How to prove that $e=3-\frac{1}{2!1\cdot 2}-\cdots-\frac{1}{n!(n-1)n}-\cdots$? How to prove that $e=3-\frac{1}{2!1\cdot 2}-\cdots-\frac{1}{n!(n-1)n}-\cdots$? I’ve proved that 
$1+1+\frac{1}{2!}+\cdots+\frac{1}{n!}+\frac{1}{n!n}= 3-\frac{1}{2!1\cdot 2}-\cdots-\frac{1}{n!(n-1)n}$. 
 A: Use $\frac{1}{n!(n-1)n}=\frac{1}{n!(n-1)}-\frac{1}{n!n}$

Thus,
$$\sum_{n=2}^{\infty}\frac{1}{n!(n-1)n}=\sum \frac{1}{n!(n-1)}-\frac{1}{n!n}$$
$$=(\frac{1}{2!\cdot 1}-\frac{1}{2!\cdot 2})+(\frac{1}{3!\cdot 2}-\frac{1}{3!\cdot 3})+...$$
$$=\frac{1}{2!\cdot 1}+(-\frac{1}{2!\cdot 2}+\frac{1}{3!\cdot 2})+(-\frac{1}{3!\cdot 3}+\frac{1}{4!\cdot 3})+....$$
$$=\frac{1}{2}+\sum_{n=2} \frac{-1}{n!n}+\frac{1}{(n+1)!n}$$
$$=\frac{1}{2}-\sum \frac{1}{(n+1)!}$$
Thus,
$$3-\sum_{n=2}^{\infty}\frac{1}{n!(n-1)n} = 1+1 +1-\frac{1}{2}+\sum_{n=2}^{\infty}\frac{1}{(n+1)!}=e$$
A: Starting with the Maclaurin expansion
$$\frac{\mathrm{e}^x-1-x}{x^2}=\sum_{n=2}^\infty \frac{x^{n-2}}{n!} ,$$
we integrate over the interval $[0,x]$ to get 
$$\text{Ei}(x )+\frac{1-e^{x }}{x }-\log (x )-\gamma +1 = \sum_{n=2}^\infty \frac{x^{n-1}}{n!(n-1)},$$
where $\text{Ei}$ is the exponential integral. A second integration yields
$$ (x-1) \text{Ei}(x)-\gamma  x+2 x-e^x-x \log (x)+\log (x)+\gamma +1=\sum_{n=2}^\infty \frac{x^n}{n!(n-1)n}, $$
and the substitution $x=1$ finishes the job.
