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.
Best 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 $0$