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What is the limit of $n \sin (2 \pi \cdot e \cdot n!)$ as $n$ goes to infinity?
In order to solve the following limit $$\lim_{n\to\infty} n\sin2\pi n!e$$ . This question is very likely to have been asked.
I remember this question and the answer is like $2\pi$ or something .
I also do remember approximating $n!e$ but somehow i don't remember and can't figure out right now .
Remember that $\mathrm e=\sum\limits_{k=0}^n1/k!+R_n$ with $R_n=\sum\limits_{k=n+1}^{+\infty}1/k!$. Hence $n!\,\mathrm e$ equals an integer plus $n!R_n$. Now $1/(n+1)!\lt R_n\lt1/(n\,n!)$, hence $n!R_n=1/n+o(1/n)$, which is all the precision you need.
Look at the usual series for $e$. Then $n!e$ is an integer plus $$\frac{1}{n+1}+\frac{1}{(n+1)(n+2)}+\frac{1}{(n+1)(n+2)(n+3)}+\cdots.\tag{$1$}$$ The sum $(1)$ is bigger than $\frac{1}{n+1}$. It is less than $$\frac{1}{n+1}+\frac{1}{(n+1)^2}+\frac{1}{(n+1)^3}+\cdots,$$ a geometric series with sum $\frac{1}{n}$.
$$ \begin{align} n!e &=n!\sum_{k=0}^\infty\frac1{k!}\\ &=\color{#C00}{\sum_{k=0}^n\frac{n!}{k!}}+\color{#090}{\sum_{k=n+1}^\infty\frac1{k!/n!}}\\ &\equiv\color{#090}{\frac1{n+1}+\frac{1}{(n+1)(n+2)}+\frac{1}{(n+1)(n+2)(n+3)}+\dots}\pmod{\color{#C00}{\mathbb{Z}}} \end{align} $$ where the last sum is greater than $\frac1{n+1}$ yet less than $\frac1n=\frac1{n+1}+\frac1{(n+1)^2}+\frac1{(n+1)^3}+\dots$
Thus, we have the bounds $$ \frac1{n+1}<n!e-\lfloor n!e\rfloor<\frac1n $$ Therefore, $$ n\sin\left(\frac{2\pi}{n+1}\right)<n\sin(2\pi n!e)<n\sin\left(\frac{2\pi}{n}\right) $$ and by the Squeeze Theorem, we get $$ \lim_{n\to\infty}n\sin(2\pi n!e)=2\pi $$
