Find the sum of the infinite series $\sum n(n+1)/n!$

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, $f''(1) = 3e$.

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 $3e$.

$$xe^x = \sum\frac{x^{n+1}}{n!}\\ \frac d{dx} x e^x = e^x + x e^x = \sum \frac{(n+1)x^n}{n!}\\ \frac {d^2}{dx^2} x e^x =2 e^x + x e^x =\sum \frac{n(n+1)x^{n-1}}{n!}$$

and as $x$ approaches $1$

$$3 e=\sum \frac{n(n+1)}{n!}$$