Infinite Series $\sum_{m=0}^\infty\sum_{n=0}^\infty\frac{m!\:n!}{(m+n+2)!}$ 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?
 A: 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,
$$ S = 2\sum_{n\geq 0}\frac{(-1)^n}{(n+1)^2} = \color{red}{\zeta(2)} = \frac{\pi^2}{6}.\tag{3}$$ 
A: One may observe that,
$$
\frac{m!}{(m+n+2)!}=\frac{m!}{(n+1)(m+n+1)!}-\frac{(m+1)!}{(n+1)(m+n+2)!}
$$ giving, by telescoping terms,
$$
\sum_{m=0}^N\frac{m!}{(m+n+2)!}=\frac1{(n+1)(n+1)!}-\frac{(N+1)!}{(n+1)(N+n+2)!}
$$ thus, as $N \to \infty$,
$$
\sum_{m=0}^\infty\frac{m!}{(m+n+2)!}=\frac1{(n+1)(n+1)!}.
$$ Then the initial series reduces to

$$
\sum_{n=0}^\infty \sum_{m=0}^\infty\frac{n!\:m!}{(m+n+2)!}=\sum_{n=0}^\infty\frac{n!}{(n+1)(n+1)!}=\sum_{n=1}^\infty\frac1{n^2}=\frac{\pi^2}6.
$$

