Proving the Hypergeometric Sequence

Question: How do you prove$$_4F_3\left[\begin{array}{c c}\frac 12n+1,n,-x,-y\\\frac 12n,x+n+1,y+n+1\end{array};-1\right]=\dfrac {\Gamma(x+n+1)\Gamma(y+n+1)}{\Gamma(n+1)\Gamma(x+y+n+1)}\tag{1}$$For $\Re(2x+2y+n+2)>0$

I'm not sure how to prove this. There are a multitude of other similar formulas. Some of which\begin{align*}_4F_3\left[\begin{array}{c c}\frac 12n+1,n,n,-x\\\frac 12n,x+n+1,1\end{array};-1\right] & =\dfrac {\Gamma(x+n+1)}{\Gamma(n+1)\Gamma(x+1)}\end{align*}\tag{2}For $\Re(2x-n+2)>0$$_3F_2\left[\begin{array}{c c}\frac 12n+1,n,-x\\\frac 12n,x+n+1\end{array};-1\right]=\dfrac {\Gamma(x+n+1)\Gamma\left(\frac 12n+\frac 12\right)}{\Gamma(n+1)\Gamma\left(x+\frac 12n+\frac 12\right)}\tag3$$ For$\Re(x)>-\frac 12$. I know that the general Hypergeometric Sequence can be written as$$_pF_q\left[\begin{array}{c c}\alpha_1,\alpha_2,\ldots,\alpha_p\\\beta_1,\beta_2,\ldots,\beta_p\end{array};x\right]=\sum\limits_{k=0}^\infty\dfrac {(\alpha_1)_k(\alpha_2)_k\cdots(\alpha_p)_k}{(\beta_1)_k(\beta_2)_k\cdots(\beta_p)_k}\dfrac {x^k}{k!}\tag4$$So$(1)$becomes$$_4F_3\left[\begin{array}{c c}\frac 12n+1,n,-x,-y\\\frac 12n,x+n+1,y+n+1\end{array};-1\right]=\sum\limits_{k=0}^{\infty}\dfrac {\left(\frac 12n+1\right)_k(n)_k(-x)_k(-y)_k}{\left(\frac 12n\right)_k(x+n+1)_k(y+n+1)_k}\dfrac {(-1)^k}{k!}\tag{5}$$However, how do you manipulate the RHS of$(5)$into the RHS of$(1)$. I think that there could be a sort of elementary transformation involved, but if so, I'm not sure what. • Is$n$supposed to be a positive integer? – David H Feb 28 '17 at 15:31 • @DavidH I don't think$n$has to be a positive integer. As long as the requirement that$\text{Re}(2x+2y+n+2)>0$is true, then$(1)$is true. – Crescendo Mar 2 '17 at 1:34 1 Answer An established identity, useful to transform a$_4F_3 $hypergeometric sequence with negative unit argument in a$_3F_2$sequence, is $$_4F_3 \left[\begin{array}{c c} a, b, c, d \\ a - b +1, a - c+1, a - d+1 \end{array};-1\right] \\ = \dfrac {\Gamma[a - b+1] \Gamma[a - c+1]}{\Gamma[a+1] \Gamma[a - b - c+1]} \, _3F_2 \left[\begin{array}{c c} a/2 - d+1, b, c \\ a/2 + 1, a - d+1 \end{array};-1\right]$$ So, rewriting your series as $$_4F_3 \left[\begin{array}{c c} n, -x, -y, \frac 12n+1\\ n+x+1, n+y+1, \frac 12n \end{array};-1\right] \\$$ we obtain that it is equal to $$\dfrac {\Gamma(n+x+1)\Gamma(n+y+1)}{\Gamma(n+1)\Gamma(n+x+y+1)} \\ _3F_2 \left[\begin{array}{c c} 0, -x, -y \\ n/2 + 1, n/2 \end{array};-1\right]$$ and since the$_3F_2$sequence, as a result of the zero among its terms, is equal to$1, we get your identity. The same method can be used to generate the second identity. Rewriting it as \begin{align*}_4F_3\left[\begin{array}{c c} n, -x, n,\dfrac 12n+1 \\ n+x+1, 1, \dfrac 12n \end{array};-1\right] & \end{align*} we get $$=\dfrac {\Gamma(n+x+1) \Gamma (1)}{\Gamma(n+1)\Gamma(x+1)} \, _3F_2 \left[\begin{array}{c c} 0, -x, n \\ n/2 + 1, n/2 \end{array};-1\right]$$ where again the_3F_2$sequence is equal to$1$, giving your identity. Lastly, another known identity can be used to express in closed form a$_3F_2$hypergeometric function with negative unit argument: $$_3F_2 \left[\begin{array}{c c} a, \dfrac a2+1,c \\ \dfrac a2,a-c+1 \end{array};-1 \right]=\dfrac {\sqrt {\pi} \,\, \Gamma(a-c+1)}{2^n \, \Gamma \left(\frac a2+1 \right)\Gamma \left(\frac{a+1}{2}-c \right)}$$ This can be used to prove the third identity of the OP. Rewriting it as follows we get $$_3F_2 \left[\begin{array}{c c} n, \frac 12n+1,-x \\ \dfrac 12n,n+x+1 \end{array};-1 \right]=\dfrac {\sqrt {\pi} \,\, \Gamma(n+x+1)}{2^n \Gamma \left(\frac n2+1 \right) \Gamma \left(\frac{n+1}{2}+x \right)}$$ Now since it is known that, for given$z $, $$\dfrac {\Gamma (z)}{\Gamma (2z)}= \dfrac {\sqrt {\pi}}{2^{2z-1}\, \Gamma \left(z+\frac{1}{2} \right)}$$ setting$z=(n+1)/2 \,\,$we have $$\dfrac {\Gamma (\frac 12n+ \frac 12 )}{\Gamma (n+1)}= \dfrac {\sqrt {\pi}}{2^{n} \Gamma \left(\frac {1}{2}n+1 \right)}$$ Substituting in the equation above we obtain $$_3F_2 \left[\begin{array}{c c} n, \frac 12n+1,-x \\ \frac 12n,n+x+1 \end{array};-1 \right]=\dfrac {\Gamma(n+x+1) \Gamma \left(\frac 12n + \frac 12 \right)}{\Gamma(n+1) \Gamma \left(\frac 12n + x+ \frac 12 \right)}$$ • Okay, can this be used to generate the other identities I gave and possible larger similarities such as$_7F_6$? – Crescendo Mar 5 '17 at 18:53 • Hi, I edited my answer to show how the other identities can be shown. Probably a similar method could be used to get larger similarities. – Anatoly Mar 6 '17 at 17:33 • Okay, cool! Thanks! – Crescendo Mar 6 '17 at 18:17 • Just wondering, but is there some transformation (or something) that you can do on the RHS of$(5)$to get$(1)\$? – Crescendo Mar 6 '17 at 18:28
• Probably such a direct transformation, if it exists, is not simple. – Anatoly Mar 7 '17 at 7:49