Evaluating the limit $\lim_{n\to\infty}n\cdot \sin(2\pi e(n!))$

Question

As I am beginner I don't have any idea about how to find limits, I want to find the limit as $n\rightarrow \infty$ of $$\lim_{n\to\infty}n\cdot \sin(2\pi e(n!)) $$ please help.

Answer 1

Let us start with the well-known formula $e=\lim_{n\to\infty}e_n$, where

$$e_n=\sum_{k=0}^n\frac{1}{k!}$$

We observe that $n!e_n$ is an integer. So, if we consider $R_n=e-e_n=\sum_{k=n+1}^\infty\frac{1}{k!}$, we get :

$$\sin(2\pi en!)=\sin(2\pi n!(e_n+R_n))=\sin(2\pi n! R_n)$$

Now, it's not difficult to prove that :

$$\forall n\in\mathbb{N},\quad\frac{1}{(n+1)!}\le R_n\le\frac{1}{n n!}$$

So $$2\pi n! R_n\sim \frac{2\pi}{n}$$

And finally, since $\sin(t)\sim t$ as $t\to 0$ :

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

Answer 2

$\displaystyle e n!=A_n+\frac{1}{n+1}+\sum_{k=n+2}^\infty \frac{n!}{k!}=A_n+\frac{1}{n+1}+O(n^{-2})$ where $A_n\in \mathbb{N}$.

So $$\lim_{n\to \infty}n\sin (2\pi e n!)=\lim_{n\to \infty}n\sin \left(\frac{2\pi}{n+1}+O(n^{-2})\right)=2\pi$$