Weiの3F2和公式

まず,
$$\begin{array}{rl}& \sum _{0\le n}\frac{\mathrm{\Gamma }(a+n)\mathrm{\Gamma }(b+n)\mathrm{\Gamma }(a+b+c+d+n-1)}{\mathrm{\Gamma }(n+1)\mathrm{\Gamma }(a+b+c+n)\mathrm{\Gamma }(a+b+d+n)}\\ & =\frac{\mathrm{\Gamma }(a)\mathrm{\Gamma }(b)\mathrm{\Gamma }(a+b+c+d-1)}{\mathrm{\Gamma }(a+b+c)\mathrm{\Gamma }(a+b+d)}{}_3{F}_2[\begin{array}{c}a,b,a+b+c+d-1\\ a+b+c,a+b+d\end{array};1]\end{array}$$
であり, Thomaeの変換公式
$$\begin{array}{rl}{}_3{F}_2[\begin{array}{c}a,b,c\\ d,e\end{array};1]& =\frac{\mathrm{\Gamma }(d)\mathrm{\Gamma }(e)\mathrm{\Gamma }(d+e-a-b-c)}{\mathrm{\Gamma }(a)\mathrm{\Gamma }(d+e-a-b)\mathrm{\Gamma }(d+e-a-c)}{}_3{F}_2[\begin{array}{c}d-a,e-a,d+e-a-b-c\\ d+e-a-b,d+e-a-c\end{array};1]\end{array}$$
において, $a\mapsto a+b+c+d-1,b\mapsto a,c\mapsto b,d\mapsto a+b+c,e\mapsto a+b+d$として,
$$\begin{array}{rl}& \frac{\mathrm{\Gamma }(a)\mathrm{\Gamma }(b)\mathrm{\Gamma }(a+b+c+d-1)}{\mathrm{\Gamma }(a+b+c)\mathrm{\Gamma }(a+b+d)}{}_3{F}_2[\begin{array}{c}a,b,a+b+c+d-1\\ a+b+c,a+b+d\end{array};1]\\ & =\frac{\mathrm{\Gamma }(a)\mathrm{\Gamma }(b)\mathrm{\Gamma }(a+b+c+d-1)}{\mathrm{\Gamma }(a+b+c)\mathrm{\Gamma }(a+b+d)}\frac{\mathrm{\Gamma }(a+b+c)\mathrm{\Gamma }(a+b+d)}{\mathrm{\Gamma }(a+b+c+d-1)\mathrm{\Gamma }(a+1)\mathrm{\Gamma }(b+1)}\\ & \cdot {}_3{F}_2[\begin{array}{c}1-c,1-d,1\\ 1+a,1+b\end{array};1]\\ & =\frac{1}{ab}\sum _{0\le n}\frac{(1-c,1-d{)}_n}{(1+a,1+b{)}_n}\end{array}$$
である. よって,
$$\begin{array}{rl}& \sum _{0\le n}\frac{\mathrm{\Gamma }(a+n)\mathrm{\Gamma }(b+n)\mathrm{\Gamma }(a+b+c+d+n-1)}{\mathrm{\Gamma }(n+1)\mathrm{\Gamma }(a+b+c+n)\mathrm{\Gamma }(a+b+d+n)}+\sum _{0\le n}\frac{\mathrm{\Gamma }(c+n)\mathrm{\Gamma }(d+n)\mathrm{\Gamma }(a+b+c+d+n-1)}{\mathrm{\Gamma }(n+1)\mathrm{\Gamma }(c+d+a+n)\mathrm{\Gamma }(c+d+b+n)}\\ & =\frac{1}{ab}\sum _{0\le n}\frac{(1-c,1-d{)}_n}{(1+a,1+b{)}_n}+\frac{1}{cd}\sum _{0\le n}\frac{(1-a,1-b{)}_n}{(1+c,1+d{)}_n}\\ & =\sum _{n\in \mathbb{Z}}\frac{(1-c,1-d{)}_n}{(1+a,1+b{)}_n}\end{array}$$
ここで, Dougallの2H2和公式より
$$\begin{array}{rl}\frac{1}{ab}\sum _{n\in \mathbb{Z}}\frac{(1-c,1-d{)}_n}{(1+a,1+b{)}_n}& =\frac{1}{ab}\frac{\mathrm{\Gamma }(1+a)\mathrm{\Gamma }(1+b)\mathrm{\Gamma }(c)\mathrm{\Gamma }(d)\mathrm{\Gamma }(a+b+c+d-1)}{\mathrm{\Gamma }(a+c)\mathrm{\Gamma }(a+d)\mathrm{\Gamma }(b+c)\mathrm{\Gamma }(b+d)}\\ & =\frac{\mathrm{\Gamma }(a)\mathrm{\Gamma }(b)\mathrm{\Gamma }(c)\mathrm{\Gamma }(d)\mathrm{\Gamma }(a+b+c+d-1)}{\mathrm{\Gamma }(a+c)\mathrm{\Gamma }(a+d)\mathrm{\Gamma }(b+c)\mathrm{\Gamma }(b+d)}\end{array}$$
だから, 定理が得られる.