TITLE: Find $\lim_\limits{n\to \infty}\{en!\}$.
QUESTION [5 upvotes]: Find the limit $\lim_\limits{n\to \infty}\{en!\}$.
$Attempt:$ $\lim_\limits{n\to \infty}\{en!\}=\lim_\limits{n\to \infty}\{(1+{1\over 1!}+{1\over 2!}+{1\over 3!}+...+{1\over n!}+...)n!\}$. The fraction part is what confuses me. Also I don't know if it has something to do with the convergence of $\sum n!$. I would appreciate your help.

REPLY [11 votes]: You can write $e = \displaystyle \sum_{k=0}^\infty \frac{1}{k!}$ so that $$en! =  \sum_{k=0}^\infty \dfrac{n!}{k!} = \sum_{k=0}^n \frac{n!}{k!} + \sum_{k=n+1}^\infty \frac{n!}{k!}$$ so that \begin{align*}\{en!\} &= \frac{1}{n+1} + \frac{1}{(n+1)(n+2)} + \frac{1}{(n+1)(n+2)(n+3)} + \cdots \\ &\le \frac{1}{n+1} + \frac{1}{(n+1)^2} + \frac{1}{(n+1)^3} + \cdots \\ &= \frac 1n \end{align*}
where the last equality comes from the geometric series formula. Thus $\{en!\} \to 0$.