二項係数付き級数

\begin{align} \sum_{n>0}\frac{\binom{2n}{n}^3}{2^{6n}n}&=\sum_{n>0}\frac{\left(\frac12\right)^3_n}{(1)_n^3n}\\ &=\sum_{n\geq 0}\frac{\left(\frac12\right)^3_{n+1}}{(1)_{n+1}^3(n+1)}\\ &=\frac18\sum_{n\geq 0}\frac{\left(\frac32\right)^3_n (1)_n^2}{(2)_n^3(2)_n n!}\\ &=\frac18{}_5F_4\left[ \begin{array}{c} \frac32,\frac32,\frac32,1,1\\ 2,2,2,2 \end{array} ;1\right]\\ \end{align}

は？。