級数一覧　𝟏

1017

主にぼくが計算した級数を列挙する。ぼくが自力で導出できなかったものにはその出処を示す。

定義・記法

　　　 $\begin{aligned} ζ (k_{1}, \dots, \overset{―}{k_{s}}, \dots, k_{a}) = \sum_{0 < n_{1} < \dots < n_{a}} \frac{(- 1)^{k_{s}}}{n_{1}^{k_{1}} \dots n_{a}^{k_{a}}} \\ ζ^{⋆} (k_{1}, \dots, k_{a}) = \sum_{0 < n_{1} \leq \dots \leq n_{a}} \frac{1}{n_{1}^{k_{1}} \dots n_{a}^{k_{a}}} \\ ζ_{N} (k_{1}, \dots, k_{a}) = \sum_{0 < n_{1} < \dots < n_{a} \leq N} \frac{1}{n_{1}^{k_{1}} \dots n_{a}^{k_{a}}} \\ ζ_{N}^{⋆} (k_{1}, \dots, k_{a}) = \sum_{0 < n_{1} \leq \dots \leq n_{a} \leq N} \frac{1}{n_{1}^{k_{1}} \dots n_{a}^{k_{a}}} \\ t (k_{1}, \dots, \overset{―}{k_{s}}, \dots, k_{a}) = \sum_{0 \leq n_{1} < \dots < n_{a}} \frac{(- 1)^{k_{s}}}{(2 n_{1} + 1)^{k_{1}} \dots (2 n_{a} + 1)^{k_{a}}} \\ β (s) = \sum_{0 \leq n} \frac{(- 1)^{n}}{(2 n + 1)^{s}} \\ η (s) = (1 - 2^{1 - s}) ζ (s) \\ {L i}_{k_{1}, \dots, k_{a}} (z) = \sum_{0 < n_{1} < \dots < n_{a}} \frac{z^{k_{a}}}{n_{1}^{k_{1}} \dots n_{a}^{k_{a}}} \\ χ_{ν} (z) = \sum_{0 \leq n} \frac{z^{2 n + 1}}{(2 n + 1)^{ν}} \\ ψ (1 + z) = - γ + \sum_{0 < n} \frac{z}{n (n + z)} \\ ψ^{'} (z) = \sum_{0 \leq n} \frac{1}{(n + z)^{2}} \\ (z)_{n} = \frac{Γ (n + z)}{Γ (z)} \\ \underset{n}{\underset{⏟}{k, \dots, k}} = {k}_{n} \\ H_{n} = \sum_{m = 1}^{n} \frac{1}{m} \\ _{p} F_{q} [\begin{array}{c} a_{1}, \dots, a_{p} \\ b_{1}, \dots, b_{q} \end{array}; z] = \sum_{0 \leq n} \frac{(a_{1}, \dots, a_{p})_{n}}{(b_{1}, \dots, b_{q})_{n}} \frac{z^{n}}{n!} \end{aligned}$

あくまでぼくが知っている範囲なので、実は計算できるというのがあれば教えてほしい。
また、間違った結果を書いている危険性があると提言するのも吝かで無い感じも僅かながら存在するので、あれば教えてほしい。
新たな結果を得たら追加していく。

$\sum_{0 < n} \frac{ζ_{n - 1} ({1}_{a - 1}) ζ_{n}^{⋆} ({1}_{b})}{n^{3}} = ζ^{⋆} (b + 1, a + 1) - \sum_{j = 1}^{a} (\binom{j + b}{b}) ζ (a - j + 1, j + b + 1)$ 　　　　(訂正後)

${L i}_{{1}_{a - 1}, b + 1} (1 - x) = \sum_{j = 0}^{b - 1} (- 1)^{j} ζ ({1}_{a - 1}, b - j + 1) {L i}_{{1}_{j}} (x) + (- 1)^{b} \sum_{0 \leq n_{1} \leq \dots \leq n_{b} \leq a} {L i}_{{1}_{n_{1}}} (1 - x) {L i}_{a - n_{b} + 1, n_{b} - n_{b - 1} + 1, \dots, n_{2} - n_{1} + 1} (x)$

$\sum_{0 < n} (\frac{1}{n^{a + 1}} - \sum_{j = 0}^{a} \frac{x^{n}}{n^{a - j + 1}} {L i}_{{1}_{j}} (1 - x)) \sum_{k = 0}^{b} (- 1)^{k} ζ_{n - 1} ({1}_{b - k}) {L i}_{{1}_{k}} (x) = {L i}_{{1}_{a - 1}, b + 2} (1 - x)$ 　　　　(出処：コメント欄)

$\sum_{0 < n_{1} < \dots < n_{a}} \frac{1}{n_{1} \dots n_{a - 1} n_{a}^{3}} \frac{n_{a}!}{(α)_{n_{a}}} = \sum_{0 \leq m < n} \frac{1}{(m + α) n^{a + 1}} - \sum_{j = 1}^{a} \sum_{0 < m \leq n} \frac{1}{m^{a - j + 1} (n + α)^{j + 1}}$

$\sum_{k = 0}^{m - 1} ζ_{n}^{⋆} ({1}_{k}) ζ_{n}^{⋆} (m - k) = m ζ_{n}^{⋆} ({1}_{m})$

　　　 $\begin{array}{r} \sum_{m = 1}^{\infty} \frac{(- 1)^{m - 1}}{m^{n + 1}} + \sum_{j = 1}^{n} \frac{1}{j!} \sum_{m = 1}^{\infty} \frac{(- 1)^{m - 1} ζ_{m - 1} ({1}_{j - 1})}{m^{n - j + 1}} = \sum_{j = 0}^{n} \frac{\ln^{j} 2}{j!} {L i}_{n - j + 1} (\frac{1}{2}) \end{array}$

　　　 $\begin{array}{r} \sum_{0 < m_{1} < \dots < m_{a} \leq n_{1} < \dots < n_{b}} \frac{2^{2 m_{a} + 2 n_{1}}}{(2 m_{1})^{2} \dots (2 m_{a})^{2} (\binom{2 m_{a}}{m_{a}}) (\binom{2 n_{1}}{n_{1}}) (2 n_{1} + 1)^{2} \dots (2 n_{b} + 1)^{2}} = \frac{1}{(2 a - 1)! (2 b)!} \int_{0}^{\frac{π}{2}} \frac{x^{2 a - 1} {(\frac{π}{2} - x)}^{2 b}}{\sin x} d x \end{array}$

　　　 $\begin{array}{r} \sum_{0 < l < m < n} \frac{2^{2 l}}{l^{2} (\binom{2 l}{l})} \frac{(\binom{2 m}{m})}{2^{2 m} m} \frac{2^{2 n}}{n^{2} (\binom{2 n}{n})} = \frac{93}{16} ζ (5) - \frac{21}{8} ζ (2) ζ (3) \end{array}$

$\sum_{0 < n} (- 1)^{n - 1} (- 1 + (2 n)^{2} \ln \frac{(2 n)^{2}}{(2 n)^{2} - 1}) = - \frac{1}{2} + \frac{8 β (2)}{π} - \frac{14 ζ (3)}{π^{2}}$

$_{2} F_{1} [\begin{matrix} \frac{1}{2}, \frac{1}{2} \\ \frac{1}{2} + α \end{matrix}; \frac{1}{2}] = \frac{1}{\sqrt{2}} \frac{Γ (\frac{α}{2}) Γ (\frac{1}{2} + α)}{Γ (\frac{1 + α}{2}) Γ (α)}$

$\sum_{0 \leq n} {(\frac{(α)_{n}}{(1 - α)_{n}})}^{2} = \frac{1}{2} + \frac{\sqrt{π}}{2} \frac{Γ^{2} (1 - α) Γ (\frac{1}{2} - 2 α)}{Γ^{2} (\frac{1}{2} - α) Γ (1 - 2 α)}$

$β (2) = \frac{1}{\sqrt{2}} \sum_{0 \leq m \leq n} \frac{(\binom{2 m}{m})}{2^{2 m} (2 m + 1)} \frac{1}{2^{n} (2 n + 1)}$

$\sum_{0 < 2 m + 1 < n} \frac{(- 1)^{n}}{(2 m + 1) (2 n + 1)} = \frac{π^{2}}{24}$

$\sum_{0 \leq 2 m < n} \frac{(\binom{2 m}{m}) (\binom{4 m}{2 m})}{2^{6 m} (2 m + 1)} \frac{2^{2 n}}{n^{2} (\binom{2 n}{n})} = 8 β (2) - π \ln 2$

$\sum_{0 \leq m < n} \frac{2^{2 m}}{(2 m + 1)^{2} (\binom{2 m}{m})} \frac{(\binom{2 n}{n})}{2^{2 n} n} = 4 β (2) \ln 2 + 8 \sum_{0 \leq m < n} \frac{1}{(2 m + 1) (4 n + 1)^{2}} - 8 \sum_{0 \leq m \leq n} \frac{1}{(2 m + 1) (4 n + 3)^{2}}$

$\sum_{0 < m < n} \frac{2^{m + n}}{m^{2} n (\binom{2 n}{n})} = \sum_{0 \leq 2 m < n} \frac{(\binom{2 m}{m}) (\binom{4 m + 2}{2 m + 1})}{2^{6 m} (2 m + 1)} \frac{2^{2 n}}{n^{2} (\binom{2 n}{n})} - \frac{π^{3}}{12} - π \ln^{2} 2$

$\sum_{0 \leq n} \frac{\sqrt{2}}{2^{n} (2 n + 1)^{2}} + \frac{π}{8} \sum_{0 \leq n} \frac{(\binom{2 n}{n}) (\binom{4 n + 2}{2 n + 1})}{2^{6 n} (2 n + 1)} = \frac{π^{2}}{4}$

$\sum_{0 \leq m < n} \frac{(- 1)^{n - 1}}{(2 m + 1) n} = \frac{π^{2}}{16}$

$\sum_{0 < n} \frac{2^{2 n}}{n^{2} (\binom{2 n}{n})} {(\frac{2 x}{1 + x^{2}})}^{2 n} = 8 \sum_{0 \leq m < n} \frac{(- 1)^{n - 1} x^{2 n}}{(2 m + 1) n}$

$\sum_{0 \leq m \leq n} \frac{2^{2 n}}{(2 m + 1) (2 n + 1)^{2} (\binom{2 n}{n})} - 2 \sum_{0 < m \leq n} \frac{(- 1)^{n - 1}}{m (2 n + 1)^{2}} = \frac{5 π^{3}}{48} - 2 β (2) \ln 2$

$\sum_{0 \leq m \leq n} \frac{{(\binom{2 m}{m})}^{2}}{2^{4 m - 2 n} (2 m + 1) (2 n + 1)^{2} (\binom{2 n}{n})} = \frac{π^{3}}{16}$

$\sum_{0 \leq n} \frac{2^{4 n}}{(2 n + 1)^{4} {(\binom{2 n}{n})}^{2}} = 2 t (4) - 8 t (1, 3) + 2 π t (1, \overset{―}{2})$

$\sum_{0 \leq m \leq n} \frac{2^{2 m + 2 n}}{(2 m + 1)^{2} (2 n + 1)^{2} (\binom{2 m}{m}) (\binom{2 n}{n})} = \sum_{0 \leq m \leq 2 n} (\frac{(- 1)^{m} π}{2 m + 1} - \frac{2}{(2 m + 1)^{2}}) \frac{1}{(2 n + 1)^{2}} + \sum_{0 \leq 2 m < n} \frac{2}{2 m + 1} (\frac{4}{(2 n + 1)^{2}} - \frac{(- 1)^{n} π}{(2 n + 1)^{3}})$

$\begin{aligned} \frac{π^{3}}{8} - π \ln^{2} 2 - \sum_{0 < n} \frac{(\binom{2 n}{n}) H_{n}}{2^{2 n} (2 n + 1)^{2}} & = 4 β (2) \ln 2 - 4 \sum_{0 < n} \frac{(- 1)^{n - 1} H_{n}}{(2 n + 1)^{2}} \\ = \sum_{0 \leq n} \frac{2^{2 n + 1}}{(2 n + 1)^{3} (\binom{2 n}{n})} \\ = π \sum_{0 \leq 2 m < n} \frac{{(\binom{2 m}{m})}^{2} (- 1)^{n - 1}}{2^{4 m} (2 m + 1) n} \\ = 8 \sum_{0 \leq m < n} \frac{(- 1)^{m} - (- 1)^{n}}{(2 m + 1) (2 n + 1)^{2}} \end{aligned}$

$\sum_{0 < n} \frac{(- 1)^{n - 1} 2^{2 n}}{n^{2} (\binom{2 n}{n})} \sum_{m = 0}^{n - 1} \frac{(\binom{2 m}{m})}{2^{2 m}} (\frac{π^{2}}{2} - \sum_{l = 1}^{m} \frac{2^{2 l}}{l^{2} (\binom{2 l}{l})}) = \sum_{0 < n} \frac{(- 1)^{n - 1} 2^{2 n}}{n^{2} (\binom{2 n}{n})} (2 H_{n} + \frac{1 - (- 1)^{n}}{n})$

$8 \sum_{0 < m < n} \frac{(\binom{2 m}{m})}{2^{2 m} m^{2}} \frac{2^{n}}{n (\binom{2 n}{n})} = π \sum_{0 < n} \frac{{(\binom{2 n}{n})}^{2}}{2^{4 n} n^{2}} + 8 \sum_{0 < m \leq n} \frac{1}{m^{2}} \frac{(- 1)^{n - 1}}{2 n + 1}$

$\sum_{0 \leq m < n} \frac{1}{(2 m + 1)^{a} n^{2}} (\frac{2^{2 n}}{(\binom{2 n}{n})} - 1) = \sum_{0 \leq m < n} ((- 1)^{a - 1} \frac{2 \ln 2 + H_{m}}{(2 m + 1)^{a}} + 2 \sum_{j = 0}^{a - 2} \frac{(- 1)^{j} t (a - j)}{(2 m + 1)^{j + 1}}) \frac{(\binom{2 n}{n})}{2^{2 n} n}$

$\sum_{0 < m < n} \frac{1}{m^{a} n^{2}} (\frac{2^{2 n}}{(\binom{2 n}{n})} - 1) = \sum_{0 < m \leq n} ((- 1)^{a - 1} \frac{H_{m}}{m^{a}} + \sum_{j = 0}^{a - 2} \frac{(- 1)^{j} ζ (a - j)}{m^{j + 1}}) \frac{(\binom{2 n}{n})}{2^{2 n} n}$

$\begin{aligned} \sum_{0 < m < n} \frac{2^{2 m}}{m^{2} (\binom{2 m}{m})} \frac{1}{(2 n - 1)^{2}} & = 2 π \sum_{0 \leq 2 l < 2 m < n} \frac{1}{(2 l + 1)^{2}} \frac{{(\binom{2 m}{m})}^{2}}{2^{4 m}} \frac{(- 1)^{n - 1}}{n} \\ = \sum_{0 \leq m < n} (\frac{π^{2}}{2 m + 1} - \frac{8}{(2 m + 1)^{3}}) \frac{(- 1)^{n - 1}}{n} \\ = \frac{π^{4}}{16} - \sum_{0 \leq m < n} \frac{1}{(2 m + 1)^{2} n^{2}} - \sum_{0 \leq m < n} (\frac{π^{2}}{4} \frac{1}{2 m + 1} - \frac{2 \ln 2 + H_{m}}{(2 m + 1)^{2}}) \frac{(\binom{2 n}{n})}{2^{2 n} n} \end{aligned}$

$\sum_{0 < m < n} \frac{2^{2 m}}{m^{2} (\binom{2 m}{m})} \frac{1}{(2 n - 1)^{2}} = 8 β (2)^{2} - \frac{π^{4}}{16}$ 　　　　( 出処 )

$\sum_{m = 0}^{n} \frac{(- 1)^{m}}{(2 m + 1)^{a + 1}} (\binom{n}{m}) = \frac{2^{2 n}}{(2 n + 1) (\binom{2 n}{n})} \sum_{0 \leq k_{1} \leq \dots \leq k_{a} \leq n} \frac{1}{(2 k_{1} + 1) \dots (2 k_{a} + 1)}$ 　　　　(出処：コメント欄)

$\sum_{0 < n} \frac{(\binom{m + n}{m})}{n (\binom{2 m + 2 n}{2 m})} = 2 \sum_{2 m \leq k} \frac{(- 1)^{k}}{k}$

$\sum_{0 < n} \frac{2^{2 n}}{n^{2} (\binom{2 n}{n})} \frac{(\binom{m + n}{m})}{(\binom{2 m + 2 n}{2 m})} = \sum_{m \leq n} \frac{1}{{(n + \frac{1}{2})}^{2}}$

$2 \sum_{0 < m \leq n} \frac{1}{m (2 n + 1)^{2}} = \sum_{0 < 2 m \leq n} \frac{2^{2 m}}{m^{2} (\binom{2 m}{m})} \frac{(- 1)^{n}}{n} = \frac{7}{2} ζ (3) - \frac{1}{2} π^{2} \ln 2$ 　　　　　(最右辺：コメント欄)

$\sum_{0 < n} (\frac{1}{n} - 2 \sum_{n < m} \frac{(- 1)^{m + n + 1}}{m}) = \ln 2$

$\begin{aligned} \sum_{0 < l < m < n} \frac{1}{l m n} \sum_{k = 0}^{n - 1} (\frac{(- 1)^{k} π}{2 k + 1} - \frac{2}{(2 k + 1)^{2}}) = & \frac{π}{4} \sum_{0 \leq n} \frac{2^{2 n}}{(\binom{2 n}{n})} (\frac{2}{(2 n + 1)^{4}} - \frac{2 \ln 2}{(2 n + 1)^{3}} + \frac{\ln^{2} 2}{(2 n + 1)^{2}}) \\ - \frac{1}{8} \sum_{0 < n} \frac{2^{4 n}}{n^{3} {(\binom{2 n}{n})}^{2}} ({(\sum_{m = 1}^{2 n - 1} \frac{(- 1)^{m}}{m})}^{2} + \sum_{2 n \leq m} \frac{(- 1)^{m}}{m^{2}}) \\ - \frac{7}{16} π^{2} ζ (3) + \frac{31}{4} ζ (5) - π β (4) \end{aligned}$

$\sum_{0 < n} \frac{2^{4 n}}{n^{3} {(\binom{2 n}{n})}^{2}} ({(\sum_{2 n \leq m} \frac{(- 1)^{m}}{m})}^{2} + \sum_{2 n \leq m} \frac{(- 1)^{m}}{m^{2}}) = 2 \sum_{0 \leq m < n} \frac{(\binom{2 n}{n})}{(2 m + 1)^{2} n^{3} 2^{2 n}}$

$\sum_{0 \leq n} \frac{2^{2 n}}{(\binom{2 n}{n})} (\frac{2}{(2 n + 1)^{4}} - \frac{2 \ln 2}{(2 n + 1)^{3}} + \frac{\ln^{2} 2}{(2 n + 1)^{2}}) = \frac{π}{2} \sum_{0 \leq n} \frac{{(\binom{2 n}{n})}^{2}}{2^{4 n} (2 n + 1)} ({(\sum_{m = 1}^{2 n} \frac{(- 1)^{m}}{m})}^{2} + \sum_{2 n < m} \frac{(- 1)^{m - 1}}{m^{2}})$

$\sum_{0 \leq m < n} \frac{(\binom{2 m}{m})}{2^{2 m} (2 m + 1)} \frac{2^{2 n}}{(2 n + 1)^{2} (\binom{2 n}{n})} = t (3)$

$\sum_{0 < m \leq n} \frac{2^{2 m + 2 n}}{m^{2} (2 n + 1)^{2} (\binom{2 m}{m}) (\binom{2 n}{n})} = 7 π ζ (3) - 24 β (4)$

$\sum_{0 \leq n_{1} \leq \dots \leq n_{a}} \frac{(2 x)^{2 n_{a}} \sqrt{1 - x^{2}}}{(2 n_{1} + 1)^{2} \dots (2 n_{a - 1} + 1)^{2} (2 n_{a} + 1) (\binom{2 n_{a}}{n_{a}})} = \sum_{0 \leq n_{1} \leq \dots \leq n_{a}} \frac{{(\binom{2 n_{1}}{n_{1}})}^{2} (2 x)^{2 n_{a}}}{2^{4 n_{1}} (2 n_{1} + 1) (2 n_{2} + 1)^{2} \dots (2 n_{a} + 1)^{2} (\binom{2 n_{a}}{n_{a}})}$

$\begin{aligned} \sum_{0 \leq m < n} \frac{(- 1)^{m}}{2 m + 1} \frac{(\binom{2 n}{n})}{2^{2 n} (2 n + 1)} & = \frac{1}{2} \sum_{0 \leq n} \frac{(- 1)^{n} (\binom{2 n}{n})}{2^{2 n} (2 n + 1)^{2}} \\ = \frac{π^{2}}{24} + \frac{1}{\sqrt{2}} \sum_{0 \leq m \leq n} \frac{1}{(2 m + 1) (2 n + 1) 2^{n}} - \frac{1}{4} \sum_{0 < m \leq n} \frac{1}{m n 2^{n}} \end{aligned}$

$\sum_{0 < n} \frac{2^{4 n}}{n^{4} {(\binom{2 n}{n})}^{2}} = \sum_{0 \leq m < n} (\frac{π (- 1)^{m}}{2 m + 1} - \frac{2}{(2 m + 1)^{2}}) \frac{8}{n^{2}} - \sum_{0 \leq m < n} \frac{8}{(2 m + 1) n^{3}} - \sum_{0 < m \leq n} \frac{16}{m} (\frac{π (- 1)^{n}}{(2 n + 1)^{2}} - \frac{2}{(2 n + 1)^{3}})$

$\sum_{0 \leq n} \frac{(- 1)^{n} (\binom{2 n}{n})}{2^{2 n} (2 n + 1)^{2}} = 2 \sqrt{2} \sum_{0 \leq m < n} \frac{(- 1)^{n - 1}}{2 m + 1} (\frac{1}{4 n - 1} + \frac{1}{4 n + 1})$

$\sum_{0 \leq n} \frac{(\binom{2 n}{n})}{2^{2 n} (2 n + 1)^{2}} (\frac{(- 1)^{n} π}{2} - \frac{1}{2 n + 1}) = π \sum_{0 < n} \frac{(\binom{4 n}{2 n})}{2^{2 n + 3} n^{2} (\binom{2 n}{n})}$

$\sqrt{2 π} \sum_{0 \leq n} \frac{2^{n} Γ^{2} (\frac{n}{2} + \frac{3}{4})}{(2 n + 1)^{2} n!} = \frac{π^{3}}{8} + \frac{π}{8} \sum_{0 < n} \frac{2^{2 n} (\binom{2 n}{n})}{n^{2} (\binom{4 n}{2 n})}$

$\sum_{0 < n} \frac{2^{2 n} (\binom{2 n}{n})}{n^{2} (\binom{4 n}{2 n})} + π \sum_{0 \leq n} \frac{(\binom{2 n}{n}) (\binom{4 n + 2}{2 n + 1})}{2^{6 n} (2 n + 1)} = π^{2}$

$\sum_{0 < n} \frac{1}{2^{2 n} n^{2}} (\frac{(\binom{4 n}{2 n})}{(\binom{2 n}{n})} - (- 1)^{n} (\binom{2 n}{n})) = ζ (2)$

$\sum_{0 \leq m < n} \frac{(\binom{2 n}{n})}{(2 m + 1) (2 n + 1)^{2} 2^{2 n}} = \frac{π \ln^{2} 2}{2} - \frac{π^{3}}{48}$

$\sum_{0 \leq m \leq n} \frac{2^{2 n}}{(2 m + 1) (2 n + 1)^{2} (\binom{2 n}{n})} - \frac{π}{4} \sum_{0 < n} \frac{{(\binom{2 n}{n})}^{2} H_{n}}{2^{4 n} (2 n + 1)} = 4 β (2) \ln 2$

$\frac{π}{2} \sum_{0 < n} \frac{H_{n}}{2^{4 n} n} {(\binom{2 n}{n})}^{2} = 16 β (2) \ln 2 - \frac{π^{3}}{3} - 16 t (1, \overset{―}{2})$

$\begin{aligned} \sum_{0 \leq n} \frac{2^{6 n}}{(2 n + 1)^{3} (\binom{2 n}{n}) (\binom{4 n + 2}{2 n + 1})} - \frac{π}{32} \sum_{0 < n} \frac{2^{2 n} (\binom{2 n}{n})}{n^{2} (\binom{4 n}{2 n})} & = \frac{π}{8} \sum_{0 \leq n} \frac{{(\binom{2 n}{n})}^{2}}{2^{4 n} (2 n + 1)^{2}} \\ = \frac{π^{3}}{16} - \sqrt{2} \sum_{0 \leq m \leq n} \frac{(\binom{2 m}{m})}{2^{2 m} (2 m + 1)} \frac{1}{2^{n} (2 n + 1)^{2}} \\ = \frac{1}{4} \sum_{0 \leq m < n} \frac{(- 1)^{m}}{2 m + 1} \frac{1}{n^{2}} + \frac{1}{2} \sum_{0 < n} \frac{(- 1)^{n - 1} H_{n}}{(2 n + 1)^{2}} \\ = 2 \sum_{0 \leq m < n} \frac{(- 1)^{m}}{(2 m + 1) (2 n + 1)^{2}} \\ = β (3) + 2 \sum_{0 \leq m < n} \frac{(- 1)^{n}}{(2 m + 1) (2 n + 1)^{2}} \\ = \frac{1}{4} \sum_{0 < m \leq 2 n + 1} \frac{(- 1)^{m - 1}}{m} \frac{2^{2 n}}{(2 n + 1)^{2} (\binom{2 n}{n})} \end{aligned}$

$\begin{aligned} \sum_{0 \leq m < n} \frac{(\binom{2 m}{m})}{2^{2 m} (2 m + 1)} \frac{(- 1)^{n - 1}}{n} & = \sum_{0 \leq n} \frac{1}{(2 n + 1)^{2}} \frac{(\binom{4 n + 2}{2 n + 1})}{2^{2 n + 2} (\binom{2 n}{n})} \\ = \frac{π}{2} \ln (- 1 + \sqrt{2}) + 4 β (2) - 4 \sum_{0 \leq n} \frac{(- 1)^{n} {(- 1 + \sqrt{2})}^{2 n + 1}}{(2 n + 1)^{2}} \end{aligned}$

$\frac{π}{8} \sum_{0 \leq n} \frac{(\binom{2 n}{n}) (\binom{4 n + 2}{2 n + 1})}{2^{6 n} (2 n + 1)^{2}} = π β (2) - \sum_{0 \leq n} \frac{(- 1)^{n} (\binom{2 n}{n})}{2^{2 n} (2 n + 1)^{3}} - 2 \sum_{0 \leq m < n} \frac{(- 1)^{m}}{(2 m + 1)^{2}} \frac{(\binom{2 n}{n})}{2^{2 n} (2 n + 1)}$

$\frac{π}{8} \sum_{0 < n} \frac{(\binom{4 n}{2 n})}{2^{2 n} n^{3} (\binom{2 n}{n})} = \frac{π}{6} \ln^{3} 2 + \frac{π^{3}}{24} \ln 2 - \frac{3}{2} π ζ (3) - \frac{π^{2}}{8} \sum_{0 \leq n} \frac{(\binom{2 n}{n}) (\binom{4 n + 2}{2 n + 1})}{2^{6 n} (2 n + 1)^{2}} + 4 \sum_{0 \leq m < n} \frac{1}{(2 m + 1)^{3}} \frac{(\binom{2 n}{n})}{2^{2 n} (2 n + 1)}$

$\frac{π}{2} \sum_{0 \leq m < n} \frac{(\binom{2 m}{m})}{2^{2 m} (2 m + 1)} \frac{(\binom{4 n}{2 n})}{2^{4 n} n} = \frac{π^{2}}{2} \ln (1 + \sqrt{2}) - \sum_{0 \leq m < n} \frac{(- 1)^{m} (\binom{2 m}{m})}{2^{2 m} (2 m + 1)} \frac{1 - (- 1)^{n}}{n^{2}}$

$\sum_{0 \leq m < n} \frac{(\binom{4 m + 2}{2 m + 1})}{2^{4 m} (2 m + 1)} \frac{2^{2 n}}{n^{2} (\binom{2 n}{n})} = 8 \sum_{0 \leq m < n} \frac{(- 1)^{m} (\binom{2 m}{m})}{2^{2 m} (2 m + 1)} \frac{1}{n^{2}}$

$\sum_{j = 0}^{a - 1} \frac{(- 1)^{a - j - 1} π^{2 j}}{(2 j + 1)!} \sum_{0 \leq m < n} \frac{(\binom{2 m}{m})}{2^{2 m} (2 m + 1)} \frac{(- 1)^{n - 1}}{n^{2 a - 2 j - 1}} = \sum_{0 < n_{1} < \dots < n_{a}} \frac{1}{n_{1}^{2} \dots n_{a - 1}^{2} (2 n_{a} - 1)^{2}} \frac{(\binom{4 n_{a} - 2}{2 n_{a} - 1})}{2^{2 n_{a}} (\binom{2 n_{a} - 2}{n_{a} - 1})}$

$π \sum_{0 \leq m < n} \frac{1}{(2 m + 1)^{2}} \frac{(\binom{2 n}{n}) (\binom{4 n}{2 n})}{2^{6 n} n} - \sum_{0 < m \leq n} \frac{1 - (- 1)^{m}}{m^{3}} \frac{(\binom{2 n}{n})}{2^{2 n} (2 n + 1)} = \frac{3}{4} π^{3} \ln 2 - 2 π^{2} β (2) - \frac{7}{8} π ζ (3)$

$\begin{aligned} \frac{Γ^{2} (\frac{1}{4})}{\sqrt{2 π}} \sum_{0 < n} \frac{2^{2 n}}{n^{2} (\binom{2 n}{n})} \frac{{(\frac{1}{4})}_{n}}{{(\frac{3}{4})}_{n}} = \frac{π}{\sqrt{2}} \sum_{0 \leq m < n} \frac{(- 1)^{m} {(\binom{2 m}{m})}^{2}}{2^{4 m} n^{2}} & - \frac{π \sqrt{π}}{Γ^{2} (\frac{1}{4})} \sum_{0 \leq n} \frac{2^{4 n + 3}}{(2 n + 1)^{2} (\binom{4 n + 2}{2 n + 1})} \frac{{(\frac{3}{4})}_{n}}{{(\frac{5}{4})}_{n}} \\ + \frac{Γ^{2} (\frac{1}{4})}{\sqrt{π}} \sum_{0 < n} \frac{2^{4 n - 4}}{n^{2} (\binom{4 n}{2 n})} \frac{{(\frac{1}{4})}_{n}}{{(\frac{3}{4})}_{n}} \end{aligned}$

$\begin{aligned} \frac{\sqrt{2}}{64} (ψ^{'} (\frac{1}{8}) + ψ^{'} (\frac{3}{8}) - ψ^{'} (\frac{5}{8}) - ψ^{'} (\frac{7}{8})) & = β (2) + \sum_{0 \leq n} \frac{2^{2 n} (\binom{2 n}{n})}{(2 n + 1)^{2} (\binom{4 n + 2}{2 n + 1})} \\ = β (2) + \frac{\sqrt{2}}{16} \sum_{0 \leq m < n} \frac{(\binom{2 m}{m}) (\binom{4 m}{2 m})}{2^{6 m}} \frac{2^{2 n}}{n^{2} (\binom{2 n}{n})} \\ = \frac{4 t (3)}{π} + \frac{\sqrt{2}}{16} \sum_{0 \leq m < n} \frac{(\binom{2 m}{m}) (\binom{4 m}{2 m})}{2^{6 m}} \frac{1}{n^{2}} \\ = \frac{4 t (3)}{π} + \frac{1}{16 π} \sum_{0 < n} \frac{2^{6 n}}{n^{3} (\binom{2 n}{n}) (\binom{4 n}{2 n})} \end{aligned}$

$\sum_{0 < n} \frac{(- 1)^{n - 1}}{n^{2 r}} \frac{{(α)}_{n}}{{(1 - α)}_{n}} + \frac{1}{2} \sum_{0 < n_{1} \leq \dots \leq n_{r}} \frac{2^{2 n_{1}}}{(\binom{2 n_{1}}{n_{1}})} \frac{{(\frac{1}{2} - α)}_{n_{1}}}{(1 - α)_{n_{1}}} \frac{1}{n_{1}^{2} \dots n_{r}^{2}} = η (2 r)$

$\sum_{0 < n} \frac{(α)_{n}}{n^{2} (1 - α)_{n}} = ζ (2) - \frac{2 \sqrt{π} Γ (1 - α)}{Γ (\frac{1}{2} - α)} \sum_{0 \leq n} \frac{{(\frac{1}{2} + α)}_{n}}{(2 n + 1)^{2} {(\frac{1}{2})}_{n}} + \frac{1}{2} \sum_{0 < n} \frac{2^{2 n}}{n^{2} (\binom{2 n}{n})} \frac{{(\frac{1}{2} - α)}_{n}}{(1 - α)_{n}}$

$\sum_{0 \leq m < n} \frac{1}{(2 m + 1) n} \frac{{(\frac{1}{4})}_{n}}{{(\frac{3}{4})}_{n}} = \frac{π^{2}}{16} + β (2)$

$\begin{aligned} \frac{1}{2} \sum_{0 < n} \frac{{(\frac{1}{2})}_{n}}{n^{2} {(\frac{3}{4})}_{n}} & = \sum_{0 < n} \frac{2^{n}}{n^{2} (\binom{2 n}{n})} - \sum_{0 < m < n} \frac{2^{n}}{m n (\binom{2 n}{n})} \\ = \frac{π \ln 2}{2} + \frac{π^{2}}{8} - 2 β (2) \end{aligned}$

$\begin{aligned} \frac{1}{2} \sum_{0 < n} \frac{{(\frac{1}{2})}_{n}}{n^{3} {(\frac{3}{4})}_{n}} = & \sum_{0 < n} \frac{2^{n}}{n^{3} (\binom{2 n}{n})} - \sum_{0 < m < n} \frac{2^{n}}{m^{2} n (\binom{2 n}{n})} - \sum_{0 < m < n} \frac{2^{n}}{m n^{2} (\binom{2 n}{n})} + \sum_{0 < l < m < n} \frac{2^{n}}{l m n (\binom{2 n}{n})} \\ = & 32 \sum_{0 < l < m \leq \frac{n}{2}} \frac{(- 1)^{n}}{(4 l - 1) (4 m - 1) (2 n + 1)} \end{aligned}$

$\begin{aligned} \frac{1}{2} \sum_{0 < n} \frac{{(\frac{1}{2})}_{n}}{n^{4} {(\frac{3}{4})}_{n}} = & \sum_{0 < n} \frac{2^{n}}{n^{4} (\binom{2 n}{n})} - \sum_{0 < m < n} \frac{2^{n}}{m^{3} n (\binom{2 n}{n})} - \sum_{0 < m < n} \frac{2^{n}}{m^{2} n^{2} (\binom{2 n}{n})} - \sum_{0 < m < n} \frac{2^{n}}{m n^{3} (\binom{2 n}{n})} \\ + \sum_{0 < l < m < n} \frac{2^{n}}{l^{2} m n (\binom{2 n}{n})} + \sum_{0 < l < m < n} \frac{2^{n}}{l m^{2} n (\binom{2 n}{n})} + \sum_{0 < l < m < n} \frac{2^{n}}{l m n^{2} (\binom{2 n}{n})} - \sum_{0 < k < l < m < n} \frac{2^{n}}{k l m n (\binom{2 n}{n})} \end{aligned}$

$\frac{1}{2} \sum_{0 < n} \frac{{(\frac{1}{2})}_{n}}{n^{r} {(\frac{3}{4})}_{n}} = \sum_{j = 1}^{r} (- 1)^{j - 1} \sum_{\begin{matrix} 0 < n_{1} < \dots < n_{j} \\ k_{1} + \dots + k_{j} = r, k_{i} \in Z_{> 0} \end{matrix}} \frac{2^{n_{j}}}{n_{1}^{k_{1}} \dots n_{j}^{k_{j}} (\binom{2 n_{j}}{n_{j}})} ($ 推測 $)$

$\sum_{0 \leq m < n} \frac{{(\frac{1}{4})}_{m}}{m!} \frac{2^{2 m}}{(2 m + 1) (\binom{2 m}{m})} \frac{(\binom{2 n}{n})}{2^{2 n} (2 n + 1)} = π (1 - \frac{4 \sqrt{2 π}}{Γ^{2} (\frac{1}{4})})$

$\sum_{0 \leq m \leq n} \frac{1}{2 m + 1} (1 + \frac{(\binom{2 m}{m})}{2^{2 m}}) \frac{2^{2 n + 2}}{(2 n + 1)^{2} (\binom{2 n}{n})} - \sum_{0 \leq m < n} \frac{1}{2 m + 1} (1 + \frac{(\binom{2 m}{m})}{2^{2 m}}) \frac{2^{2 n}}{n^{2} (\binom{2 n}{n})} = 8 β (2) \ln 2$

$2 β (2)^{2} = \sum_{0 \leq m < n} (\frac{1}{(2 m + 1)^{2}} \frac{1 - (- 1)^{n}}{n^{2}} + \frac{1}{2 m + 1} \frac{1 - (- 1)^{n}}{n^{3}} - \frac{π^{2}}{2} \frac{1}{2 m + 1} \frac{(- 1)^{n - 1}}{n})$

$6 \sum_{0 < n} \frac{2^{4 n}}{(2 n + 1)^{3} {(\binom{2 n}{n})}^{2}} \sum_{m = 0}^{n - 1} \frac{1}{2 m + 1} \sum_{k = 1}^{n} \frac{1}{k^{2}} = \sum_{0 \leq m < n} (\frac{π^{3} (- 1)^{m}}{(2 m + 1)^{2}} - \frac{24 π (- 1)^{m}}{(2 m + 1)^{4}} + \frac{48}{(2 m + 1)^{5}}) \frac{(- 1)^{n - 1}}{n}$

$\sum_{0 < m \leq n} \frac{(- 1)^{n - 1}}{m^{2} (2 n + 1)} + 4 \sqrt{2} \sum_{0 \leq n} \frac{(\binom{2 n}{n})}{2^{3 n} (2 n + 1)^{3}} = \frac{π^{3}}{6} + \frac{5}{4} π \ln^{2} 2 - 2 β (2) \ln 2 + \frac{π}{8} \sum_{0 < n} \frac{{(\binom{2 n}{n})}^{2}}{2^{4 n} n^{2}}$

$\sum_{0 < m < n} \frac{1}{m n} \frac{2^{n} (\binom{2 m}{m})}{2^{m} (\binom{2 n}{n})} = \frac{π \ln 2}{2}$

$\sqrt{π} \sum_{0 < n} \frac{1}{2^{n} n^{2}} \frac{n!}{Γ^{2} (\frac{5}{4} + \frac{n}{2})} = 2 \sum_{0 < m + 1 < n} \frac{{(\binom{2 m}{m})}^{2}}{2^{5 m} (m + 1)} \frac{1}{n^{2}}$

$\frac{(\binom{2 m}{m})}{2^{2 m}} \sum_{m < k} \frac{2^{2 k}}{k (k + n) (\binom{2 k}{k})} + \frac{(\binom{2 n}{n})}{2^{2 n}} \sum_{n < k} \frac{2^{2 k}}{k (k + m) (\binom{2 k}{k})} = π^{2} \frac{(\binom{2 m}{m}) (\binom{2 n}{n})}{2^{2 m + 2 n}}$

$\frac{2^{2 m}}{m (\binom{2 m}{m})} \sum_{k = 0}^{m - 1} \frac{(\binom{2 k}{k})}{2^{2 k} (k + n)} + \frac{2^{2 n}}{n (\binom{2 n}{n})} \sum_{k = 0}^{n - 1} \frac{(\binom{2 k}{k})}{2^{2 k} (k + m)} = \frac{2^{2 m + 2 n}}{m n (\binom{2 m}{m}) (\binom{2 n}{n})}$

$\sum_{0 < m \leq n} \frac{2^{2 m}}{m^{2} (\binom{2 m}{m})} \frac{(\binom{2 n}{n})}{2^{2 n} n^{2}} = π^{2} \ln^{2} 2 - \sum_{0 < m \leq n} \frac{2^{2 n}}{m n^{3} (\binom{2 n}{n})}$

$\begin{aligned} \frac{4 β (2)^{2}}{π} + \sum_{0 \leq m < n} \frac{2^{4 m}}{(2 m + 1)^{2} {(\binom{2 m}{m})}^{2}} \frac{{(\binom{2 n}{n})}^{2}}{2^{4 n} (2 n + 1)} & = 2 β (2) \ln 2 + \frac{1}{2} \sum_{0 \leq n} \frac{2^{2 n} H_{n}}{(2 n + 1)^{2} (\binom{2 n}{n})} \\ = 2 β (3) + \sum_{0 < m \leq n} \frac{2^{2 m}}{m^{2} (\binom{2 m}{m})} \frac{(\binom{2 n}{n})}{2^{2 n} (2 n + 1)} \end{aligned}$

$\sum_{0 < m \leq n} \frac{2^{2 m}}{m^{2} (\binom{2 m}{m})} \frac{(\binom{2 n}{n})}{2^{2 n} (2 n + 1)^{2}} + 2 \sum_{0 \leq n} \frac{2^{2 n} (H_{n} + 2 \ln 2)}{(2 n + 1)^{3} (\binom{2 n}{n})} = 8 t (2) (β (2) - 2 π \ln 2)$

$\frac{2^{2 n}}{(\binom{2 n}{n})} \sum_{m < k} \frac{2^{2 k}}{k^{2} (\binom{2 k}{k})} \frac{k! n!}{(k + n)!} = \frac{2^{2 m}}{(\binom{2 m}{m})} \sum_{n < k} \frac{2^{2 k}}{k^{2} (\binom{2 k}{k})} \frac{k! m!}{(k + m)!}$

$\sum_{0 \leq m < n} \frac{2^{2 n}}{(2 m + 1) n^{3} (\binom{2 n}{n})} = - \frac{π^{4}}{24} - 32 t (\overset{―}{3}, \overset{―}{1})$

$\sum_{0 \leq m < n} \frac{(\binom{2 m}{m})}{2^{2 m} (2 m + 1)} \frac{2^{2 n}}{(2 n + 1)^{3} (\binom{2 n}{n})} = - 4 t (\overset{―}{1}, \overset{―}{3})$

$\sum_{0 < 2 m \leq n} \frac{1}{(2 m - 1)^{2}} \frac{(- 1)^{n}}{n} = \frac{1}{8} \sum_{0 < m < n} \frac{(\binom{2 m}{m})}{2^{2 m} m} \frac{2^{2 n}}{n^{2} (\binom{2 n}{n})} = \frac{π^{2} \ln 2}{8} - \frac{7}{16} ζ (3)$

$16 \sum_{0 \leq m < n} \frac{(\binom{2 n}{n})}{(2 m + 1)^{2} (2 n + 1)^{2} 2^{2 n}} = 2 π \sum_{0 < 2 m < n} \frac{(- 1)^{n - 1}}{m^{2} n} = π \sum_{0 < m < n} \frac{(\binom{2 m}{m})}{2^{2 m} m n^{2}} = \frac{π^{3} \ln 2}{3} - \frac{3 π ζ (3)}{2}$

$\sum_{0 \leq m < n} \frac{(- 1)^{m + n - 1}}{(2 m + 1)^{2} n} = 2 t (\overset{―}{3}) + 2 t (2, \overset{―}{1}) + 2 t (1, \overset{―}{2}) - t (\overset{―}{2}) \ln 2$

$\sum_{0 < 2 m + 1 < n} \frac{2^{4 m}}{(2 m + 1)^{2} {(\binom{2 m}{m})}^{2}} \frac{(- 1)^{n - 1}}{n} = - 2 t (\overset{―}{3}) - 4 t (2, \overset{―}{1}) - 4 t (1, \overset{―}{2}) + 2 t (\overset{―}{2}) \ln 2$

$\begin{aligned} π β (2) \ln 2 = & - 2 t (4) + 2 π t (\overset{―}{1}, 2) - 4 π t (1, \overset{―}{2}) + 8 t (1, 3) \\ + \sum_{0 \leq m < n} (\frac{π}{2} \frac{(- 1)^{m}}{2 m + 1} - \frac{1}{(2 m + 1)^{2}}) \frac{1}{n^{2}} - \sum_{0 < m \leq n} \frac{4}{m} (\frac{π}{2} \frac{(- 1)^{n}}{(2 n + 1)^{2}} - \frac{1}{(2 n + 1)^{3}}) \end{aligned}$

$\sum_{0 \leq n} \frac{{(\frac{1}{2} + α)}_{n}}{{(n + \frac{1}{2})}^{2} {(\frac{1}{2})}_{n}} = \sum_{0 < n} \frac{2^{2 n}}{n (n - α) (\binom{2 n}{n})}$

$\sum_{0 \leq n} \frac{2^{2 n}}{(2 n + 1)^{2} (\binom{2 n}{n})} = 2 β (2)$

$\begin{aligned} \sum_{0 \leq n} \frac{2^{2 n}}{(2 n + 1)^{3} (\binom{2 n}{n})} & = 2 β (2) \ln 2 - 2 \sum_{0 < m \leq n} \frac{(- 1)^{n - 1}}{m (2 n + 1)^{2}} \\ = 4 t (\overset{―}{1}, 2) - 4 t (1, \overset{―}{2}) \\ = 2 t (\overset{―}{3}) + 4 t (2, \overset{―}{1}) \end{aligned}$

$\begin{aligned} \sum_{0 \leq n} \frac{2^{2 n}}{(2 n + 1)^{4} (\binom{2 n}{n})} & = \frac{\ln 2}{2} \sum_{0 \leq n} \frac{2^{2 n}}{(2 n + 1)^{3} (\binom{2 n}{n})} + \frac{7}{8} π ζ (3) - \frac{1}{8} π^{3} \ln 2 - 2 \sum_{0 < l \leq m \leq n} \frac{(- 1)^{n}}{l (2 m + 1)^{2} (2 n + 1)} - 2 \sum_{0 < l \leq m < n} \frac{(- 1)^{n}}{l (2 m + 1) (2 n + 1)^{2}} \\ = 4 t (2) t (\overset{―}{2}) - 2 t (\overset{―}{4}) - 8 t (\overset{―}{2}, 2) + 8 t (1, 1, \overset{―}{2}) - 8 t (1, \overset{―}{1}, 2) \\ = 2 t (\overset{―}{4}) + 4 t (2, \overset{―}{2}) + 4 t (3, \overset{―}{1}) + 8 t (2, 1, \overset{―}{1}) \end{aligned}$

$\sum_{0 \leq n} \frac{2^{2 n}}{(2 n + 1)^{a} (\binom{2 n}{n})} = \sum_{\begin{matrix} w t (k) = a \\ k_{1} \geq 2 \end{matrix}} 2^{j} t (k_{1}, \dots, k_{j - 1}, \overset{―}{k_{j}})$ 　　　　　(訂正後。出処：コメント欄)

$\sum_{0 < n} \frac{2^{2 n}}{n^{2} (\binom{2 n}{n})} = 4 t (2)$

$\sum_{0 < n} \frac{2^{2 n}}{n^{3} (\binom{2 n}{n})} = - 4 t (3) - 6 ζ (2) ζ (\overset{―}{1})$

$\sum_{0 < n} \frac{2^{2 n}}{n^{4} (\binom{2 n}{n})} = ζ (2, 2) + 4 ζ (\overset{―}{2}, 2) + 4 ζ (\overset{―}{1}, 3) - 4 ζ (1, \overset{―}{3}) + 12 ζ (2) ζ (1, \overset{―}{1})$

$\sum_{0 < m < n} \frac{2^{2 n}}{m n^{2} (\binom{2 n}{n})} = 7 ζ (3)$

$\sum_{0 < m < n} \frac{2^{2 n}}{m^{2} n^{2} (\binom{2 n}{n})} = 4 t (4)$

$\sum_{0 < m < n} \frac{2^{2 n}}{m n^{3} (\binom{2 n}{n})} = 14 ζ (2, 2) - 4 ζ (\overset{―}{2}, 2) - 4 ζ (\overset{―}{1}, 3) + 4 ζ (1, \overset{―}{3}) + 4 ζ (3, \overset{―}{1}) - 4 ζ (\overset{―}{3}, \overset{―}{1})$

$\sum_{n = 1}^{\infty} \frac{{(\binom{2 n}{n})}^{2}}{2^{4 n} n^{2}} (8 β (2) + \sum_{m = 1}^{n} \frac{2^{4 m}}{m^{2} {(\binom{2 m}{m})}^{2}}) = 2 π \sum_{n = 0}^{\infty} \frac{{(\binom{2 n}{n})}^{2}}{2^{4 n}} (\frac{ζ (2)}{2 n + 1} - \frac{4}{(2 n + 1)^{2}} (\sum_{m = 0}^{n} \frac{1}{2 m + 1} - \ln 2))$

$\sum_{n = 0}^{\infty} \frac{{(\binom{2 n}{n})}^{2}}{2^{4 n} (2 n + 1)^{2}} (8 β (2) + \sum_{m = 1}^{n} \frac{2^{4 m}}{m^{2} {(\binom{2 m}{m})}^{2}}) = 2 π t (3) + 4 t (2) (π \ln 2 - 2 β (2)) - \frac{π}{2} \sum_{n = 1}^{\infty} \frac{{(\binom{2 n}{n})}^{2}}{2^{4 n} n^{2}} \sum_{m = 0}^{n - 1} \frac{1}{2 m + 1}$

$\begin{aligned} \sum_{n = 1}^{\infty} \frac{(\binom{2 n}{n})}{2^{2 n} n} \sum_{k = 0}^{n - 1} \frac{1}{2 k + 1} \sum_{l = 0}^{n - 1} \frac{1}{(2 l + 1)^{2}} & = 2 t (4) + t (2) \sum_{n = 1}^{\infty} \frac{{(\binom{2 n}{n})}^{2} H_{n}}{2^{4 n} n} \\ = \frac{π}{8} \sum_{n = 1}^{\infty} \frac{1}{n^{2}} \sum_{m = 0}^{n - 1} \frac{(\binom{2 m}{m})}{2^{2 m} (2 m + 1)} + \frac{π}{8} \sum_{n = 1}^{\infty} \frac{(\binom{2 n}{n})}{2^{2 n} (2 n + 1)} \sum_{m = 1}^{n} \frac{2^{2 m}}{m^{2} (\binom{2 m}{m})} \\ = π \sum_{n = 1}^{\infty} \frac{(- 1)^{n - 1}}{n^{2}} \sum_{m = 0}^{n - 1} \frac{(- 1)^{m}}{2 m + 1} \\ = \frac{π}{8} \sum_{n = 1}^{\infty} \frac{2^{4 n}}{n^{3} {(\binom{2 n}{n})}^{2}} \sum_{m = 0}^{n - 1} \frac{{(\binom{2 m}{m})}^{2}}{2^{4 m}} \end{aligned}$

$\frac{π x}{2} \sum_{n = 0}^{\infty} \frac{{(\binom{2 n}{n})}^{2}}{2^{4 n} (n + x)} = \sum_{n = 0}^{\infty} \frac{1}{2 n + 1} \frac{{(\frac{1}{2} + x)}_{n}}{{(1 + x)}_{n}}$

$_{3} F_{2} [\begin{matrix} \frac{1}{2}, \frac{1}{2}, α \\ \frac{3}{2}, β \end{matrix}; 1] = \frac{Γ (β) Γ (\frac{1}{2} - α + β)}{Γ (β - α) Γ (\frac{1}{2} + α)}_{3} F_{2} [\begin{matrix} 1, 1, α \\ \frac{3}{2}, \frac{1}{2} + β \end{matrix}; 1]$

$\begin{aligned} \sum_{n = 0}^{\infty} \frac{{(\binom{2 n}{n})}^{3}}{2^{6 n}} (3 ζ (2) + \sum_{m = 1}^{n} \frac{2^{2 m}}{m^{2} (\binom{2 m}{m})}) & = \frac{1}{π} \sum_{n = 1}^{\infty} \frac{2^{6 n}}{n^{3} {(\binom{2 n}{n})}^{3}} \sum_{m = 0}^{n - 1} \frac{{(\binom{2 m}{m})}^{2}}{2^{4 m}} \\ = 4 \sum_{n = 0}^{\infty} \frac{2^{2 n}}{(2 n + 1)^{2} (\binom{2 n}{n})} \sum_{m = 0}^{n} \frac{{(\binom{2 m}{m})}^{3}}{2^{6 m}} \end{aligned}$

$\begin{aligned} \sum_{n = 1}^{\infty} \frac{{(\binom{2 n}{n})}^{3}}{2^{6 n} n} (3 ζ (2) + \sum_{m = 1}^{n} \frac{2^{2 m}}{m^{2} (\binom{2 m}{m})}) & = 18 ζ (2) \ln 2 - 4 t (3) - 4 π β (2) - \frac{1}{π} \sum_{n = 1}^{\infty} \frac{2^{6 n}}{n^{4} {(\binom{2 n}{n})}^{3}} \sum_{m = 0}^{n - 1} \frac{{(\binom{2 m}{m})}^{2}}{2^{4 m}} \end{aligned}$

$8 β (2) = \sum_{n = 1}^{\infty} \frac{2^{2 n}}{n^{2} (\binom{2 n}{n})} \sum_{m = 0}^{n - 1} \frac{{(\binom{2 m}{m})}^{2}}{2^{4 m}}$

$\begin{aligned} 2 ζ (2) β (2) + \sum_{n = 1}^{\infty} \frac{1}{n^{2}} \sum_{m = 0}^{n - 1} \frac{2^{4 m}}{(2 m + 1)^{2} {(\binom{2 m}{m})}^{2}} & = \sum_{n = 0}^{\infty} \frac{2^{2 n + 2}}{(2 n + 1)^{2} (\binom{2 n}{n})} \sum_{m = 0}^{n} (\frac{π}{2} \frac{(\binom{2 m}{m})}{2^{2 m} (2 m + 1)} - \frac{1}{(2 m + 1)^{2}}) \\ = 2 \sum_{n = 1}^{\infty} \frac{(\binom{2 n}{n}) (2 \ln 2 + H_{n})}{2^{2 n} (2 n + 1)} \sum_{m = 0}^{n - 1} \frac{1}{(2 m + 1)^{2}} \\ = 4 π t (3) - 8 β (4) \end{aligned}$

$\frac{π}{2} \frac{1}{\sqrt{1 - x^{2}}} \sum_{n = 0}^{\infty} \frac{{(\binom{2 n}{n})}^{2}}{2^{4 n}} x^{2 n} = \sum_{n = 0}^{\infty} \frac{2^{2 n} x^{2 n}}{(2 n + 1) (\binom{2 n}{n})} \sum_{m = 0}^{n} \frac{{(\binom{2 m}{m})}^{2}}{2^{4 m} x^{2 m}}$

$\begin{aligned} χ_{2} (\frac{1}{\sqrt{2}}) & = \frac{5}{6} t (2) - \frac{1}{8} \sum_{n = 1}^{\infty} \frac{(\binom{4 n}{2 n})}{2^{2 n} n^{2} (\binom{2 n}{n})} \\ = t (2) - \frac{π}{16} \sum_{n = 0}^{\infty} \frac{(\binom{2 n}{n}) (\binom{4 n + 2}{2 n + 1})}{2^{6 n} (2 n + 1)} \end{aligned}$

$\sum_{n = 1}^{\infty} \frac{2^{2 n}}{n^{2} (\binom{2 n}{n})} \sum_{m = 0}^{n - 1} \frac{{(\binom{2 m}{m})}^{2}}{2^{4 m}} = 2 π \ln 2 + 2 \sum_{n = 0}^{\infty} \frac{(\binom{2 n}{n}) H_{n}}{2^{2 n} (2 n + 1)}$

$\frac{π}{2} \frac{\tanh^{- 1} x}{\sqrt{1 - x^{2}}} + \frac{2}{1 + x} \sum_{n = 0}^{\infty} \frac{{(- \frac{1 - x}{1 + x})}^{n}}{(2 n + 1)^{2}} = \sum_{n = 0}^{\infty} \frac{(\binom{2 n}{n})}{2^{2 n}} x^{2 n} (2 β (2) + \frac{1}{4} \sum_{m = 1}^{n} \frac{2^{4 m}}{m^{2} {(\binom{2 m}{m})}^{2}})$

$2 \sum_{n = 1}^{\infty} \frac{(α)_{n}}{n^{2} (2 α)_{n}} = ζ (2) + ψ^{'} (2 α) - {(ψ (α) - ψ (2 α))}^{2}$

$\sum_{n = 0}^{\infty} \frac{{(\binom{2 n}{n})}^{3}}{2^{6 n}} \sum_{m = 1}^{n} \frac{2^{4 m}}{m^{2} {(\binom{2 m}{m})}^{2}} = (π^{2} - 8 β (2)) \sum_{n = 0}^{\infty} \frac{{(\binom{2 n}{n})}^{3}}{2^{6 n}}$

$ζ (2) + \sum_{n = 1}^{\infty} \frac{{(\binom{2 n}{n})}^{2} H_{n}}{2^{4 n} n} = \frac{1}{π} \sum_{n = 1}^{\infty} \frac{2^{4 n}}{n^{3} {(\binom{2 n}{n})}^{2}} \sum_{m = 0}^{n - 1} \frac{{(\binom{2 m}{m})}^{2}}{2^{4 m}}$

$ζ (3) + \sum_{n = 1}^{\infty} \frac{{(\binom{2 n}{n})}^{2}}{2^{4 n} n} (ζ (2) - \frac{H_{n}}{n}) = \frac{1}{π} \sum_{n = 1}^{\infty} \frac{2^{4 n}}{n^{4} {(\binom{2 n}{n})}^{2}} \sum_{m = 0}^{n - 1} \frac{{(\binom{2 m}{m})}^{2}}{2^{4 m}}$

$\sum_{n = 1}^{\infty} \frac{2^{2 n}}{n^{3} (\binom{2 n}{n})} \sum_{m = 0}^{n - 1} \frac{{(\binom{2 m}{m})}^{2}}{2^{4 m}} = 4 π \sum_{n = 0}^{\infty} \frac{{(\binom{2 n}{n})}^{2}}{2^{4 n} (2 n + 1)^{2}} - 16 β (2) \ln 2$

$\sum_{m = 1}^{\infty} \frac{(\binom{2 m}{m})}{2^{2 m + 1} m} \sum_{l = 0}^{m - 1} \frac{2^{4 l}}{(2 l + 1)^{3} {(\binom{2 l}{l})}^{2}} - \sum_{m = 0}^{n - 1} \frac{2^{4 m}}{(2 m + 1)^{2} {(\binom{2 m}{m})}^{2}} (\frac{π^{2}}{24} - \frac{1}{2 m + 1} (- \ln 2 + \sum_{k = 0}^{m} \frac{1}{2 k + 1})) = \frac{2^{4 n}}{{(\binom{2 n}{n})}^{2}} \sum_{m = 1}^{\infty} \frac{2^{4 m}}{(2 m)^{3} {(\binom{2 m}{m})}^{2}} \sum_{l = 0}^{m - 1} \frac{{(\binom{2 l}{l})}^{2}}{2^{4 l} (2 l + 2 n + 1)}$

　　　 $\sum_{n = 1}^{\infty} {(\frac{2 x}{1 + x^{2}})}^{2 n} \frac{(\binom{2 n}{n})}{2^{2 n} n} \sum_{m = 1}^{n} \frac{2^{2 m}}{m^{2} (\binom{2 m}{m})} = 4 χ_{3} (x^{2})$

　　　 $\sum_{0 \leq m < n} \frac{2^{2 n}}{n^{2} (\binom{2 n}{n})} \frac{{(\binom{2 m}{m})}^{2}}{2^{4 m}} ((- 1)^{m} + \frac{1}{2^{m + \frac{1}{2}}}) = \sqrt{\frac{π}{2}} \frac{Γ^{2} (\frac{1}{4})}{2}$

$\sum_{0 < m \leq n} \frac{2^{2 m}}{m^{2} (\binom{2 m}{m})} \frac{(- 1)^{n - 1} {(\binom{2 n}{n})}^{2}}{2^{4 n}} = \sum_{0 \leq m < n} \frac{{(\binom{2 m}{m})}^{2}}{2^{4 m} n^{2}} ((- 1)^{m} - \frac{1}{2^{m + \frac{1}{2}}})$

　　　 $2 \sum_{0 < m \leq n} \frac{2^{4 m}}{m^{2} {(\binom{2 m}{m})}^{2}} \frac{(\binom{2 n}{n})}{2^{2 n}} (- 1)^{m + n} = \sum_{0 \leq m < n} \frac{{(\binom{2 m}{m})}^{2}}{2^{5 m}} \frac{2^{2 n}}{n^{2} (\binom{2 n}{n})}$

　　　 $\sum_{0 < m < n} \frac{1}{m^{3} n^{2} (\binom{2 n}{n})} = \frac{58}{9} ζ (5) - \frac{4}{9} ζ (2) ζ (3) - 2 π \sum_{0 < n} \frac{1}{n^{4}} \sin \frac{π n}{3}$

　　　 $\sum_{0 < m < n} \frac{2^{2 m}}{m^{2} (\binom{2 m}{m})} \frac{1}{n^{2}} = \sum_{0 < m < n} \frac{(\binom{2 m}{m})}{2^{2 m} m^{2}} \frac{2^{2 n}}{n^{2} (\binom{2 n}{n})}$

　　　 $\sum_{j = 0}^{r - 1} \sum_{0 < m < n} \frac{(- 1)^{m - 1}}{m^{j + 1} n^{r - j}} (\frac{(- 1)^{n}}{m} - \frac{1}{n}) = \sum_{0 < m < n} \frac{(- 1)^{n - 1}}{m n} (\frac{1}{m^{r}} - \frac{1}{n^{r}})$

　　　 $\sum_{m = 0}^{n - 1} \frac{(32 m + 17) 2^{6 m}}{(4 m + 1) (\binom{2 m}{m}) (\binom{4 m}{2 m})} = \frac{(4 n - 1) 2^{6 n}}{(\binom{2 n}{n}) (\binom{4 n}{2 n})} + 1$

$\begin{aligned} \frac{Γ (\frac{1}{2}) Γ (1 + a) Γ^{2} (b)}{Γ (a + b) Γ (\frac{1}{2} + b)} \frac{(1 + a)_{n}}{(a + b)_{n}} \sum_{0 \leq m} \frac{(\binom{2 m}{m}) (b)_{m}}{{(\frac{1}{2} + b)}_{m} 2^{2 m} (a + b + m + n)} \\ = & \frac{{(\frac{1}{2} + a)}_{n}}{{(\frac{1}{2} + a + b)}_{n}} (\sum_{0 \leq m} \frac{(1 - b)_{m}}{m! (\frac{1}{2} + a + m)} (γ + ψ (1 + a) + 2 \ln 2 + \sum_{k = 0}^{m - 1} \frac{1}{k + 1 + a}) + \frac{2}{1 + 2 a} \frac{Γ (1 + a) Γ (b)}{Γ (1 + a + b)} \sum_{m = 0}^{n - 1} \frac{{(\frac{1}{2} + a + b)}_{m} (1 + a)_{m}}{{(\frac{3}{2} + a)}_{m} (1 + a + b)_{m}}) \end{aligned}$
　　　　　　（ただし $a > - \frac{1}{2}$ ）

$\sum_{m = 0}^{\infty} \frac{(\binom{2 m}{m})}{2^{2 m}} \frac{{(\frac{1}{2} + a)}_{m}}{(1 + a)_{m} (a + m + n)} = \frac{Γ (a)}{Γ (\frac{1}{2}) Γ (\frac{1}{2} + a)} \frac{2^{2 n}}{n (\binom{2 n}{n})} \frac{(a)_{n}}{{(\frac{1}{2} + a)}_{n}} \sum_{m = 0}^{n - 1} \frac{(\binom{2 m}{m})}{2^{2 m}} \frac{{(\frac{1}{2} + a)}_{m}}{(1 + a)_{m}}$

$\underset{―}{}$

$Z_{C} (s) = \sum_{n = 0}^{\infty} \frac{(\binom{2 n}{n})}{2^{2 n} {(n + \frac{1}{2})}^{s}}, Z_{C^{2}} (s) = \sum_{n = 0}^{\infty} \frac{{(\binom{2 n}{n})}^{2}}{2^{4 n} {(n + \frac{1}{2})}^{s}}, T_{1} (s) = \sum_{n = 1}^{\infty} \frac{2^{2 n}}{n^{s} (\binom{2 n}{n})} \sum_{m = 0}^{n - 1} \frac{{(\binom{2 m}{m})}^{2}}{2^{4 m}}$