Evaluate in closed form: $ \sum_{m=0}^\infty \sum_{n=0}^\infty \sum_{p=0}^\infty\frac{m!n!p!}{(m+n+p+2)!}$ 
Evaluate in closed form:
  $$ \sum_{m=0}^\infty \sum_{n=0}^\infty \sum_{p=0}^\infty\frac{m!n!p!}{(m+n+p+2)!}$$ 

I tried to use same method for similar question and two variables.Was not able to get the final answer. Question proposed by Jalil Hajimir --
Infinite Series $\sum_{m=0}^\infty\sum_{n=0}^\infty\frac{m!\:n!}{(m+n+2)!}$
 A: Playing with Mathematica a bit and you will see that the sum is
$$
S(3)=\sum_{p=0}^\infty  \sum_{m=0}^\infty  \sum_{n=0}^\infty\frac{m!n!p!}{(m+n+p+3)!}
=
\frac{1}{12} \left(3 \zeta (3)+2 \pi ^2 \log (2)\right)+
\sum _{n=1}^{\infty } \left(\frac{\psi ^{(1)}(n)}{4 n (2 n-1)}-\frac{\psi ^{(1)}\left(n+\frac{1}{2}\right)}{4 n (2 n-1)}\right)
$$
where $\psi$ is the digamma function. It does not look like it can be further simplified.

Curiously though, a slightly different sum do has a closed form
$$
S(2)=\sum_{p=0}^\infty  \sum_{m=0}^\infty  \sum_{n=0}^\infty\frac{m!n!p!}{(m+n+p+2)!}
=\pi^2/4
$$
To see this, one can check
$$
S(2)
=\sum_{p=0}^\infty  \sum_{m=0}^\infty\frac{m! p!}{(m+p+1)^2 (m+p)!}
=\sum_{p=0}^\infty\frac{\, _3F_2(1,1,p+1;p+2,p+2;1)}{(p+1)^2}
=:\sum_{p=0}^\infty a_p
$$
Mathematica cannot simplify this anymore, but look at the first few terms
$$
\frac{\pi ^2}{6},\frac{1}{6} \left(12-\pi ^2\right),\frac{1}{6} \left(\pi ^2-9\right),\frac{1}{18} \left(31-3 \pi ^2\right),\frac{1}{72} \left(12 \pi ^2-115\right),\frac{3019-300 \pi ^2}{1800},\frac{1}{600} \left(100 \pi ^2-973\right),\frac{48877-4900 \pi ^2}{29400}
$$
The pattern is quite obvious, we should have
$$
a_{2p}+a_{2p+1}= \frac{2}{(2 p+1)^2}, \quad p \ge 0.
$$
So 
$$
S(2)=\sum_{p \ge 0} \frac{2}{(2 p+1)^2} = \pi^2/4.
$$
A: In the book (Almost) Impossible Integrals, Sums, and Series, page $533$, the result in $(6.291)$, it is easily shown by transforming the Gamma summand into a summand involving a product of Beta function (which we can further deal with using the integral representation of Beta function) that
$$\sum_{i=1}^{\infty}\left( \sum_{j=1}^{ \infty}  \frac{\Gamma(i)\Gamma(j)\Gamma(x)}{\Gamma(i+j+x)}\right)=\frac{1}{2}\left(\psi^{(1)}\left(\frac{x}{2}\right)-\psi^{(1)}\left(\frac{1+x}{2}\right)\right).$$
Now, replacing $x$ by $k$, considering the summation from $k=1$ to $\infty$, passing from Gamma function to factorials and reindexing the series, we get in the left-hand side the series the Op is interested in, and in the right-hand side a simple telescoping sum involving Trigamma function. The answer is indeed $\pi^2/4$.
