$\lim_{n \to +\infty } \left \{ en! \right \}$

You have $n!e=n!(1+\frac{1}{2!}+...+\frac{1}{n!})+\frac{1}{n+1}+\frac{1}{(n+1)(n+2)}+....+\frac{1}{(n+1)(n+2)....(n+p)}+...$.

We have $\frac{1}{n+1}+\frac{1}{(n+1)(n+2)}+....+\frac{1}{(n+1)(n+2)....(n+p)}+...$ $\leq \frac{1}{n+1}+\frac{1}{(n+1)^2}+...+\frac{1}{(n+1)^p}+...=$ $=\frac{1}{n+1}(1+\frac{1}{n+1}+...+\frac{1}{(n+1)^p}+..)=\frac{1}{n+1}\frac{1}{1-\frac{1}{n+1}}=\frac{1}{n}$.

This means that $\{n!e\}\leq \frac{1}{n}$, and the conclusion follows.

Use the fact that $$ en! = n! \cdot\biggl(1 + \frac{1}{1!} + \frac{1}{2!} + \cdots + \frac{1}{n!} \biggr) + n! \cdot \biggl( \frac{1}{(n+1)!} + \frac{1}{(n+2)!} + \cdots \biggr)$$