(asked previously in MSE here)

In the course of a calculation, I have met the following complicated identity. Let $A$ and $a$ be positive integers. Then I believe that $$ \sum_{B\ge A,b\ge a} (-1)^{a+b}{b\choose a} {B-1\choose A-1}\frac{(B-1)!b!}{(B+b)(x-b)^{(B+b)}}=\frac{(A-1)!a!}{(A+a)(x-A+1)^{(A+a)}},$$ where $x$ is some variable and $(x)^{(a)}=x(x+1)\cdots(x+a-1)$ is the rising factorial. This is what I would like to prove.

Notice how the left hand side in principle has poles at all integer values of $x$, while the right hand side only has poles at integers smaller than $A-1$.

I think the double sum on the left hand side is convergent only for large enough $x$, and it is in this regime that it agrees with the right hand side. Take the case $A=a=1$, for example. If I sum $B$ and $b$ both from 1 to 2 I get $$ \frac{1}{2}\frac{x^2-3x-8}{x(x-2)(x^2-1)}$$ for the left hand side, which is $$ \frac{1}{2x^2}- \frac{1}{2x^3}+O\left(\frac{1}{x^4}\right)$$ for large $x$. If I sum $B$ and $b$ up to larger values, more terms in the large $x$ expansion of the left hand side agree with $$\frac{1}{2x^2}-\frac{1}{2x^3}+\frac{1}{2x^4}-\frac{1}{2x^5}+\cdots,$$ which is the large $x$ series of $\frac{1}{2x(x+1)}$, the right hand side.