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.

Define $f$ by $$f(x) = \sum_{n=0}^\infty \frac{x^{n+1}}{n!}$$ for $x\in\mathbb{R}$. (It is easy to check that the radius of convergence of this function is infinite.)
In particular:


*

*For all $x\in\mathbb{R}$, $f''(x) = \sum_{n=1}^\infty \frac{(n+1)n}{n!}x^{n-1}$, so you are looking for $f''(1)$;

*For all $x\in\mathbb{R}$, $f(x) = x e^x$ using the known power series for $\exp$, so that $f''(x) = (x+2)e^x$.
Therefore, $f''(1) = 3e$.
