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(Previously asked in MSE, no answer even with bounty offer)

In the course of a calculation, I arrived at the quantity $$ f(x,y,a,b)= \sum_{n,m,i,r,q,l\ge 0}\sum_{k=0}^{n+m} K_{n,m,i,r,l,q,k}\frac{(x)^{(i)}(y)^{(r)}(x)_{(l)}(y)_{(q)}}{(x+y+1)^{(n+m+r+i-k)}(x+y-1)_{(q+l+k)}} a^{n+i+l} b^{m+r+q}, $$ where $(x)^{(i)}=x(x+1)\cdots(x+i-1)$ is the rising factorial, $(x)_{(i)}=x(x-1)\cdots(x-i+1)$ is the falling factorial and $$ K_{n,m,i,r,l,q,k}= (-1)^k\frac{(n+m+r+i-k)!(k+q+l)!}{i!r!l!q!(n+m+r+i+q+l+1)}. $$

Based on some previous experience, I suspect/hope this expression can be simplified, as a series in powers of $a,b$.

For example, there are a priori 10 terms corresponding to $a^1b^1$ in this sum, but they can be arranged in the form $$\frac{2(x-1)(y-1)}{3(x+y-2)(x+y-1)}+\frac{2xy-1}{3(x+y-1)(x+y+1)}+\frac{2(x+1)(y+1)}{3(x+y+1)(x+y+2)}$$

What have I tried? Well, according to folklore, everytime you find multiple sums, play with their order. I played and played, but I keep getting tangled up in the summation limits.

Part of the motivation for trying to simplify the expression is to prove that, when $a=b$, we surprisingly have nothing more than $$f(x,y,a,a)=\frac{1}{(1-a)^2}-\frac{(x+y)^2a}{(x+y-1)(x+y+1)(1-a)^2}$$

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  • $\begingroup$ your expression is already a series in powers of $a,b$; isn't that what you want? a more compact expression than what you have written down seems unlikely. $\endgroup$
    – Carlo Beenakker
    Commented Feb 1, 2022 at 18:08
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    $\begingroup$ @CarloBeenakker It is in the format I want, but currently I have 6 infinite sums. I suspect some of them might be carried out explicitly. For instance, one might use the fact that $\sum_{i=0}^j\frac{(x)^{(i)}}{i!}=\frac{(x+1)^{(j)}}{j!}$. Notice the example I gave where 10 terms can be written as 3. $\endgroup$
    – Marcel
    Commented Feb 1, 2022 at 18:16
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    $\begingroup$ Rephrasing it with $\binom{n}{k} = \frac{n_{(k)}}{k!}$ as $$f(x,y,a,b)= \sum_{m,n \geq 0} a^n b^m \sum_{i,r,q,l\ge 0}\frac{1}{(n+m+i+r+q+l+1)} \binom{x+i-1}{i}a^i \binom{y+r-1}{r}b^r \binom{y}{q} b^q \binom{x}{l}a^l \sum_{k=0}^{m+n} (-1)^k \binom{x+y+n+m+r+i-k}{n+m+r+i-k}^{-1} \binom{x+y+q+l+k-2}{q+l+k}^{-1}$$ doesn't give any immediate gain, as the sum on the right doesn't seem to be Zeilberger-summable, but might at least be a more recognisable form. $\endgroup$
    – Peter Taylor
    Commented Feb 1, 2022 at 22:51

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When $a=b$ it reduces to $$f(x,y,a,a)=\sum_{mrq\ge 0}\sum_{k=0}^m\frac{(-1)^k(m+1)(m-k+r)!(k+q)!(x+y)^{(r)}(x+y)_{(q)}a^{m+r+q}}{r!q!(m+r+q+1)(x+y+1)^{(m-k+r)}(x+y-1)_{(k+q)}}.$$

Three of the infinite sums are gone. But it is still not quite what you would like, I suppose.

I have used $$\sum_{i=0}^r\frac{(x)^{(i)}(y)^{(r-i)}}{i!(r-i)!}=\frac{(x+y)^{(r)}}{r!}$$ and a similar identity for the falling factorials.

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