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?
On
Because $$n!e=n!\left(2+\frac{1}{2!}+...+\frac{1}{n!}\right)+\frac{1}{n+1}+\frac{1}{(n+1)(n+2)}+...$$ and $$\frac{1}{n+1}+\frac{1}{(n+1)(n+2)}+...<\frac{1}{n+1}+\frac{1}{(n+1)^2}+...=\frac{\frac{1}{n+1}}{1-\frac{1}{n+1}}=\frac{1}{n}<1$$
On
$$n!e=n!\sum_{k=0}^\infty\frac1{k!}=\sum_{k=0}^\infty\frac{n!}{k!}$$
Since $k!$ divides $n!$ for $k\le n$, the sum $$\sum_{k=0}^n\frac{n!}{k!}$$ is an integer, say $N$. Then $$n!e=N+\sum_{k=n+1}^\infty\frac{n!}{k!}=\sum_{k=1}^\infty\frac1{\prod_{j=1}^k(n+k)}<\sum_{k=1}^\infty\frac1{(n+1)^k}=\frac1n$$
On
Write
\begin{align} a_n &=\sum_{k\geq 1}\frac{1}{\prod_{1\leq i \leq k}(n+i)}\\ &=\sum_{k\geq 1}\frac{n!}{(n+k)!}\\ &=n!\,\sum_{k\geq 1}\frac1{(n+k)!}\\ &=n!\,\sum_{j\geq n+1}\frac1{j!}\\ &=n!\,\left[\left(\sum_{j\geq 0}\frac1{j!}\right) -\left(\sum_{j=0}^n\frac1{j!}\right)\right]\\ &=n!\,\left[e -\left(\sum_{j=0}^n\frac1{j!}\right)\right]=n!e-n!\left(\sum_{j=0}^n\frac1{j!}\right) \end{align}
Do you think you can conclude?
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.