Does this limit exist finitely?

Question

Find the value of the following (if it exists):

$$\lim_{n\to\infty}n\sin(2\pi en!)$$

Does it exist? I think that it doesn't exist but I can't prove it. Please help me.

Answer 1

Here is a complete proof:

For given $n\geq2$ one has $$e\cdot n!=n!\sum_{t=0}^\infty{1\over t!}=n!\left(\sum_{t=0}^n{1\over t!}+\sum_{t=n+1}^\infty{1\over t!}\right)=m_n+r_n$$ 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 $$\lim_{n\to\infty}a_n=2\pi\lim_{n\to\infty}\bigl(n\> r_n\bigr)=2\pi\ .$$

Answer 2

Hint: If we truncate the series for $e$ just after the term $\frac{1}{n!}$, the truncation error is $$\frac{1}{(n+1)!}+\frac{1}{(n+2)!}+\frac{1}{(n+3)!}+\cdots.$$ This is less than $$\frac{1}{(n+1)!}\left(1+\frac{1}{n+2}+\frac{1}{(n+2)^2}+\cdots\right).$$ Thus for some integer $Q_n$, we have $$Q_n+\frac{n!}{(n+1)!}\lt en!\lt Q_n+\frac{n!}{(n+1)!} \frac{1}{1-\frac{1}{n+2}}.$$