Factorial Summation

To start with, $\sum_{k=0}^\infty \frac{k!}{(n+r+k+2)!} \; = \; \frac{1}{(n+r+1) \times (n+r+1)!} \hspace{2cm} n,r \ge 0 \;,$ so that the required sum is $S \; = \; \sum_{n,r=0}^\infty \frac{1}{n+r+1} \times \frac{n! r!}{(n+r+1)!} \; = \; \sum_{r,n=0}^\infty \frac{1}{n+r+1} \int_0^1 x^n(1-x)^r\,dx$ Next we use the series identity $\sum_{r,n=0}^\infty \frac{x^n (1-x)^r}{n+r+1} \; = \; \frac{\ln(1-x) - \ln x}{1-2x} \hspace{2cm} 0 \le x \le 1$ to deduce that $S \; = \; \int_0^1 \frac{\ln(1-x) - \ln x}{1-2x}\,dx \; = \; \tfrac14\pi^2$ making the answer $2+4=\boxed{6}$ .

$\displaystyle S = \sum_{a,b,c\in\mathbb{N}} \beta(a,b,c) = \sum_{a,b,c\in\mathbb{N}} \beta(a,b)\times \beta(a+b,c)$

$\displaystyle = \int_{0}^{1}\int_{0}^{1}\sum_{a,b,c\in\mathbb{N}} x^{a-1}(1-x)^{b-1}y^{a+b-1}(1-y)^{c-1} \space dx\space dy$

$\displaystyle = \int_{0}^{1}\int_{0}^{1} \frac{dx\space dy}{(1-xy)(1+xy-y)}$

A little simplification gives, $\displaystyle S = -2\int_{0}^{1}\frac{\ln y}{1-y^2}dy = -2 \sum_{n\ge 0} \int_{0}^{1} y^{2n}\ln y \space dy = 2\times \sum_{n\ge 0} \frac{1}{(2n+1)^2} = \frac{\pi^2}{4}$

$\displaystyle \text{Note : } \beta(a,b,c)= \frac{\Gamma(a)\Gamma(b)\Gamma(c)}{\Gamma(a+b+c)}$ which is a Multivariate Beta function