<s>[INST] How to prove $\{n!e\} = \frac{1}{n +1} + \frac{1}{(n+1)(n+2)} + \frac{1}{(n+1)(n+2)(n+3)} + \dots$ I have the definition $a_n = \frac{1}{n +1} + \frac{1}{(n+1)(n+2)} + \frac{1}{(n+1)(n+2)(n+3)} + \dots$
I need to show that:
$a)$ $0 < a_n < \frac{1}{n}$
$b)$ $a_n = n!e - \lfloor{n!e}\rfloor$
So I know that
$\frac{1}{n} = \frac{1}{n+1}+\frac{1}{(n+1)^2}+\frac{1}{(n+1)^3}+\dots$
$e = \sum_{n=0}^{+\infty}\frac{1}{n!}$
It's clear that $\{n!e\}=a_n<\frac{1}{n}$, but how to show it? 
 [/INST] $$n!e=n!\sum_{k=0}^\infty\frac1{k!}=\sum_{k=0}^\infty\frac{n!}{k!}$$
Since $k!$ divides $n!$ for $k\le n$, the sum
$$\sum_{k=0}^n\frac{n!}{k!}$$
is an integer, say $N$. Then
$$n!e=N+\sum_{k=n+1}^\infty\frac{n!}{k!}=\sum_{k=1}^\infty\frac1{\prod_{j=1}^k(n+k)}<\sum_{k=1}^\infty\frac1{(n+1)^k}=\frac1n$$
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