分母に二項係数の3乗が付いている級数

1211

はじめに

どうも, 色数です.
今回は最近導出した面白い級数を紹介します.

内容

$ζ (2) = \sum_{n = 1}^{\infty} \frac{2^{4 n}}{n^{2} {(\binom{2 n}{n})}^{3}} - \frac{1}{3} \sum_{n = 1}^{\infty} \frac{2^{4 n}}{n^{3} {(\binom{2 n}{n})}^{3}}$

~~weight が一致していなくて不思議です.~~
余余余さんによると, 右辺をまとめることで分子が $(3 n - 1)$ となるため整合性はあるらしいです.

証明

$f_{k} (n) : = \sum_{0 < m} \frac{(a)_{m}^{2}}{(a + m - 1)^{k} (a)_{n + m}^{2}}$ とおく.
$\begin{aligned} \frac{1}{(a)_{n}^{2}} & = \sum_{0 < m} (\frac{(a)_{m - 1}^{2}}{(a)_{n + m - 1}^{2}} - \frac{(a)_{m}^{2}}{(a)_{n + m}^{2}}) \\ = \sum_{0 < m} \frac{(a)_{m}^{2}}{(a)_{n + m}^{2}} (\frac{(a + n + m - 1)^{2}}{(a + m - 1)^{2}} - 1) \\ = \sum_{0 < m} \frac{(a)_{m}^{2}}{(a)_{n + m}^{2}} (\frac{2 a n + 2 m n + n^{2} - 2 n}{(a + m - 1)^{2}}) \\ = \sum_{0 < m} \frac{(a)_{m}^{2}}{(a)_{n + m}^{2}} (\frac{2 n (a + m - 1) + n^{2}}{(a + m - 1)^{2}}) \\ = 2 n f_{1} (n) + n^{2} f_{2} (n) \end{aligned}$
$\begin{aligned} \frac{a}{(a)_{n}^{2}} & = \sum_{0 < m} (\frac{(a)_{m - 1} (a)_{m}}{(a)_{n + m - 1}^{2}} - \frac{(a)_{m} (a)_{m + 1}}{(a)_{n + m}}) \\ = \sum_{0 < m} \frac{(a)_{m} (a)_{m + 1}}{(a)_{n + m}^{2}} (\frac{(a + n + m - 1)^{2}}{(a + m - 1) (a + m)} - 1) \\ = \sum_{0 < m} \frac{(a)_{m}}{(a)_{n + m}^{2}} (\frac{(2 n - 1) (a + m - 1) + n^{2}}{(a + m - 1)}) \\ = (2 n - 1) f_{0} (n) + n^{2} f_{1} (n) \end{aligned}$
$\begin{aligned} f_{0} (n) & = \sum_{0 < m} \frac{(a)_{m}^{2}}{(a)_{n + m}^{2}} \\ = \sum_{0 \leq m} \frac{(a)_{m}^{2}}{(a)_{n + m}^{2}} - \frac{1}{(a)_{n}^{2}} \\ = \sum_{0 < m} \frac{(a)_{m}^{2}}{(a + m - 1)^{2} (a)_{n + m - 1}^{2}} - \frac{1}{(a)_{n}^{2}} \\ = f_{2} (n - 1) - \frac{1}{(a)_{n}^{2}} \end{aligned}$
まとめると,
$\begin{aligned} 2 n f_{1} (n) + n^{2} f_{2} (n) & = \frac{1}{(a)_{n}^{2}} \\ (2 n - 1) f_{0} (n) + n^{2} f_{1} (n) & = \frac{a}{(a)_{n}^{2}} \\ f_{2} (n - 1) - \frac{1}{(a)_{n}^{2}} & = f_{0} (n) \end{aligned}$
となる.
これにより,
$\begin{aligned} f_{1} (n) & = \frac{a}{n^{2} (a)_{n}^{2}} - \frac{2 n - 1}{n^{2}} f_{0} (n) \\ = \frac{a}{n^{2} (a)_{n}^{2}} - \frac{2 n - 1}{n^{2}} (f_{2} (n - 1) - \frac{1}{(a)_{n}^{2}}) \\ = \frac{a + 2 n - 1}{n^{2} (a)_{n}^{2}} - \frac{2 n - 1}{n^{2}} f_{2} (n - 1) \end{aligned}$
$\begin{aligned} f_{1} (n) & = \frac{1}{2 n (a)_{n}^{2}} - \frac{n}{2} f_{2} (n) \\ f_{2} (n) - \frac{2 (2 n - 1)}{n^{3}} f_{2} (n - 1) & = \frac{1}{n^{2} (a)_{n}^{2}} - \frac{2 a + 4 n - 2}{n^{3} (a)_{n}^{2}} \end{aligned}$
両辺に $n!^{4} / (2 n)!$ をかけて和を取り, $a = 1 / 2$ とすれば上記を得る.