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$$
前の記事(
ガンマ関数で表せる二項係数の4乗が付いた級数
)で
\begin{align}
\sum_{n=0}^{\infty}\frac{\binom{2n}n^4}{2^{8n}(4n+1)}=\frac{\pi^3}{6\Gamma\left(\frac 34\right)^8}
\end{align}
を示した. 今回はこの等式を, 古典的な超幾何級数の変換公式と和公式のみを用いて示したいと思う.
導出
まず, 元の和は超幾何級数により,
\begin{align}
\sum_{n=0}^{\infty}\frac{\binom{2n}n^4}{2^{8n}(4n+1)}&=\F54{\frac 12,\frac 12,\frac 12,\frac 12,\frac 14}{1,1,1,\frac 54}1\\
&=\F76{\frac 12,\frac 54,\frac 12,\frac 12,\frac 12,\frac 14,\frac 14}{\frac 14,1,1,1,\frac 54,\frac 54}1\\
\end{align}
と表される. これはvery-well-poised${}_7F_6$である. ${}_7F_6$の二項変換公式(
前の記事
の定理4)において, $a=\frac 12,b=c=d=\frac 12,e=f=\frac 14$とすると, $v=1$となり,
\begin{align}
&\F76{\frac 12,\frac 54,\frac 12,\frac 12,\frac 12,\frac 14,\frac 14}{\frac 14,1,1,1,\frac 54,\frac 54}1\\
&=\frac 1{\Gamma\left(\frac 12\right)\Gamma\left(\frac 32\right)^3}\F76{1,\frac 32,1,\frac 12,\frac 12,\frac 34,\frac 34}{\frac 12,1,\frac 32,\frac 32,\frac 54,\frac 54}1\\
&=\frac{8}{\pi^2}\F43{1,\frac 12,\frac 34,\frac 34}{\frac 32,\frac 54,\frac 54}1
\end{align}
となる. ここで,
non-terminating Whippleの変換公式
において, $a=d=1,b=e=\frac 12,c=f=\frac 34$とすると,
\begin{align}
&\F43{1,\frac 12,\frac 34,\frac 34}{\frac 32,\frac 54,\frac 54}1\\
&=\frac{\Gamma\left(\frac 32\right)\Gamma\left(\frac 54\right)\Gamma\left(-\frac 14\right)}{\Gamma\left(\frac 12\right)\Gamma\left(\frac14\right)\Gamma\left(\frac 34\right)}\F43{1,\frac 12,\frac 34,\frac 34}{\frac 32,\frac 54,\frac 54}1\\
&\qquad+\frac{\Gamma\left(\frac 32\right)^2\Gamma\left(\frac 54\right)^2\Gamma\left(\frac 14\right)\Gamma\left(\frac 12\right)}{\Gamma\left(\frac 12\right)\Gamma\left(\frac 34\right)^2\Gamma\left(\frac 54\right)}\F32{\frac 12,\frac 14,\frac 12}{\frac 54,1}1\\
&=-\frac 12\F43{1,\frac 12,\frac 34,\frac 34}{\frac 32,\frac 54,\frac 54}1+\frac{\pi^3}{8\Gamma\left(\frac 34\right)^4}\F32{\frac 12,\frac 14,\frac 12}{\frac 54,1}1
\end{align}
となる. よって,
\begin{align}
\F43{1,\frac 12,\frac 34,\frac 34}{\frac 32,\frac 54,\frac 54}1&=\frac{\pi^3}{12\Gamma\left(\frac 34\right)^4}\F32{\frac 12,\frac 14,\frac 12}{\frac 54,1}{1}
\end{align}
を得る. ここで,
Dixonの和公式
より
\begin{align}
\F32{\frac 12,\frac 14,\frac 12}{\frac 54,1}{1}&=\frac{\Gamma\left(\frac 54\right)^2\Gamma\left(\frac 12\right)}{\Gamma\left(\frac 32\right)\Gamma\left(\frac 34\right)^2}\\
&=\frac{\pi^2}{4\Gamma\left(\frac 34\right)^4}
\end{align}
であるから, これを代入して
\begin{align}
\F43{1,\frac 12,\frac 34,\frac 34}{\frac 32,\frac 54,\frac 54}1&=\frac{\pi^5}{48\Gamma\left(\frac 34\right)^8}
\end{align}
を得る. よって,
\begin{align}
\sum_{n=0}^{\infty}\frac{\binom{2n}n^4}{2^{8n}(4n+1)}
&=\frac{8}{\pi^2}\frac{\pi^5}{48\Gamma\left(\frac 34\right)^8}=\frac{\pi^3}{6\Gamma\left(\frac 34\right)^8}
\end{align}
となって示すべき等式が得られる. 途中の等式
\begin{align}
\F43{1,\frac 12,\frac 34,\frac 34}{\frac 32,\frac 54,\frac 54}1&=\frac{\pi^5}{48\Gamma\left(\frac 34\right)^8}
\end{align}
の証明は
pisco氏の解答
により与えられたものである.
