TITLE: Infinite Series $\sum_{m=0}^\infty\sum_{n=0}^\infty\frac{m!\:n!}{(m+n+2)!}$
QUESTION [7 upvotes]: Evaluating
$$\sum_{m=0}^\infty \sum_{n=0}^\infty\frac{m!n!}{(m+n+2)!}$$
involving binomial coefficients.
My attempt: $$\frac{1}{(m+1)(n+1)}\sum_{m=0}^\infty \sum_{n=0}^\infty\frac{(m+1)!(n+1)!}{(m+n+2)!}=\frac{1}{(m+1)(n+1)} \sum_{m=0}^\infty \sum_{n=0}^\infty\frac{1}{\binom{m+n+2}{m+1}}=?$$
Is there any closed form of this expression?

REPLY [5 votes]: Another (way less subtle) approach. We have:
$$ S=\sum_{m\geq 0}\sum_{n\geq 0}\frac{\Gamma(m+1)\,\Gamma(n+1)}{(m+n+2)\,\Gamma(m+n+2)}=\sum_{m,n\geq 0}\iint_{(0,1)^2} x^m(1-x)^n y^{m+n+1}\,dx\,dy \tag{1}$$
hence:
$$ S = \iint_{(0,1)^2}\frac{y\,dx\,dy}{(1-xy)(1-y+xy)}=2\int_{0}^{1}\frac{-\log(1-y)}{2-y}\,dy=2\int_{0}^{1}\frac{-\log(t)}{1+t}\,dt\tag{2} $$
and by expanding $\frac{1}{1+t}$ as a geometric series,
$$ S = 2\sum_{n\geq 0}\frac{(-1)^n}{(n+1)^2} = \color{red}{\zeta(2)} = \frac{\pi^2}{6}.\tag{3}$$