分母に二項係数の3乗が付いている級数

$\displaystyle f_k(n)\coloneqq\sum_{0< m}\frac{(a)_m^2}{(a+m-1)^k(a)_{n+m}^2}$とおく.
\begin{align} \frac{1}{(a)_{n}^2}&=\sum_{0< m}\left(\frac{(a)_{m-1}^2}{(a)_{n+m-1}^2}-\frac{(a)_{m}^2}{(a)_{n+m}^2}\right)\\ &=\sum_{0< m}\frac{(a)_{m}^2}{(a)_{n+m}^2}\left(\frac{(a+n+m-1)^2}{(a+m-1)^2}-1\right)\\ &=\sum_{0< m}\frac{(a)_{m}^2}{(a)_{n+m}^2}\left(\frac{2an+2mn+n^2-2n}{(a+m-1)^2}\right)\\ &=\sum_{0< m}\frac{(a)_{m}^2}{(a)_{n+m}^2}\left(\frac{2n(a+m-1)+n^2}{(a+m-1)^2}\right)\\ &=2nf_1(n)+n^2f_2(n) \end{align}
\begin{align} \frac{a}{(a)_{n}^2}&=\sum_{0< m}\left(\frac{(a)_{m-1}(a)_m}{(a)_{n+m-1}^2}-\frac{(a)_m(a)_{m+1}}{(a)_{n+m}}\right)\\ &=\sum_{0< m}\frac{(a)_m(a)_{m+1}}{(a)_{n+m}^2}\left(\frac{(a+n+m-1)^2}{(a+m-1)(a+m)}-1\right)\\ &=\sum_{0< m}\frac{(a)_m}{(a)_{n+m}^2}\left(\frac{(2n-1)(a+m-1)+n^2}{(a+m-1)}\right)\\ &=(2n-1)f_0(n)+n^2f_1(n) \end{align}
\begin{align} f_0(n)&=\sum_{0< m}\frac{(a)_m^2}{(a)_{n+m}^2}\\ &=\sum_{0\le m}\frac{(a)_m^2}{(a)_{n+m}^2}-\frac{1}{(a)_n^2}\\ &=\sum_{0< m}\frac{(a)_m^2}{(a+m-1)^2(a)_{n+m-1}^2}-\frac{1}{(a)_n^2}\\ &=f_2(n-1)-\frac{1}{(a)_n^2} \end{align}
まとめると,
\begin{align} 2nf_1(n)+n^2f_2(n)&=\frac{1}{(a)_n^2}\\ (2n-1)f_0(n)+n^2f_1(n)&=\frac{a}{(a)_n^2}\\ f_2(n-1)-\frac{1}{(a)_n^2}&=f_0(n) \end{align}
となる.
これにより,
\begin{align} f_1(n)&=\frac{a}{n^2(a)_n^2}-\frac{2n-1}{n^2}f_0(n)\\ &=\frac{a}{n^2(a)_n^2}-\frac{2n-1}{n^2}\left(f_2(n-1)-\frac{1}{(a)_n^2}\right)\\ &=\frac{a+2n-1}{n^2(a)_n^2}-\frac{2n-1}{n^2}f_2(n-1) \end{align}
\begin{align} f_1(n)&=\frac{1}{2n(a)_n^2}-\frac{n}{2}f_2(n)\\ f_2(n)-\frac{2(2n-1)}{n^3}f_2(n-1)&=\frac{1}{n^2(a)_n^2}-\frac{2a+4n-2}{n^3(a)_n^2} \end{align}
両辺に$n!^4/(2n)!$をかけて和を取り, $a=1/2$ とすれば上記を得る.