Abel's lemma の可能性？？　 ζ(2), ζ(3), β(2) などの加速級数の例 30 個以上

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$§ 1.$ 序言

$A b e l^{'} s l e m m a$

\begin{array}{r} \sum_{n = 0}^{\infty} B_{n} (A_{n} - A_{n - 1}) = lim_{n \to \infty} A_{n} B_{n + 1} - A_{- 1} B_{0} + \sum_{n = 0}^{\infty} A_{n} (B_{n} - B_{n + 1}) \end{array}

を基に，様々な等式を導出する。

$§ 2.$ 定義

$Re (1 + 2 a - b - c - d - e) > 0$ に対し

\begin{array}{r} Ω (a | b, c, d, e) = \sum_{n = 0}^{\infty} \frac{(a + 2 n) (b, c, d, e)_{n}}{(1 + a - b, 1 + a - c, 1 + a - d, 1 + a - e)_{n}} \end{array}

$§ 3.$ $C a s e$ $1$

\begin{aligned} A_{n} & = \frac{(- a + b + c + d, 1 + e)_{n}}{(2 + 2 a - b - c - d, 1 + a - e)_{n}} \\ B_{n} & = \frac{(b, c, d, 2 + 2 a - b - c - d)_{n}}{(1 + a - b, 1 + a - c, 1 + a - d, - 1 - a + b + c + d)_{n}} \end{aligned}

とし， $A b e l^{'} s l e m m a$ にあてはめれば

$\underset{―}{補題 1}$

\begin{array}{r} Ω (a | b, c, d, e) = \frac{(a - e) (1 + 2 a - b - c - d)}{1 + 2 a - b - c - d - e} + \frac{e (1 + a - b - c) (1 + a - b - d) (1 + a - c - d)}{(1 + a - b) (1 + a - c) (1 + a - d) (1 + 2 a - b - c - d - e)} Ω (1 + a | b, c, d, 1 + e) \end{array}

を得る。この繰り返しにより

\begin{aligned} Ω (a | b, c, d, e) & = \sum_{n = 0}^{m - 1} \frac{(a - e) (1 + 2 a - b - c - d + 2 n)}{1 + 2 a - b - c - d - e} \frac{(1 + a - b - c, 1 + a - b - d, 1 + a - c - d, e)_{n}}{(1 + a - b, 1 + a - c, 1 + a - d, 2 + 2 a - b - c - d - e)_{n}} \\ + \frac{(1 + a - b - c, 1 + a - b - d, 1 + a - c - d, e)_{m}}{(1 + a - b, 1 + a - c, 1 + a - d, 1 + 2 a - b - c - d - e)_{m}} Ω (m + a | b, c, d, m + e) \end{aligned}

を得る。いま， $m \to \infty$ に対して

\begin{aligned} \frac{(1 + a - b - c, 1 + a - b - d, 1 + a - c - d, e)_{m}}{(1 + a - b, 1 + a - c, 1 + a - d, 1 + 2 a - b - c - d - e)_{m}} & \approx C o n s t . \times m^{- 1 - 2 a + 2 e} \\ Ω (m + a | b, c, d, m + e) & \approx m + a \end{aligned}

ので， $Re (a - e) > 0$ の場合に第二項は $0$ に収束し

$\underset{―}{定理 1}$

\begin{array}{r} Ω (a | b, c, d, e) = \frac{a - e}{1 + 2 a - b - c - d - e} Ω (1 + 2 a - b - c - d | 1 + a - b - c, 1 + a - b - d, 1 + a - c - d, e) \end{array}

を得る。

$§ 4.$ $C a s e$ $2$

\begin{aligned} A_{n} & = \frac{(1 + c, 1 + e)_{n}}{(1 + a - c, 1 + a - e)_{n}} \\ B_{n} & = \frac{(b, d)_{n}}{(1 + a - b, 1 + a - d)_{n}} \end{aligned}

とすれば，

$\underset{―}{補題 2}$

\begin{array}{r} Ω (a | b, c, d, e) = \frac{(a - c) (a - e)}{a - c - e} + \frac{c e (1 + a - b - d)}{(1 + a - b) (1 + a - d) (- a + c + e)} Ω (1 + a | b, 1 + c, d, 1 + e) \end{array}

を得る。

$§ 5.$ $C a s e$ $3$

\begin{aligned} A_{n} & = \frac{(1 + b, 1 + c, 1 + d, 1 + 2 a - b - c - d)_{n}}{(1 + a - b, 1 + a - c, 1 + a - d, 1 - a + b + c + d)_{n}} \\ B_{n} & = \frac{(1 - a + b + c + d, e)_{n}}{(2 a - b - c - d, 1 + a - e)_{n}} \end{aligned}

とすれば，

$\underset{―}{補題 3}$

\begin{array}{r} Ω (a | b, c, d, e) = \frac{(a - b) (a - c) (a - d) (a - b - c - d)}{(a - b - c) (a - b - d) (a - c - d)} + \frac{b c d (2 a - b - c - d - e)}{(a - b - c) (a - b - d) (a - c - d) (1 + a - e)} Ω (1 + a | 1 + b, 1 + c, 1 + d, e) \end{array}

を得る。

$§ 6.$ 定理 $2$

対称性より

\begin{array}{r} Ω (1 + a | b, c, d, 1 + e) = Ω (1 + a | b, c, 1 + e, d) \end{array}

左辺に補題 $1$ ，右辺に補題 $2$ を用いれば

\begin{aligned} Ω (1 + a | b, c, d, 1 + e) & = - \frac{(1 + a - b) (1 + a - c) (1 + a - d) (a - e) (1 + 2 a - b - c - d)}{e (1 + a - b - c) (1 + a - b - d) (1 + a - c - d)} + \frac{(1 + a - b) (1 + a - c) (1 + a - d) (1 + 2 a - b - c - d - e)}{e (1 + a - b - c) (1 + a - b - d) (1 + a - c - d)} Ω (a | b, c, d, e) \\ Ω (1 + a | b, c, 1 + e, d) & = \frac{(1 + a - c) (1 + a - d)}{1 + a - c - d} - \frac{c d (1 + a - b - e)}{(2 + a - b) (1 + a - e) (1 + a - c - d)} Ω (2 + a | b, 1 + c, 1 + d, 1 + e) \end{aligned}

より

$\underset{―}{補題 4}$

\begin{aligned} Ω (a | b, c, d, e) & = \frac{(1 + 2 a - b - c - d, a - e)_{1}}{(1 + 2 a - b - c - d - e)_{1}} + \frac{(e, 1 + a - b - c, 1 + a - b - d)_{1}}{(1 + a - b, 1 + 2 a - b - c - d - e)_{1}} \\ - \frac{(c, d, e, 1 + a - b - c, 1 + a - b - d, 1 + a - b - e)_{1}}{(1 + a - c, 1 + a - d, 1 + a - e, 1 + 2 a - b - c - d - e)_{1}} \frac{Ω (2 + a | b, 1 + c, 1 + d, 1 + e)}{(1 + a - b)_{2}} \end{aligned}

$\underset{―}{定理 2}$

\begin{array}{r} Ω (a | b, c, d, e) = \sum_{n = 0}^{\infty} \frac{(- 1)^{n} (c, d, e, 1 + a - b - c, 1 + a - b - d, 1 + a - b - e)_{n}}{(1 + a - c, 1 + a - d, 1 + a - e, 1 + 2 a - b - c - d - e)_{n} (1 + a - b)_{2 n}} P_{n} \end{array}

\begin{array}{r} (P_{n} = \frac{(1 + 2 a - b - c - d + 2 n) (a - e + n)}{1 + 2 a - b - c - d - e + n} + \frac{(1 + a - b - c + n) (1 + a - b - d + n) (e + n)}{(1 + a - b + 2 n) (1 + 2 a - b - c - d - e + n)}) \end{array}

$§ 7.$ 定理 $3$

対称性より

\begin{array}{r} Ω (1 + a | b, c, d, 1 + e) = Ω (1 + a, b, 1 + e, d, c) \end{array}

左辺と右辺に補題 $1$ を用いれば

\begin{aligned} Ω (1 + a | b, c, d, 1 + e) & = - \frac{(1 + a - b) (1 + a - c) (1 + a - d) (a - e) (1 + 2 a - b - c - d)}{e (1 + a - b - c) (1 + a - b - d) (1 + a - c - d)} + \frac{(1 + a - b) (1 + a - c) (1 + a - d) (1 + 2 a - b - c - d - e)}{e (1 + a - b - c) (1 + a - b - d) (1 + a - c - d)} Ω (a | b, c, d, e) \\ Ω (1 + a | b, 1 + e, d, c) & = \frac{(1 + a - c) (2 + 2 a - b - d - e)}{2 + 2 a - b - c - d - e} + \frac{c (2 + a - b - d) (1 + a - b - e) (1 + a - d - e)}{(2 + a - b) (2 + a - d) (1 + a - e) (2 + 2 a - b - c - d - e)} Ω (2 + a | b, 1 + c, d, 1 + e) \end{aligned}

より

$\underset{―}{補題 5}$

\begin{aligned} Ω (a | b, c, d, e) & = \frac{(1 + 2 a - b - c - d) (a - e)}{1 + 2 a - b - c - d - e} + \frac{e (1 + a - b - c) (1 + a - b - d) (1 + a - c - d) (2 + 2 a - b - d - e)}{(1 + a - b) (1 + a - d) (1 + 2 a - b - c - d - e)_{2}} \\ + \frac{c e (1 + a - b - c) (1 + a - b - e) (1 + a - c - d) (1 + a - d - e) (1 + a - b - d)_{2}}{(1 + a - c) (1 + a - e) (1 + a - b, 1 + a - d, 1 + 2 a - b - c - d - e)_{2}} Ω (2 + a | b, 1 + c, d, 1 + e) \end{aligned}

$\underset{―}{定理 3}$

\begin{array}{r} Ω (a | b, c, d, e) = \sum_{n = 0}^{\infty} \frac{(c, e, 1 + a - b - c, 1 + a - b - e, 1 + a - c - d, 1 + a - d - e)_{n} (1 + a - b - d)_{2 n}}{(1 + a - c, 1 + a - e)_{n} (1 + a - b, 1 + a - d, 1 + 2 a - b - c - d - e)_{2 n}} P_{n} \end{array}

\begin{array}{r} (P_{n} = \frac{(1 + 2 a - b - c - d + 3 n) (a - e + n)}{1 + 2 a - b - c - d - e + 2 n} + \frac{(e + n) (1 + a - b - c + n) (1 + a - b - d + 2 n) (1 + a - c - d + n) (2 + 2 a - b - d - e + 3 n)}{(1 + a - b + 2 n) (1 + a - d + 2 n) (1 + 2 a - b - c - d - e + 2 n)_{2}}) \end{array}

$§ 8.$ アルゴリズム

上述のように，対称性と補題 $1$ , $2$ などを用いて $Ω (a | b, c, d, e) = A (a, b, c, d, e) + B (a, b, c, d, e) Ω (p + a | q + b, r + c, s + d, t + e)$ のかたちを作ることができ，

\begin{aligned} Ω (a | b, c, d, e) & = \sum_{n = 0}^{m - 1} A (p n + a, q n + b, r n + c, s n + d, t n + e) \prod_{k = 0}^{n - 1} B (p k + a, q k + b, r k + c, s k + d, t k + e) \\ + Ω (p m + a | q m + b, r m + c, s m + d, t m + e) \prod_{k = 0}^{m - 1} B (p k + a, q k + b, r k + c, s k + d, t k + e) \end{aligned}

および

\begin{aligned} Ω (a | b, c, d, e) & = \sum_{n = 0}^{\infty} A (p n + a, q n + b, r n + c, s n + d, t n + e) \prod_{k = 0}^{n - 1} B (p k + a, q k + b, r k + c, s k + d, t k + e) \\ + lim_{m \to \infty} Ω (p m + a | q m + b, r m + c, s m + d, t m + e) \prod_{k = 0}^{m - 1} B (p k + a, q k + b, r k + c, s k + d, t k + e) \end{aligned}

を得る。

$§ 9.$ 定理のまとめ

$\underset{―}{定理 9.0}$

\begin{array}{r} Ω (a | b, c, d, e) = \frac{a - e}{1 + 2 a - b - c - d - e} Ω (1 + 2 a - b - c - d | 1 + a - b - c, 1 + a - b - d, 1 + a - c - d, e) \end{array}

$\underset{―}{定理 9.1}$

\begin{array}{r} Ω (a | b, c, d, e) = \sum_{n = 0}^{\infty} \frac{(- 1)^{n} (c, d, e, 1 + a - b - c, 1 + a - b - d, 1 + a - b - e)_{n}}{(1 + a - c, 1 + a - d, 1 + a - e, 1 + 2 a - b - c - d - e)_{n} (1 + a - b)_{2 n}} R_{n}^{(1)} \end{array}

\begin{array}{r} (R_{n}^{(1)} = \frac{(1 + 2 a - b - c - d + 2 n) (a - e + n)}{1 + 2 a - b - c - d - e + n} + \frac{(1 + a - b - c + n) (1 + a - b - d + n) (e + n)}{(1 + a - b + 2 n) (1 + 2 a - b - c - d - e + n)}) \end{array}

$\underset{―}{定理 9.2}$

\begin{array}{r} Ω (a | b, c, d, e) = \sum_{n = 0}^{\infty} \frac{(c, e, 1 + a - b - c, 1 + a - b - e, 1 + a - c - d, 1 + a - d - e)_{n} (1 + a - b - d)_{2 n}}{(1 + a - c, 1 + a - e)_{n} (1 + a - b, 1 + a - d, 1 + 2 a - b - c - d - e)_{2 n}} R_{n}^{(2)} \end{array}

\begin{array}{r} (R_{n}^{(2)} = \frac{(1 + 2 a - b - c - d + 3 n) (a - e + n)}{1 + 2 a - b - c - d - e + 2 n} + \frac{(e + n) (1 + a - b - c + n) (1 + a - b - d + 2 n) (1 + a - c - d + n) (2 + 2 a - b - d - e + 3 n)}{(1 + a - b + 2 n) (1 + a - d + 2 n) (1 + 2 a - b - c - d - e + 2 n)_{2}}) \end{array}

$\underset{―}{定理 9.3}$

\begin{array}{r} Ω (a | b, c, d, e) = \sum_{n = 0}^{\infty} \frac{(- 1)^{n} (c, d, 1 + a - b - e, 1 + a - c - d)_{n} (e, 1 + a - b - c, 1 + a - b - d)_{2 n}}{(1 + a - e)_{n} (1 + a - c, 1 + a - d, 1 + 2 a - b - c - d - e)_{2 n} (1 + a - b)_{3 n}} R_{n}^{(3)} \end{array}

\begin{array}{r} (R_{n}^{(3)} = \frac{(1 + 2 a - b - c - d + 4 n) (a - e + n)}{1 + 2 a - b - c - d - e + 2 n} + \frac{(e + 2 n) (1 + a - b - c + 2 n) (1 + a - b - d + 2 n) (1 + a - c - d + n) (2 + 2 a - b - d - e + 3 n)}{(1 + a - b + 3 n) (1 + a - d + 2 n) (1 + 2 a - b - c - d - e + 2 n)_{2}} + \frac{(c + n) (e + 2 n) (1 + a - b - c + 2 n) (1 + a - b - e + n) (1 + a - c - d + n) (1 + a - b - d + 2 n)_{2}}{(1 + a - c + 2 n) (1 + a - d + 2 n) (1 + a - b + 3 n, 1 + 2 a - b - c - d - e + 2 n)_{2}}) \end{array}

$\underset{―}{定理 9.4}$

\begin{array}{r} Ω (a | b, c, d, e) = \sum_{n = 0}^{\infty} \frac{(1 + a - b - c, 1 + a - b - e, 1 + a - c - d, 1 + a - d - e)_{n} (c, e)_{2 n} (1 + a - b - d)_{3 n}}{(1 + a - c, 1 + a - e, c + e - a)_{n} (1 + 2 a - b - c - d - e)_{2 n}} R_{n}^{(4)} \end{array}

\begin{array}{r} (R_{n}^{(4)} = \frac{(1 + 2 a - b - c - d + 4 n) (a - e + n)}{1 + 2 a - b - c - d - e + 2 n} + \frac{(e + 2 n) (1 + a - b - c + n) (1 + a - b - d + 3 n) (1 + a - c - d + n) (2 + 2 a - b - d - e + 4 n)}{(1 + a - b + 3 n) (1 + a - d + 3 n) (1 + 2 a - b - c - d - e + 2 n)_{2}} - \frac{(c + 2 n) (e + 2 n) (1 + a - b - c + n) (1 + a - b - e + n) (1 + a - c - d + n) (1 + a - d - e + n) (1 + a - b - d + 3 n)_{2}}{(c + e - a + n) (1 + a - b + 3 n, 1 + a - d + 3 n, 1 + 2 a - b - c - d - e + 2 n)_{2}}) \end{array}

$\underset{―}{定理 9.5}$

\begin{array}{r} Ω (a | b, c, d, e) = \sum_{n = 0}^{\infty} \frac{(- 1)^{n} (c, 1 + a - b - d, 1 + a - b - e)_{n} (d, e, 1 + a - b - c)_{2 n}}{(d + e - a, 1 + a - d, 1 + a - e, 1 + 2 a - b - c - d - e)_{n} (1 + a - c)_{2 n} (1 + a - b)_{3 n}} R_{n}^{(5)} \end{array}

\begin{array}{r} (R_{n}^{(5)} = \frac{(1 + 2 a - b - c - d + 3 n) (a - e + n)}{1 + 2 a - b - c - d - e + n} + \frac{(e + 2 n) (1 + a - b - c + 2 n) (1 + a - b - d + n)}{(1 + a - b + 3 n) (1 + 2 a - b - c - d - e + n)} + \frac{(c + n) (d + 2 n) (e + 2 n) (1 + a - b - c + 2 n) (1 + a - b - d + n) (1 + a - b - e + n)}{(- a + d + e + n) (1 + a - c + 2 n) (1 + 2 a - b - c - d - e + n) (1 + a - b + 3 n)_{2}}) \end{array}

$\underset{―}{定理 9.6}$

\begin{aligned} Ω (a | b, c, d, e) & = \frac{Γ (1 + a - b) Γ (1 + a - c) Γ (1 + a - d) Γ (1 + a - e) Γ (- a + b + e) Γ (- a + c + e) Γ (- a + d + e) Γ (1 + 2 a - b - c - d - e)}{Γ (b) Γ (c) Γ (d) Γ (e) Γ (1 + a - b - c) Γ (1 + a - b - d) Γ (1 + a - c - d)} \\ + \sum_{n = 0}^{\infty} \frac{(b, c, d, 1 + a - b - c, 1 + a - b - d, 1 + a - c - d)_{n} (e)_{3 n}}{(- a + b + e, - a + c + e, - a + d + e)_{n} (1 + a - b, 1 + a - c, 1 + a - d)_{2 n}} R_{n}^{(6)} \end{aligned}

\begin{array}{r} (R_{n}^{(6)} = \frac{(a - c + 2 n) (a - e)}{a - c - e - n} - \frac{(c + n) (e + 3 n) (a - e) (1 + a - b - d + n)}{(1 + a - d + 2 n) (a - b - e - n) (a - c - e - n)} + \frac{(b + n) (c + n) (e + 3 n) (1 + e + 3 n) (a - e) (1 + a - b - d + n) (1 + a - c - d + n)}{(1 + a - b + 2 n) (1 + a - c + 2 n) (1 + a - d + 2 n) (a - b - e - n) (a - c - e - n) (a - d - e - n)}) \end{array}

$\underset{―}{定理 9.7}$

\begin{array}{r} Ω (a | b, c, d, e) = \sum_{n = 0}^{\infty} \frac{(- 1)^{n} (b, c, d, e, 1 + a - b - c, 1 + a - b - d, 1 + a - b - e, 1 + a - c - d, 1 + a - c - e, 1 + a - d - e)_{n}}{(1 + a - b, 1 + a - c, 1 + a - d, 1 + a - e, 1 + 2 a - b - c - d - e)_{2 n}} R_{n}^{(7)} \end{array}

\begin{array}{r} (R_{n}^{(7)} = \frac{(1 + 2 a - b - c - d + 3 n) (a - e + 2 n)}{1 + 2 a - b - c - d - e + 2 n} + \frac{(e + n) (1 + a - b - c + n) (1 + a - b - d + n) (1 + a - c - d + n) (2 + 2 a - b - d - e + 3 n)}{(1 + a - b + 2 n) (1 + a - d + 2 n) (1 + 2 a - b - c - d - e + 2 n)_{2}} + \frac{(c + n) (e + n) (1 + a - b - c + n) (1 + a - b - d + n) (1 + a - b - e + n) (1 + a - c - d + n) (1 + a - d - e + n)}{(1 + a - b + 2 n) (1 + a - c + 2 n) (1 + a - d + 2 n) (1 + a - e + 2 n) (1 + 2 a - b - c - d - e + 2 n)_{2}}) \end{array}

$§ 10.$ 系のまとめ

$\underset{―}{系 10.0}$

\begin{array}{r} Ω (\frac{3}{2} | 1, 1, 1, \frac{1}{2}) = 2 Ω (1 | \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}) = \frac{7}{4} ζ (3) \end{array}

\begin{array}{r} Ω (\frac{3}{2} | 1, 1, 1, - \infty) = Ω (1 | \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, - \infty) = β (2) \end{array}

$\underset{―}{系 10.1}$

$∙ (a, b, c, d, e) = (1, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2})$

\begin{array}{r} \frac{21}{2} ζ (3) = \sum_{n = 0}^{\infty} \frac{(- 1)^{n}}{2^{2 n}} \frac{n!^{2}}{(\frac{5}{4}, \frac{7}{4})_{n}} \frac{20 n^{2} + 32 n + 13}{(n + 1) (2 n + 1)^{2}} \end{array}

$∙ (a, b, c, d, e) = (\frac{3}{2}, \frac{1}{2}, 1, 1, 1)$

\begin{array}{r} \frac{7}{2} ζ (3) = \sum_{n = 0}^{\infty} \frac{(- 1)^{n}}{2^{2 n}} \frac{n!^{5}}{(\frac{3}{2})_{n}^{5}} (10 n^{2} + 14 n + 5) \end{array}

$∙ (a, b, c, d, e) = (1, 1, \frac{1}{2}, \frac{1}{2}, \frac{1}{2})$

\begin{array}{r} \frac{3}{5} ζ (2) = \sum_{n = 0}^{\infty} \frac{(- 1)^{n}}{2^{2 n}} \frac{(\frac{1}{2})_{n}^{3}}{n! (\frac{3}{2})_{n}^{2}} \end{array}

$∙ (a, b, c, d, e) = (2, 2, \frac{1}{2}, \frac{1}{2}, \frac{1}{2})$

\begin{array}{r} \frac{243}{16} ζ (2) = \sum_{n = 0}^{\infty} \frac{(- 1)^{n}}{2^{2 n}} \frac{(\frac{1}{2})_{n}^{5}}{n! (\frac{5}{2})_{n}^{4}} (20 n^{2} + 44 n + 25) \end{array}

$∙ (a, b, c, d, e) = (\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2})$

\begin{array}{r} \frac{4}{3 ζ (2)} = \sum_{n = 0}^{\infty} \frac{(- 1)^{n}}{2^{2 n}} \frac{(\frac{1}{2})_{n}^{5}}{n!^{5}} (20 n^{2} + 8 n + 1) \end{array}

$∙ (a, b, c, d, e) = (\frac{3}{2}, \frac{3}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2})$

\begin{array}{r} \frac{64}{3 ζ (2)} = \sum_{n = 0}^{\infty} \frac{(- 1)^{n}}{2^{2 n}} \frac{(\frac{1}{2})_{n}^{5}}{n!^{5}} \frac{20 n^{2} + 32 n + 13}{(n + 1)^{4}} \end{array}

$∙ (a, b, c, d, e) = (1, \frac{3}{4}, 1, \frac{1}{4}, \frac{1}{2})$

\begin{array}{r} \frac{5 π}{8} = \sum_{n = 0}^{\infty} \frac{(- 1)^{n}}{2^{2 n}} \frac{n! (\frac{1}{4})_{n}^{2}}{(\frac{3}{2}, \frac{9}{8}, \frac{13}{8})_{n}} (5 n + 2) \end{array}

$∙ (a, b, c, d, e) = (\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, - \infty)$

\begin{array}{r} \frac{4}{π} = \sum_{n = 0}^{\infty} \frac{1}{2^{2 n}} \frac{(\frac{1}{2})_{n}^{3}}{n!^{3}} (6 n + 1) \end{array}

$∙ (a, b, c, d, e) = (\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{4}, \frac{3}{4})$

\begin{array}{r} \frac{8}{π} = \sum_{n = 0}^{\infty} \frac{(- 1)^{n}}{2^{2 n}} \frac{(\frac{1}{2}, \frac{1}{4}, \frac{3}{4})_{n}}{n!^{3}} (20 n + 3) \end{array}

$\underset{―}{系 10.2}$

$∙ (a, b, c, d, e) = (2, 1, 1, 1, 1)$

\begin{array}{r} 16 ζ (3) = \sum_{n = 0}^{\infty} \frac{1}{2^{4 n}} \frac{n!^{2}}{(\frac{3}{2})_{n}^{2}} \frac{30 n + 19}{(n + 1) (2 n + 1)} \end{array}

$∙ (a, b, c, d, e) = (\frac{3}{2}, 1, 1, 1, \frac{1}{2})$

\begin{array}{r} \frac{7}{4} ζ (3) = \sum_{n = 0}^{\infty} \frac{1}{2^{2 n}} \frac{n!^{2}}{(\frac{3}{2})_{n}^{2}} \frac{3 n + 2}{(n + 1) (2 n + 1)} \end{array}

$∙ (a, b, c, d, e) = (1, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2})$

\begin{array}{r} \frac{63}{2} ζ (3) = \sum_{n = 0}^{\infty} \frac{1}{2^{4 n}} \frac{n!^{4}}{(\frac{5}{4}, \frac{7}{4})_{n}^{2}} \frac{60 n^{2} + 94 n + 37}{2 n + 1} \end{array}

$∙ (a, b, c, d, e) = (1, \frac{1}{2}, 1, \frac{1}{2}, \frac{1}{2})$

\begin{array}{r} \frac{3}{2} ζ (2) = \sum_{n = 0}^{\infty} \frac{1}{2^{2 n}} \frac{n!^{3}}{(\frac{3}{2})_{n}^{3}} (3 n + 2) \end{array}

$∙ (a, b, c, d, e) = (\frac{3}{2}, \frac{1}{2}, 1, \frac{1}{2}, 1)$

\begin{array}{r} 16 ζ (2) = \sum_{n = 0}^{\infty} \frac{1}{2^{4 n}} \frac{n!^{3} (\frac{3}{4}, \frac{5}{4})_{n}}{(\frac{3}{2})_{n}^{5}} (60 n^{2} + 77 n + 25) \end{array}

$∙ (a, b, c, d, e) = (\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2})$

\begin{array}{r} \frac{16}{3 ζ (2)} = \sum_{n = 0}^{\infty} \frac{1}{2^{4 n}} \frac{(\frac{1}{2})_{n}^{3} (\frac{1}{4}, \frac{3}{4})_{n}}{n!^{5}} (120 n^{2} + 34 n + 3) \end{array}

$∙ (a, b, c, d, e) = (1, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, - \infty)$

\begin{array}{r} 18 β (2) = \sum_{n = 0}^{\infty} \frac{(- 1)^{n}}{2^{2 n}} \frac{n!^{3} (\frac{1}{2})_{n}}{(\frac{5}{4}, \frac{7}{4})_{n}^{2}} (40 n^{2} + 56 n + 19) \end{array}

$\underset{―}{系 10.3}$

$∙ (a, b, c, d, e) = (2, 1, 1, 1, 1)$

\begin{array}{r} 24 ζ (3) = \sum_{n = 0}^{\infty} \frac{(- 1)^{n}}{3^{3 n}} \frac{n!^{2}}{(\frac{4}{3}, \frac{5}{3})_{n}} \frac{56 n^{2} + 80 n + 29}{(n + 1) (2 n + 1)^{2}} \end{array}

$∙ (a, b, c, d, e) = (\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2})$

\begin{array}{r} \frac{128}{3 ζ (2)} = \sum_{n = 0}^{\infty} \frac{(- 1)^{n}}{3^{3 n}} \frac{(\frac{1}{2})_{n} (\frac{1}{4}, \frac{3}{4})_{n}^{3}}{n!^{5} {(\frac{4}{3}, \frac{5}{3})}_{n}} (7168 n^{4} + 8832 n^{3} + 3376 n^{2} + 492 n + 27) \end{array}

$\underset{―}{系 10.4}$

$∙ (a, b, c, d, e) = (\frac{3}{2}, \frac{1}{2}, 1, \frac{1}{2}, 1)$

\begin{array}{r} 24 ζ (2) = \sum_{n = 0}^{\infty} \frac{2^{2 n}}{3^{3 n}} \frac{n!^{3} (\frac{5}{6}, \frac{7}{6})_{n}}{(\frac{3}{2})_{n} {(\frac{4}{3}, \frac{5}{3})}_{n}^{2}} (69 n^{2} + 98 n + 35) \end{array}

$∙ (a, b, c, d, e) = (2, 1, 1, 1, - \infty)$

\begin{array}{r} 18 ζ (2) = \sum_{n = 0}^{\infty} \frac{(- 1)^{n} 2^{2 n}}{3^{3 n}} \frac{n! (\frac{1}{2})_{n}}{{(\frac{4}{3}, \frac{5}{3})}_{n}} \frac{279 n^{3} + 504 n^{2} + 302 n + 61}{(n + 1) (3 n + 1) (3 n + 2)} \end{array}

$\underset{―}{系 10.5}$

$∙ (a, b, c, d, e) = (\frac{3}{2}, 1, \frac{1}{2}, 1, 1)$

\begin{array}{r} \frac{105}{2} ζ (3) = \sum_{n = 0}^{\infty} \frac{(- 1)^{n} 2^{4 n}}{3^{3 n}} \frac{n!^{2}}{(\frac{7}{6}, \frac{11}{6})_{n}} \frac{86 n^{2} + 151 n + 67}{(n + 1) (2 n + 1)^{2}} \end{array}

$∙ (a, b, c, d, e) = (\frac{3}{2}, \frac{1}{2}, 1, 1, 1)$

\begin{array}{r} 63 ζ (3) = \sum_{n = 0}^{\infty} \frac{(- 1)^{n} 2^{4 n}}{3^{3 n}} \frac{n!^{5}}{(\frac{3}{2}, \frac{4}{3}, \frac{5}{3}, \frac{5}{4}, \frac{7}{4})_{n}} (172 n^{2} + 269 n + 106) \end{array}

$∙ (a, b, c, d, e) = (2, 1, 1, 1, \frac{3}{2})$

\begin{array}{r} 24 ζ (2) = \sum_{n = 0}^{\infty} \frac{(- 1)^{n} 2^{4 n}}{3^{3 n}} \frac{n! (\frac{3}{4}, \frac{5}{4})_{n}}{(\frac{3}{2}, \frac{4}{3}, \frac{5}{3})_{n}} \frac{86 n^{2} + 124 n + 45}{(n + 1) (2 n + 1)} \end{array}

$∙ (a, b, c, d, e) = (\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2})$

\begin{array}{r} \frac{32}{3 ζ (2)} = \sum_{n = 0}^{\infty} \frac{(- 1)^{n} 2^{4 n}}{3^{3 n}} \frac{(\frac{1}{2})_{n} (\frac{1}{4}, \frac{3}{4})_{n}^{3}}{n!^{5} (\frac{4}{3}, \frac{5}{3})_{n}} (1376 n^{4} + 1808 n^{3} + 784 n^{2} + 138 n + 9) \end{array}

$∙ (a, b, c, d, e) = (\frac{3}{2}, 1, - \infty, 1, 1)$

\begin{array}{r} 30 β (2) = \sum_{n = 0}^{\infty} \frac{2^{4 n}}{3^{3 n}} \frac{n!^{2}}{(\frac{7}{6}, \frac{11}{6})_{n}} \frac{22 n + 21}{2 n + 1} \end{array}

$\underset{―}{系 10.6}$

$∙ (a, b, c, d, e) = (\frac{3}{2}, \frac{1}{2}, 1, 1, 1)$

\begin{array}{r} \frac{27}{2} ζ (2) = \sum_{n = 0}^{\infty} \frac{3^{3 n}}{2^{6 n}} \frac{n!^{3} (\frac{1}{3}, \frac{2}{3})_{n}}{(\frac{3}{2})_{n} (\frac{5}{4}, \frac{7}{4})_{n}^{2}} (74 n^{3} + 138 n^{2} + 83 n + 16) \end{array}

$∙ (a, b, c, d, e) = (2, 1, 1, 1, \frac{3}{2})$

\begin{array}{r} 32 ζ (2) = \sum_{n = 0}^{\infty} \frac{3^{3 n}}{2^{6 n}} \frac{n!^{3} (\frac{5}{6}, \frac{7}{6})_{n}}{(\frac{3}{2})_{n}^{5}} (74 n^{2} + 101 n + 35) \end{array}

$∙ (a, b, c, d, e) = (\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, 0)$

\begin{array}{r} \frac{8}{ζ (2)} = \sum_{n = 0}^{\infty} \frac{3^{3 n}}{2^{6 n}} \frac{(\frac{1}{2})_{n}^{3} (\frac{1}{3}, \frac{2}{3})_{n}}{n!^{5}} (74 n^{2} + 27 n + 3) \end{array}

$∙ (a, b, c, d, e) = (1, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, - \infty)$

\begin{array}{r} 2 β (2) = \sum_{n = 0}^{\infty} \frac{(- 1)^{n}}{2^{3 n}} \frac{n!^{3}}{(\frac{3}{2})_{n}^{3}} (3 n + 2) \end{array}

$\underset{―}{系 10.7}$

$∙ (a, b, c, d, e) = (2, 1, 1, 1, 1)$

\begin{array}{r} 64 ζ (3) = \sum_{n = 0}^{\infty} \frac{(- 1)^{n}}{2^{10 n}} \frac{n!^{5}}{(\frac{3}{2})_{n}^{5}} (205 n^{2} + 250 n + 77) \end{array}

$∙ (a, b, c, d, e) = (1, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2})$

\begin{array}{r} \frac{567}{2} ζ (3) = \sum_{n = 0}^{\infty} \frac{(- 1)^{n}}{2^{10 n}} \frac{n!^{5} (\frac{1}{2})_{n}^{3}}{(\frac{5}{4}, \frac{7}{4})_{n}^{4}} (3280 n^{5} + 10888 n^{4} + 14236 n^{3} + 9122 n^{2} + 2844 n + 341) \end{array}

$∙ (a, b, c, d, e) = (2, 1, 1, 1, \frac{3}{2})$

\begin{array}{r} 144 ζ (2) = \sum_{n = 0}^{\infty} \frac{(- 1)^{n}}{2^{10 n}} \frac{n!^{3} (\frac{1}{2})_{n}}{(\frac{5}{4}, \frac{7}{4})_{n}^{2}} (410 n^{2} + 623 n + 237) \end{array}

$∙ (a, b, c, d, e) = (2, 1, 1, 1, - \infty)$

\begin{array}{r} 8 ζ (2) = \sum_{n = 0}^{\infty} \frac{1}{2^{6 n}} \frac{n!^{3}}{(\frac{3}{2})_{n}^{3}} (21 n + 13) \end{array}

$∙ (a, b, c, d, e) = (\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2})$

\begin{array}{r} \frac{64}{3 ζ (2)} = \sum_{n = 0}^{\infty} \frac{(- 1)^{n}}{2^{10 n}} \frac{(\frac{1}{2})_{n}^{5}}{n!^{5}} (820 n^{2} + 180 n + 13) \end{array}

投稿日：2022年11月23日

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Abel's lemma の可能性？？　 ζ(2), ζ(3), β(2) などの加速級数の例 30 個以上

$\S 1.$序言
$\S 2.$定義
$\S 3.$$\rm Case$$1$
$\S 4.$$\rm Case$$2$
$\S 5.$$\rm Case$$3$
$\S 6.$定理$2$
$\S 7.$定理$3$
$\S 8.$アルゴリズム
$\S 9.$定理のまとめ
$\S 10.$系のまとめ

Abel's lemma の可能性？？ ζ(2), ζ(3), β(2) などの加速級数の例 30 個以上

§§1.序言

§§2.定義

§§3.Case1

§§4.Case2

§§5.Case3

§§6.定理2

§§7.定理3

§§8.アルゴリズム

§§9.定理のまとめ

§§10.系のまとめ

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Abel's lemma の可能性？？　 ζ(2), ζ(3), β(2) などの加速級数の例 30 個以上

$§ 1.$ 序言

$§ 2.$ 定義

$§ 3.$ $C a s e$ $1$

$§ 4.$ $C a s e$ $2$

$§ 5.$ $C a s e$ $3$

$§ 6.$ 定理 $2$

$§ 7.$ 定理 $3$

$§ 8.$ アルゴリズム

$§ 9.$ 定理のまとめ

$§ 10.$ 系のまとめ