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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 $$f(x,y,a,a)=\frac{1}{(1-a)^2}-\frac{(x+y)^2a}{(x+y-1)(x+y+1)(1-a)^2}$$

(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 $$f(x,y,a,a)=\frac{1}{(1-a)^2}-\frac{(x+y)^2a}{(x+y-1)(x+y+1)(1-a)^2}$$

(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 $$f(x,y,a,a)=\frac{1}{(x+y)(1-a)^2}-\frac{(x+y)a}{(x+y-1)(x+y+1)(1-a)^2}$$

(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 $$f(x,y,a,a)=\frac{1}{(1-a)^2}-\frac{(x+y)^2a}{(x+y-1)(x+y+1)(1-a)^2}$$

(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 $$f(x,y,a,a)=\frac{xy}{x+y}-\frac{xy(x+y)a}{(x+y)^2-1}$$

(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 $$f(x,y,a,a)=\frac{1}{(x+y)(1-a)^2}-\frac{(x+y)a}{(x+y-1)(x+y+1)(1-a)^2}$$