$\sum_{m\ge 0}\sum_{n\ge 0}\sum_{p\ge 0}\frac{m!n!p!}{(m+n+p+3)!}$ Evaluate: $\sum_{m\ge 0}\sum_{n\ge 0}\sum_{p\ge 0}\frac {m!n!p!}{(m+n+p+3)!}$
my working :
Claim:$\sum_{k\ge0}\frac{k!}{(k+l)!}=\frac{1}{(l-1)((l-1)!)}$
proof:$\frac{k!}{(k+l)!}=\frac{1}{l-1}\left(\frac{k!}{(l+k-1)!}-\frac{(k+1)!}{(l+k)!}\right)$
$\implies\sum_{k\ge0}\frac{k!}{(k+l)!}=\frac{1}{(l-1)((l-1)!)}$
now $\sum_{m\ge 0}\sum_{n\ge 0}\sum_{p\ge 0}\frac {m!n!p!}{(m+n+p+3)!}=\sum_{m\ge 0}\sum_{n\ge 0}m!n!\sum_{p\ge 0}\frac {p!}{(m+n+p+3)!}$
$=\sum_{m\ge 0}\sum_{n\ge 0}m!n!\frac {1}{(m+n+2)((m+n+2)!)}$
 A: This is a partial answer, I can only reduce the sum to a single integral.
For any $d \ge 1$, let $\Delta_d$ be the simplex
$$\Delta_d = \{ (x_1,\ldots,x_d) \in [0,\infty)^d : x_1 + \cdots x_d \le 1 \}$$
The summands at hand can be represented as integrals over $\Delta_3$.
$$\frac{m!n!p!}{(m+n+p+3)!} = \int_{\Delta_3} x^m y^n z^p dxdydz$$
This leads to
$$\begin{align}
\mathcal{S} \stackrel{def}{=} & \sum_{m,n,p \ge 0}\frac{m!n!p!}{(m+n+p+3)!}
= \int_{\Delta_3}\frac{dxdydz}{(1-x)(1-y)(1-z)}\\
= &\int_{\Delta_2}  \left(\int_0^{1-y-z}\frac{dx}{1-x}\right) \frac{dydz}{(1-y)(1-z)}
\\
= & -\int_{\Delta_2} \frac{\log(y+z)}{(1-y)(1-z)} dydz\\
= & -\int_{\Delta_2} \left(\frac{1}{1-y} + \frac{1}{1-z}\right)\frac{\log(y+z) dydz}{2 - y - z}\\
= & -2\int_{\Delta_2} \frac{\log(y+z)dydz}{(1-y)(2-y-z)}
\end{align}$$
Change variable to $(y,\lambda) = (y,y+z)$, we obtain
$$\begin{align}\mathcal{S}
&= -2\int_0^1 \int_0^\lambda \frac{\log \lambda}{(1-y)(2-\lambda)} dy d\lambda
= 2\int_0^1 \frac{\log\lambda\log(1-\lambda)}{2-\lambda} d\lambda\\
&\approx 0.4861407
\end{align}
$$
This is as far as I can go.
