Op.3
はじめに
本記事では次を示す.
$$\sum_{n=0}^{\infty}\dfrac{(3n)!(2n)!}{(n!)^53^{3n}2^{2n}}=\dfrac{9\Gamma\bigl(\frac{1}{3}\bigl)^{6}}{16\sqrt[3]{4}\pi^4}$$
準備
$$_3F_2\!\left(\begin{matrix} 2a,2b,a+b \\ 2a+2b,a+b+\frac{1}{2} \end{matrix};\, z\right)=\Biggl[{}_2F_1\!\left(\begin{matrix} a,b \\ a+b+\frac{1}{2} \end{matrix};\, z\right)\Biggl]^2$$
後々単体であげます.今は略.
$${}_2F_1\!\left(\begin{matrix}
a,b \\
c
\end{matrix};\, 1\right)=\dfrac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}$$
($(\mathrm{Re}(c-a-b)>0)$ を満たす.)
後々単体であげます.今は略
$$\Gamma\left(\frac{2}{3}\right)=\dfrac{2\pi}{\sqrt{3}\Gamma\bigl(\frac{1}{3}\bigl)},\quad \Gamma\left(\frac{5}{6}\right)=\dfrac{4\pi\sqrt{\pi}}{\sqrt{3}\sqrt[3]{4}\Gamma\bigl(\frac{1}{3}\bigl)^2}$$
左から.
$\Gamma(z)\Gamma(1-z)=\dfrac{\pi}{\sin{\pi z}}$より,$z=\frac{2}{3}$とすると,
$$\Gamma\left(\frac{1}{3}\right)\Gamma\left(\frac{2}{3}\right)=\dfrac{2\pi}{\sqrt{3}} $$
より,得る.
右は
$\Gamma(z)\Gamma\left(z+\dfrac{1}{2}\right)=2^{1-2z}\sqrt{\pi}\,\Gamma(2z)$より,$z=\frac{1}{3}$とすれば,
$$\Gamma\left(\frac{5}{6}\right)=2^{\frac{1}{3}}\sqrt{\pi}\,\dfrac{\Gamma\left(\frac{2}{3}\right)}{\Gamma\left(\frac{1}{3}\right)} $$
先程のものを代入し,整理すれば得る.
本題
$$\sum_{n=0}^{\infty}\dfrac{(3n)!(2n)!}{(n!)^53^{3n}2^{2n}}$$
$(3n)!=3^{3n}\left(\dfrac{1}{3}\right)_n\left(\dfrac{2}{3}\right)_n n!$
$(2n)!=2^{2n}\left(\dfrac{1}{2}\right)_n n!$
これらを元の級数に代入すると,
$$\sum_{n=0}^{\infty}\dfrac{(3n)!(2n)!}{(n!)^53^{3n}2^{2n}}=\sum_{n=0}^{\infty}\dfrac{(\frac{1}{3})_n(\frac{2}{3})_n(\frac{1}{2})_n}{(1)_n(1)_n n!}={}_3F_2\!\left(\begin{matrix}
\frac{1}{3},\frac{2}{3},\frac{1}{2} \\
1,1
\end{matrix};\, 1\right) $$
補題$1$より,
$${}_3F_2\!\left(\begin{matrix}
\frac{1}{3},\frac{2}{3},\frac{1}{2} \\
1,1
\end{matrix};\, 1\right) =\Biggl[{}_2F_1\!\left(\begin{matrix}
\frac{1}{6},\frac{1}{3} \\
1
\end{matrix};\, 1\right)\Biggl]^2$$
そして補題$2$より,
$$_2F_1\!\left(\begin{matrix}
\frac{1}{6},\frac{1}{3} \\
1
\end{matrix};\, 1\right)=\dfrac{\Gamma(1)\Gamma(1-\frac{1}{6}-\frac{1}{3})}{\Gamma(1-\frac{1}{6})\Gamma(1-\frac{1}{3})}$$
($(\mathrm{Re}(\frac{1}{2})>0)$ を満たす.)
$$_2F_1\!\left(\begin{matrix}
\frac{1}{6},\frac{1}{3} \\
1
\end{matrix};\, 1\right)=\dfrac{\Gamma(\frac{1}{2})}{\{\Gamma(\frac{5}{6})\Gamma(\frac{2}{3})\}}=\dfrac{\sqrt{\pi}}{\{\Gamma(\frac{5}{6})\Gamma(\frac{2}{3})\}}$$
$$\Biggl[{}_2F_1\!\left(\begin{matrix}
\frac{1}{6},\frac{1}{3} \\
1
\end{matrix};\, 1\right)\Biggl]^2=\dfrac{\pi}{\{\Gamma(\frac{5}{6})\Gamma(\frac{2}{3})\}^2}$$
補題$3$より, $\Gamma(\frac{2}{3})=\dfrac{2\pi}{\sqrt{3}\Gamma(\frac{1}{3})},\Gamma(\frac{5}{6})=\dfrac{4\pi\sqrt{\pi}}{\sqrt{3}\sqrt[3]{4}\Gamma(\frac{1}{3})^2}$を用いて,
$$\sum_{n=0}^{\infty}\dfrac{(3n)!(2n)!}{(n!)^53^{3n}2^{2n}}=\pi\cdot \dfrac{3\Gamma(\frac{1}{3})^2}{4\pi^2}\cdot \dfrac{3\cdot \sqrt[3]{4^2}\Gamma(\frac{1}{3})^4}{16\pi^3}
$$
整理すれば,
$$\sum_{n=0}^{\infty}\dfrac{(3n)!(2n)!}{(n!)^53^{3n}2^{2n}}=\dfrac{9\Gamma\bigl(\frac{1}{3}\bigl)^{6}}{16\sqrt[3]{4}\pi^4}$$
を得る.
最後に
めちゃお気に入り級数です.
ちなみに$\Gamma\left(\frac{1}{3}\right)$って定数、かなり好きです.
