Double sum Inverse product

The entire thing can be written as :- $\Large \frac{1}{6}\sum_{k=0}^{\infty} \sum_{m=0}^{\infty} \frac{(k+m+7)-(k+m+1)}{\prod_{\alpha=1}^{\alpha=7}(k+m+\alpha)}$

$\Large \frac{1}{6}\sum_{k=0}^{\infty} \sum_{m=0}^{\infty} \frac{1}{\prod_{\alpha=1}^{\alpha=6}(k+m+\alpha)}-\frac{1}{\prod_{\alpha=2}^{\alpha=7} (k+m+\alpha)}$

If $\Large V_{m} = \frac{1}{\prod_{\alpha=1}^{\alpha=6}(k+m+\alpha)}$

Then we are looking at nothing but

$\Large \frac{1}{6}\sum_{k=0}^{\infty} (\sum_{m=0}^{\infty} (V_{m} - V_{m+1}))$

$\Large = \frac{1}{6}\sum_{k=0}^{\infty} V_{0}$ (As $V_{m}$ tend to $0$ as $m\to\infty$ )

$\Large = \frac{1}{6}\sum_{k=0}^{\infty} \frac{1}{\prod_{\alpha=1}^{\alpha=6}(k+\alpha)}$

This entire thing can be written as :-

$\Large = \frac{1}{6\cdot 5}\sum_{k=0}^{\infty} \frac{(k+6)-(k+1)}{\prod_{\alpha=1}^{\alpha=6}(k+\alpha)}$

$\Large = \frac{1}{6\cdot 5}\sum_{k=0}^{\infty}( \frac{1}{\prod_{\alpha=1}^{\alpha=5}(k+\alpha)}-\frac{1}{\prod_{\alpha=2}^{\alpha=6}(k+\alpha)})$

If $\Large T_{k} = \frac{1}{\prod_{\alpha=1}^{\alpha=5}(k+\alpha)}$

Then we are looking at:-

$\Large = \frac{1}{6\cdot 5}\sum_{k=0}^{\infty} (T_{k} - T_{k+1})$

$\Large = \frac{1}{6\cdot 5} T_{0}$ (As $T_{k}$ tends to $0$ as $k\to\infty$ )

$T_{0} = \frac{1}{5!}$

So our answer is $\frac{1}{5 \cdot 6 \cdot 5!} = \frac{1}{3600}$

Relevant Wiki :- Telescopic Series

We have, changing summation variables to $X = k+m$ and $k$ , $\begin{aligned} \sum_{k=0}^\infty \sum_{m=0}^\infty \frac{(k+m)!}{(k+m+p)!} & = \; \sum_{X=0}^\infty\sum_{k=0}^X \frac{X!}{(X+p)!} \; = \; \sum_{X=0}^\infty \frac{(X+1)!}{(X+p)!} \; = \; \sum_{X=0}^\infty \frac{\Gamma(X+2)}{\Gamma(X+p+1)} \\ & =\; \frac{1}{\Gamma(p-1)}\sum_{X=0}^\infty B(X+2,p-1) \; =\; \frac{1}{\Gamma(p-1)}\sum_{X=0}^\infty\int_0^1 u^{X+1}(1-u)^{p-2}\,du \\ & = \; \frac{1}{\Gamma(p-1)}\int_0^1 u(1-u)^{p-3}\,du \; = \; \frac{1}{\Gamma(p-1)}B(2,p-2) \; =\; \frac{\Gamma(2)\Gamma(p-2)}{\Gamma(p-1)\Gamma(p)} \; = \; \frac{1}{(p-2)(p-1)!} \end{aligned}$ for any integer $p \ge 3$ , without needing to use hypergeometrics. Putting $p=7$ we obtain $A = 5 \times 6! = \boxed{3600}$ .

In this solution, I'll be generalizing the given sum to be $\sum _{ k=0 }^{ \infty }{ \sum _{ m=0 }^{ \infty }{ \frac { 1 }{ \prod _{ \alpha =1 }^{ n }{ \left( k+m+\alpha \right) } } } } =\sum _{ k=0 }^{ \infty }{ \sum _{ m=0 }^{ \infty }{ \frac { 1 }{ \left( k+m+1 \right) \left( k+m+2 \right) .........\left( k+m+n \right) } } } = X$ (here $X$ is the value for the time being) . we can again simplify it and write it as $\sum _{ k=0 }^{ \infty }{ \sum _{ m=0 }^{ \infty }{ \frac { k!m!\left( k+m \right) ! }{ \left( k+m+n \right) !k!m! } } } =\; X$ in terms of gamma function we can write the sum as $\sum _{ k=0 }^{ \infty }{ \sum _{ m=0 }^{ \infty }{ \frac { \Gamma \left( k+1 \right) \Gamma \left( m+1 \right) \Gamma \left( k+m+1 \right) }{ \Gamma \left( k+m+n-1 \right) k!m! } } } = X$ we"ll multiply both sides of the equation by $\Gamma\left(n-1\right)$ . i.e $\sum _{ k=0 }^{ \infty }{ \sum _{ m=0 }^{ \infty }{ \frac { \Gamma \left( k+1 \right) \Gamma \left( m+1 \right) \Gamma \left( k+m+1 \right) \Gamma \left( n-1 \right) }{ \Gamma \left( k+m+n-1 \right) k!m! } } } \; =\quad X\Gamma \left( n-1 \right)$ . The left hand sum can be expressed in terms of pochhammer symbol for uprising factorial i.e $\sum _{ k=0 }^{ \infty }{ \sum _{ m=0 }^{ \infty }{ \frac { { \left( 1 \right) }_{ k }{ \left( 1 \right) }_{ m }{ \left( 1 \right) }_{ m+k } }{ { \left( n-1 \right) }_{ m+k }k!m! } } } \; =\quad X\Gamma \left( n-1 \right)$ The left hand sum is a Appells hypergeometric series which is of the form ${ F }_{ 1 }\left( a,{ b }_{ 1 },{ b }_{ 2 };c,x,y \right) =\sum _{ k=0 }^{ \infty }{ \sum _{ m=0 }^{ \infty }{ \frac { { \left( a \right) }_{ m+k }{ \left( { b }_{ 1 } \right) }_{ k }{ \left( { b }_{ 2 } \right) }_{ m } }{ { \left( c \right) }_{ m+k }k!m! } } } { x }^{ k }{ y }^{ m }$ we came to know that $a={b}_{1}={b}_{2}=1$ and $c=n-1$ . On substituting the values in Euler integral for Appell series we get ${ F }_{ 1 }\left( 1,1,1;n-1,1,1 \right) =\frac { \Gamma \left( n-1 \right) }{ \Gamma \left( 1 \right) \Gamma \left( n-2 \right) } \int _{ 0 }^{ 1 }{ \\ } { \left( 1-t \right) }^{ n-4-1 }dt =\frac { \Gamma \left( n-1 \right) }{ \Gamma \left( 1 \right) \Gamma \left( n-2 \right) } \frac { \Gamma \left( 1 \right) \Gamma \left( n-4 \right) }{ \Gamma \left( n-3 \right) }$ from the 5th part of this solution we can isolate the value of $X$ to be $X =\frac { 1 }{ \Gamma \left( n-1 \right) } { F }_{ 1 }\left( 1,1,1;n-1,1,1 \right) =\frac { 1 }{ \Gamma \left( n-1 \right) } \frac { \Gamma \left( n-1 \right) }{ \Gamma \left( 1 \right) \Gamma \left( n-2 \right) } \frac { \Gamma \left( 1 \right) \Gamma \left( n-4 \right) }{ \Gamma \left( n-3 \right) }$ on further simplification we get $X =\frac { 1 }{ \left( n-2 \right) \left( n-1 \right) ! }$ . this implies $\boxed{\sum _{ k=0 }^{ \infty }{ \sum _{ m=0 }^{ \infty }{ \frac { 1 }{\displaystyle \prod _{ \alpha =1 }^{ n }{ \left( k+m+\alpha \right) } } } } =\frac { 1 }{ \left( n-2 \right) \left( n-1 \right) ! } }$ .so we get $\sum _{ k=0 }^{ \infty }{ \sum _{ m=0 }^{ \infty }{ \frac { 1 }{ \displaystyle\prod _{ \alpha =1 }^{ 7 }{ \left( k+m+\alpha \right) } } } } =\frac { 1 }{ 3600 }$ Therefore $\boxed{A = 3600}$