# Compute a sum involving binomial coefficients

Let $0 < a < b$ and $p_1 >0$ and $p_2>0$ be integers. The question is to prove the following identity:

$$\sum\limits_{j=a}^b \left(\begin{array}{c} j \\ p_1 \end{array} \right) \left(\begin{array}{c} j \\ p_2 \end{array} \right) = \sum\limits_{q=0}^{p_2} (-1)^{q+1} \left[ \left(\begin{array}{c} a+q \\ p_1+q+1 \end{array} \right) \left(\begin{array}{c} a \\ p_2-q \end{array} \right) - \left(\begin{array}{c} b+1+q \\ p_1+q+1 \end{array} \right) \left(\begin{array}{c} b+1 \\ p_2-q \end{array} \right) \right]$$

Can this identity be generalised for $p_1,p_2$ being real?

• What sense do you give to the binomial coefficients for real numbers ? – zozoens Feb 3 '14 at 12:26
• Well, the binomial coefficient can be defined for any two complex numbers as $$\left( \begin{array}{c} n \\ p \end{array} \right) = \frac{\Gamma[n+1]}{\Gamma[p+1] \Gamma[n-p+1]}$$. As such it makes sense to consider generic real or even complex values for $p1,p2$. I am not quite sure how to rewrite the right hand side in case of $p_1,p_2$ being real. – Przemo Feb 3 '14 at 12:27

Let us prove a more generic statement. Define: $${\mathcal S}_{p_1,p_2}^{q_1,q_2}(a,b) := \sum\limits_{j=a}^b \left(\begin{array}{c} j+q_1\\ p_1 \end{array} \right) \left(\begin{array}{c} j+q_2\\ p_2 \end{array} \right)$$ Then we have: \begin{eqnarray} (1) {\mathcal S}_{p_1,p_2}^{q_1,q_2}(a,b) = \prod\limits_{\xi=1}^2 \left(\begin{array}{c} a+q_\xi\\ p_\xi \end{array} \right) \cdot F_{3,2}\left[ \begin{array}{ccc} 1 & 1+a+q_1 & 1+a +q_2 \\ & 1+a-p_1+q_1 & 1+a-p_2+q_2 \end{array};1 \right] - \\ \prod\limits_{\xi=1}^2 \left(\begin{array}{c} 1+b+q_\xi\\ p_\xi \end{array} \right) \cdot F_{3,2}\left[ \begin{array}{ccc} 1 & 2+b+q_1 & 2+b +q_2 \\ & 2+b-p_1+q_1 & 2+b-p_2+q_2 \end{array};1 \right] \end{eqnarray} If both $p_1$ and $p_2$ are positive integers the above equation reduces to: $$(2) {\mathcal S}_{p_1,p_2}^{q_1,q_2}(a,b) = \sum\limits_{l=0}^{p_2} (-1)^{l+1} \left[ \left(\begin{array}{c} a+q_1 +l \\ p_1+l+1 \end{array} \right) \left(\begin{array}{c} a+q_2 \\ p_2-l \end{array} \right) - \left(\begin{array}{c} 1+b+q_1 +l \\ p_1+l+1 \end{array} \right) \left(\begin{array}{c} 1+b+q_2 \\ p_2-l \end{array} \right) \right]$$
We prove equation (1). \begin{eqnarray} (3){\mathcal S}_{p_1,p_2}^{q_1,q_2}(a,b) &:=& \left.\frac{1}{p_1!p_2!} \frac{d^{p_1}}{d x_1^{p_1}} \frac{d^{p_2}}{d x_1^{p_2}} \left(\sum\limits_{j=a}^b x_1^{q_1+j} x_2^{q_2+j}\right)\right|_{x_1=1,x_2=1}\\ &=& \left.\frac{1}{p_1!p_2!} \frac{d^{p_1}}{d x_1^{p_1}} \frac{d^{p_2}}{d x_1^{p_2}} x_1^{q_1} x_2^{q_2} \frac{(x_1 x_2)^a - (x_1 x_2)^{b+1}}{1 - x_1 x_2}\right|_{x_1=1,x_2=1} \\ &=& \left.\frac{1}{p_1!p_2!} \frac{d^{p_1}}{d x_1^{p_1}} \frac{d^{p_2}}{d x_1^{p_2}} \left( \sum\limits_{n=0}^\infty x_1^{a+q_1+n} x_2^{a+q_2+n} - x_1^{1+b+q_1+n} x_2^{1+b+q_2+n} \right)\right|_{x_1=1,x_2=1}\\ &=& \left.\left( \sum\limits_{n=0}^\infty \frac{(a+q_1+n)_{(p_1)}}{p_1!}\frac{(a+q_2+n)_{(p_2)}}{p_2!} - \frac{(1+b+q_1+n)_{(p_1)}}{p_1!}\frac{(1+b+q_2+n)_{(p_2)}}{p_2!} \right)\right|_{x_1=1,x_2=1} \end{eqnarray} Now we use the identity $$(n+a)_{(p_1)} = \frac{(1+a)^{(n)}}{(1+a-p_1)^{(n)}} a_{(p_1)}$$ and arrive at the right-hand-side of equation (1) in a straightforward way.
Now, we prove equation (2). Let us start from the second equality in (3). We have: \begin{eqnarray} {\mathcal S}_{p_1,p_2}^{q_1,q_2}(a,b) =\\ \left. \frac{(-1)^{p_1+p_2}}{p_1!p_2!} \frac{d^{p_1}}{d x_1^{p_1}} \frac{d^{p_2}}{d x_1^{p_2}} \frac{(1-x_1)^{a+q_1}(1-x_2)^{a+q_2} - (1-x_1)^{b+1+q_1}(1-x_2)^{b+1+q_2}}{1 - (1-x_1)(1-x_2)} \right|_{x_1=0,x_2=0} \end{eqnarray} Now, we will compute the derivatives by using the chain rule. Note, that it is only here that we assume that both $p_1$ and $p_2$ are integers. The derivative over $x_2$ at $x_2=0$ reads: $$(-1)^{p_2}\sum\limits_{l_2=0}^{p_2} \left(\begin{array}{c} p_2 \\ l_2 \end{array} \right)l_2! \frac{\left[ (1 - x_1)^{a+q_1+l_2} (a+q_2)_{(p_2-l_2)} - (1-x_1)^{b+1+q_1+l_2} (b+1+q_2)_{(p_2-l_2)} \right]}{x_1^{l_2+1}}$$ Now we expand the result in a Taylor series in $x_1$. We have $$(-1)^{p_2}\sum\limits_{l_2=0}^{p_2} \left(\begin{array}{c} p_2 \\ l_2 \end{array} \right)l_2! \sum\limits_{l_1=0} (-1)^{l_1} \left( \frac{(a+q_1+l_2)_{(l_1)}}{l_1!} (a+q_2)_{(p_2-l_2)} - \frac{(b+1+q_1+l_2)_{(l_1)}}{l_1!} (b+1+q_2)_{(p_2-l_2)} \right) x_1^{l_1-l_2-1}$$ Obviously, the only term that contributes to the result is the one where $l_1-l_2-1=p_1$. Extracting that term out of the sum over $l_1$ and then dividing the whole result by $(-1)^{p_1+p_2}/(p_1! p_2!)$ produces equation (2).