I have the following summation that when expanded represents the factorial, for example:
$$\sum_{n=2}^{\infty} \sum_{m=1}^n \frac{\prod_{i=0}^n(i+2)}{n+m}=(2\cdot4)+(2\cdot3\cdot5)+(2\cdot3\cdot4\cdot6)+(2\cdot3\cdot4\cdot5\cdot6)+\cdots+(2\cdots n\cdot(n+2))$$ I want to express this summation by the following method $$\sum_{n=0}^{\infty}\sum_{k=0}^n A_{k,n}=\sum_{n=0}^{\infty}\sum_{k=0}^{\infty} A_{k,n+k}$$
But notice the index for $n$ is $2$ and for $m$ its $1$, so I then have $$\sum_{n=2}^{\infty} \sum_{m=1}^n \frac{\prod_{i=0}^n(i+2)}{n+m}=\sum_{n=1}^{\infty} \sum_{m=1}^n \frac{\prod_{i=0}^{n+1}(i+2)}{(n+1)+m}$$ And so to complete the summation equality, I equate the two RHS $$\sum_{n=1}^{\infty} \sum_{m=1}^n \frac{\prod_{i=0}^{n+1}(i+2)}{(n+1)+m}=\sum_{n=1}^{\infty} \sum_{m=1}^\infty \frac{\prod_{i=0}^{\infty}(i+2)}{(n+m)+m}$$
Looping back here because I have answered what I wanted to gain from this post of mine. So we know the following,
So taking only a single part of the equation: $$\sum_{n=-\infty}^{\infty} \prod_{i=-\infty}^{n}(i+a)$$ We have the upper and lower triangular elements of a factorial arranged into a determinant. Then the diagonal elements for an infinite determinant is denoted by $1+a_{ii}$, so the above diagonal entries becomes: $$\prod_{i=-\infty}^{\infty}\left(1+\sum_{n=-\infty}^{\infty} \prod_{i=-\infty}^{n}(i+a) \right)$$
When I look at the summaton part, its obvious the sum of the factorial is divergent, therefore the diagonals are also divergent. : $$\sum_{n=-\infty}^{\infty}\Gamma(n+a)$$