<s>[INST] 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.
 [/INST] $$xe^x = \sum\frac{x^{n+1}}{n!}\\
\frac d{dx} x e^x = e^x + x e^x = \sum \frac{(n+1)x^n}{n!}\\
\frac {d^2}{dx^2} x e^x =2 e^x + x e^x =\sum \frac{n(n+1)x^{n-1}}{n!}$$
and as $x$ approaches $1$
$$3 e=\sum \frac{n(n+1)}{n!}$$
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