Find the sum of the infinite series $\sum n(n+1)/n!$ How do find the sum of the series till infinity?
$$ \frac{2}{1!}+\frac{2+4}{2!}+\frac{2+4+6}{3!}+\frac{2+4+6+8}{4!}+\cdots$$
I know that it gets reduced to $$\sum\limits_{n=1}^∞ \frac{n(n+1)}{n!}$$
But I don't know how to proceed further.

Note that for $n\ge 2$ we have
$$\frac{n(n+1)}{n!}= \frac{1}{(n-2)!} + \frac{2}{(n-1)!}. $$
This lets us rewrite the sum as
$$
\frac{2}{0!} + \left(\frac{1}{0!} + \frac{2}{1!} \right)
+ \left(\frac{1}{1!} + \frac{2}{2!} \right)
+ \left(\frac{1}{2!} + \frac{2}{3!} \right)
+ \left(\frac{1}{3!} + \frac{2}{4!} \right)
+ \cdots
$$
Rearranging in the obvious way gives the sum as $3e$.
