A multivariate sum that yields a closed form expression

Question

Let $d\ge 2$ be a an integer. Let $b_1,b_2,\cdots,b_d$ be positive integers. As a by product of certain calculations I have discovered that: \begin{equation} \sum\limits_{q_2=0}^{b_2} \cdots \sum\limits_{q_d=0}^{b_d} \frac{(\sum\limits_{j=2}^d q_j)!}{\prod\limits_{j=2}^d q_j!} \cdot \frac{\prod\limits_{j=2}^d (-b_j)^{(q_j)}}{(2+b_1)^{(\sum\limits_{j=2}^d q_j)}} = \frac{1+b_1}{1+\sum\limits_{j=1}^d b_j} \end{equation} Here $b^{(n)} := \prod\limits_{j=0}^{n-1} (b+j)$ is the Pochhammer symbol.

As a sanity check we take $d=2$. Then \begin{equation} lhs = F_{2,1} \left[\{1,-b_2\},\{2+b_1\};1\right] = \frac{\Gamma(2+b_1)\Gamma(1+b_1+b_2)}{\Gamma(1+b_1) \Gamma(2+b_1+b_2)} = \frac{1+b_1}{1+b_1+b_2} \end{equation} where we used the Gauss' theorem.

The question is how do we prove the said identity in the generic case ?

Answer 1

We prove the result in case $d=3$. From the proof it will be readily seen that the generalization to an arbitrary value of $d$ is straightforward. \begin{eqnarray} lhs &=& \sum\limits_{q_2=0}^\infty \sum\limits_{q_3=0}^\infty \binom{q_2+q_3}{q_2} \frac{\Gamma(-b_2+q_2)}{\Gamma(-b_2)} \cdot \frac{\Gamma(-b_3+q_3)}{\Gamma(-b_3)} \cdot \frac{\Gamma(2+b_1)}{\Gamma(2+b_1+q_2+q_3)}\\ &=& \frac{\Gamma(2+b_1)}{\Gamma(-b_2) \Gamma(-b_3) \Gamma(b_1+b_2+b_3+2)} \cdot \sum\limits_{q_2=0}^\infty \sum\limits_{q_3=0}^\infty \binom{q_2+q_3}{q_2} \int\limits_{0\le t_1\le t_2\le 1} t_1^{-b_2+q_2-1} (t_2-t_1)^{-b_3+q_3-1} (1-t_2)^{b_1+b_2+b_3+1} dt_1 d t_2 \\ &=& \frac{\Gamma(2+b_1)}{\Gamma(-b_2) \Gamma(-b_3) \Gamma(b_1+b_2+b_3+2)} \cdot \int\limits_{0\le t_1\le t_2\le 1} t_1^{-b_2-1} (t_2-t_1)^{-b_3-1} (1-t_2)^{b_1+b_2+b_3} dt_1 d t_2 \\ &=& \frac{\Gamma(2+b_1)}{\Gamma(-b_2) \Gamma(-b_3) \Gamma(b_1+b_2+b_3+2)} \cdot \frac{\Gamma(-b_2) \Gamma(-b_3) \Gamma(b_1+b_2+b_3+1)}{\Gamma(b_1+1)} \\ &=& \frac{1+b_1}{1+b_1+b_2+b_3} \end{eqnarray} In the first line we used the identity $a^{(n)} = \Gamma(a+n)/\Gamma(a)$. In the second line we used the integral representation of the generalized Beta function: \begin{equation} B(x,y,z) := \frac{\Gamma(x) \Gamma(y) \Gamma(z)}{\Gamma(x+y+z)} = \int\limits_{0\le t_1\le t_2\le 1} t_1^{x-1} (t_2-t_1)^{y-1} (1-t_2)^{z-1} dt_1 d t_2 \end{equation} In the next line we did the double sum by using the binomial expansion formula and the geometric series. Finally we used the integral representation of the generalized beta function again and then we simplified the result.