frac{1}{n}\right)=2\pi.$$
A: (Added fix recommended by Craig in comments, and complete rewrite for clarity.)
We will use the following: $\lim_{x\rightarrow 0} {\frac{\sin x}{x}}=1$.
Lemma: If $\{x_n\}$ is a sequence (of non-zero values) that converges to $0$, then 
$$\lim_{n\rightarrow\infty}{n \sin{x_n}} = \lim_{n\rightarrow\infty} nx_n$$
Proof: Rewrite $n\sin{x_n} = n x_n \frac{\sin{x_n}}{x_n}$.  The lemma follows since $\sin{x_n}/x_n \rightarrow 1$ by above.
Now, let $[[x]]$ be the fractional part of $x$.  Let $e_n = [[n!e]]$.
Lemma: For $n>1$, $e_n\in (\frac{1}{n+1}, \frac{1}{n-1})$
Proof:
$$n!e = K + \sum_{m=n+1}^\infty \frac{n!}{m!}$$
Where $K$ is an integer.
But for $m>n$, $\frac{n!}{m!} = \frac{1}{(n+1)(n+2)...m} < n^{n-m}$.
So $$\frac{1}{n+1}<\sum_{m=n+1}^\infty \frac{n!}{m!} < \sum_{m=n+1}^\infty n^{n-m} = \sum_{k=1}^\infty n^{-k}$$
But the right hand side is a geometric series whose sum is $\frac{1}{n-1}$.
So $n!e-K\in(\frac{1}{n+1}, \frac{1}{n-1})$, and, since $K$ is an integer, it must be $e_n=n!e-K$.
Theorem: $\lim_{n\rightarrow \infty} n \sin(2\pi n! e) = 2\pi$
Proof: By periodicity of $\sin$, $\sin(2\pi n! e) = \sin(2\pi e_n)$.
Letting $x_n = 2\pi e_n$, we see, from our first lemma:
$$\lim n \sin x_n = \lim n x_n$$
But $nx_n = 2\pi ne_n$, and, since $ne_n\in(\frac{n}{n+1},\frac{n}{n-1})$, we see that $ne_n\rightarrow 1$.  So our limit is $2\pi$.
A: For given $n\geq2$ one has
$$e\cdot n!=n!\sum_{k=0}^\infty{1\over k!}=n!\left(\sum_{k=0}^n{1\over k!}+\sum_{k=n+1}^\infty{1\over k!}\right)=m_n+r_n$$
with $m_n\in{\mathbb Z}$ and
$${1\over n+1}<r_n={1\over n+1}+{1\over (n+1)(n+2)}+\ldots<{1\over n}+{1\over n^2}+\ldots={1\over n-1}\ .$$
Since
$$a_n:=n\>\sin\left(2\pi\cdot e\cdot n!\right)=n\>\sin(2\pi r_n)=n\ \ 2\pi r_n\ {\sin(2\pi r_n)\over 2\pi r_n}$$
and $r_n\to 0$ it follows that
$$\lim_{n\to\infty}a_n=2\pi\lim_{n\to\infty}\bigl(n\> r_n\bigr)=2\pi\ .$$
