# Evaluate $\lim_{n\rightarrow\infty}\sin(2\pi\cdot n!\cdot e)$

Evaluate $$\lim_{n\rightarrow\infty}\sin(2\pi\cdot n!\cdot e)$$

I can't think of a theorem to solve this. I think first we have to convert it into the form of standard limit. Also by just putting $$\infty$$ into the limit we get $$\sin(\infty)$$ which I think is indeterminate. Any help is greatly appreciated.

• "convert it into the form of standard limit " What do you mean by that? Commented Sep 10, 2022 at 10:39
• like we have a $$\lim_{n\rightarrow a} \frac{x^n-a^n}{x-a}=na^{n-1}$$ this is just an example of a standard limit....i meant something like that Commented Sep 10, 2022 at 10:42
• I don't think, in this form, it is solvable by standard limits. Are you sure for the given expression?
– user
Commented Sep 10, 2022 at 11:01
• In my opinion, it is really an undetermined limit. Commented Sep 10, 2022 at 11:05

## 1 Answer

Note that $$n!e =n!\sum_{k=0}^\infty \frac{1}{k!}$$ and so it equals an integer plus $$\sum_{k=n+1}^\infty \frac{n!}{k!}=\frac{1}{n+1}+\frac{1}{(n+1)(n+2)}+....$$ Since $$\sin(x)=\sin(x+2m\pi)$$ for any $$m\in \mathbb{Z}$$, using the monotonicity of $$\sin$$ in $$[0,\frac{\pi}{2}]$$ and the estimates $$\frac{1}{n+1} < \sum_{k=n+1}^\infty \frac{n!}{k!} < \sum_{k=n+1}^\infty \frac{1}{(n+1)^{k-n}}=\frac{n+1}{n}-1=\frac{1}{n}$$ we conclude that $$\sin(\frac{2\pi}{n+1})< \sin(2\pi n!e) < \sin(\frac{2\pi}{n})$$ and so by confrontation the limit is $$0$$

• Nice answer, I like it very much ! Commented Sep 10, 2022 at 12:08
• when you did $$n!\sum_{k=0}^{\infty}\frac{1}{k!}$$ shouldn't this come $$\sum_{k=0}^{\infty}\frac{n!}{k!}$$ and not $$\sum_{k=n+1}^{\infty}\frac{n!}{k!}$$ How did you come with that?? Commented Sep 10, 2022 at 15:48
• Read carefully. I didn't say that these two quantities are equal. I said that their difference is an integer Commented Sep 10, 2022 at 16:23
• @DarkMagician can you pls add why? Commented Sep 10, 2022 at 16:27
• Of course. Their difference is $$\sum_{k=0}^n \frac{n!}{k!}$$ and $k!$ divides $n!$ if $k\leq n$ Commented Sep 10, 2022 at 16:30