(2/3)_nが入った級数について

まず,
$$\begin{array}{rl}\sum _{0\le n}\frac{{\left(\frac{2}{3}\right)}_n^2}{n{!}^2{(n+\frac{1}{3})}^2}& =\frac{1}{\mathrm{\Gamma }\left(\frac{1}{3}\right)\mathrm{\Gamma }\left(\frac{2}{3}\right)}{\int }_0^1\sum _{0\le n}\frac{{\left(\frac{2}{3}\right)}_n}{n!{(n+\frac{1}{3})}^2}{u}^{n+\frac{1}{3}}\cdot \frac{du}{(u(1-u){)}^{\frac{2}{3}}}\\ & =\frac{1}{\mathrm{\Gamma }\left(\frac{1}{3}\right)\mathrm{\Gamma }\left(\frac{2}{3}\right)}{\int }_{0<s<t<u<1}\frac{ds}{(s(1-s){)}^{\frac{2}{3}}}\frac{dt}{t}\frac{du}{(u(1-u){)}^{\frac{2}{3}}}\\ & =\frac{1}{\mathrm{\Gamma }\left(\frac{1}{3}\right)\mathrm{\Gamma }\left(\frac{2}{3}\right)}{\int }_{0<s<t<u<1}\frac{ds}{(s(1-s){)}^{\frac{2}{3}}}\frac{dt}{1-t}\frac{du}{(u(1-u){)}^{\frac{2}{3}}}(s,t,u)\mapsto (1-u,1-t,1-s)\\ & =\frac{1}{\mathrm{\Gamma }\left(\frac{1}{3}\right)\mathrm{\Gamma }\left(\frac{2}{3}\right)}{\int }_0^1\sum _{0\le n<m}\frac{{\left(\frac{2}{3}\right)}_n}{n!(n+\frac{1}{3})}\frac{u^{m+\frac{1}{3}}}{m+\frac{1}{3}}\frac{du}{(u(1-u){)}^{\frac{2}{3}}}\\ & =\sum _{0\le n<m}\frac{{\left(\frac{2}{3}\right)}_n}{n!(n+\frac{1}{3})}\frac{{\left(\frac{2}{3}\right)}_m}{m!(m+\frac{1}{3})}\\ & =\frac{1}{2}({\left(\sum _{0\le n}\frac{{\left(\frac{2}{3}\right)}_n}{n!(n+\frac{1}{3})}\right)}^2-\sum _{0\le n}\frac{{\left(\frac{2}{3}\right)}_n^2}{n{!}^2{(n+\frac{1}{3})}^2})\end{array}$$
である. よって,
$$\begin{array}{rl}\sum _{0\le n}\frac{{\left(\frac{2}{3}\right)}_n^2}{n{!}^2{(n+\frac{1}{3})}^2}& =\frac{1}{3}{\left(\sum _{0\le n}\frac{{\left(\frac{2}{3}\right)}_n}{n!(n+\frac{1}{3})}\right)}^2\end{array}$$
が得られる. ここで,
$$\begin{array}{rl}\sum _{0\le n}\frac{{\left(\frac{2}{3}\right)}_n}{n!(n+\frac{1}{3})}& ={\int }_0^1\frac{dt}{(t(1-t){)}^{\frac{2}{3}}}\\ & =\frac{\mathrm{\Gamma }{\left(\frac{1}{3}\right)}^2}{\mathrm{\Gamma }\left(\frac{2}{3}\right)}\end{array}$$
であるから,
$$\begin{array}{rl}\sum _{0\le n}\frac{{\left(\frac{2}{3}\right)}_n^2}{n{!}^2{(3n+1)}^2}& =\frac{\mathrm{\Gamma }{\left(\frac{1}{3}\right)}^4}{27\mathrm{\Gamma }{\left(\frac{2}{3}\right)}^2}\end{array}$$
を得る.