TITLE: Evaluating the sum $\sum_{n=1}^{\infty}\frac{1}{n!(n+2)}$
QUESTION [6 upvotes]: The sum again is
$$\sum_{n=1}^{\infty}\frac{1}{n!(n+2)}$$
It looks like it should be amenable to some modification to get towards the exponential power series or something close to it, But I really can't get anything to pop out. 
Wolfram alpha says it sums to 1/2 and my friend verified this by partial sums but I am curious if there is a slick way to evaluate this, maybe something like partial fractions that can deal with the factorial or a product rule type thing?

REPLY [4 votes]: $$e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}$$
$$xe^x=\sum_{n=0}^{\infty}\frac{x^{n+1}}{n!}$$
$$\int_{0}^{1}xe^xdx=\int_{0}^{1}\sum_{n=0}^{\infty}\frac{x^{n+1}}{n!}dx$$
$$1=\sum_{n=0}^{\infty}\frac{1}{n!(n+2)}$$
$$1=\frac{1}{2}+\sum_{n=1}^{\infty}\frac{1}{n!(n+2)}$$
so
$$\sum_{n=1}^{\infty}\frac{1}{n!(n+2)}=\frac{1}{2}$$