Identity involving double sum with binomials (Edit: This question has now been answered in MathOverflow, 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. It is like a generalization of the identity  $$\sum_{a=0}^\infty \frac{1}{x^a}=\frac{x}{x-1}.$$
Notice a kind of miracle: 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 $A-1$.
Following the criticism in the answer by Paul Sinclair, I think the meaning of the equality is a bit unclear. I have in mind a large $x$ expansion. 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.
(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. So finding different residues at $x=0$ does not invalidate the identity.)
 A: In the case $A = 1, a = 1$, and letting $n = B + b -1$, this reduces to
$$-\sum_{n=1}^\infty\sum_{b=1}^n (-1)^bb\dfrac{(n-b)!b!}{(n+1)(x-b)^{(n+1)}} = \dfrac 1{2x(x+1)}$$
But that cannot be. The residue at $0$ can be found by multiplying both sides by $x$ and taking the limit as $x \to 0$. The right-hand side easily gives $1$. The left-hand is as little harder. $x-b \to -b$ and since $b \le n$
$$(x-b)^{(n+1)} = (x-b)(x-b+1)\cdots(x-1)x(x+1)\cdots(x+n-b)$$
So $$\lim_{x\to 0} \frac 1x (x-b)^{(n+1)} = (-b)(-b+1)\cdots(-1)(1)\cdots(n-b) = (-1)^bb!(n-b)!$$
Which means the left-hand side reduces to $$-\sum_{n=1}^\infty\sum_{b=1}^n \frac{b}{n+1}= -\sum_{n=1}^\infty\frac n2=-\infty$$
It seems clear tome that even if $x$ is only near $0$, this series is still going to diverge. Thus it cannot be equal to the much better behaved right-hand side.
Either you've made a mistake in your calculations that led you to this conclusion, or else in pulling this particular "equality" out as the cause your larger calculation is true.
