<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] By comparison with the Geometric series we have  $$0<a_n<\frac 1{n+1}+\frac 1{(n+1)^2}+\cdots=\frac 1{n+1}\times \frac 1{1-\frac 1{n+1}}=\frac 1n$$
Now assume $n>1$ (so $0<a_n<\frac 1n<1$).  We write: $$n!e=n!+ \frac {n!}{2!}+\cdots \frac {n!}{n!}+a_n$$  and it is clear that the part preceding $a_n$ is an integer (each term in the sum is an integer) so we are done.
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