Evaluating the sum $\sum_{n=1}^{\infty}\frac{1}{n!(n+2)}$

HINT:

$$\frac{1}{n!(n+2)}=\frac{1}{(n+1)!}-\frac{1}{(n+2)!}$$

Work via generating functions: $e^x=\sum_{n=0}^\infty\dfrac{x^n}{n!}$, so $xe^x=\sum_{n=0}^\infty\dfrac{x^{n+1}}{n!}$. Now integrate this with respect to $x$; you should find that the RHS becomes $\sum_{n=0}^\infty\dfrac{x^{n+2}}{(n+2)n!}$ (and the LHS is easily integrable by parts). Finally, set $x=1$ (and note that your sum starts at $n=1$ and not $n=0$, so there's a term from this series that you're missing).

$$e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}$$ $$xe^x=\sum_{n=0}^{\infty}\frac{x^{n+1}}{n!}$$ $$\int_{0}^{1}xe^xdx=\int_{0}^{1}\sum_{n=0}^{\infty}\frac{x^{n+1}}{n!}dx$$ $$1=\sum_{n=0}^{\infty}\frac{1}{n!(n+2)}$$ $$1=\frac{1}{2}+\sum_{n=1}^{\infty}\frac{1}{n!(n+2)}$$

so $$\sum_{n=1}^{\infty}\frac{1}{n!(n+2)}=\frac{1}{2}$$