二項係数の3乗が入った級数その2

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この記事では, 二項係数の3乗が入った2つの級数

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

に証明を与える. 用いる定理は以下である.

Dixonの恒等式

$\begin{array}{r} \sum_{n = 0}^{\infty} \frac{(a)_{n} (b)_{n} (c)_{n}}{n! (1 + a - b)_{n} (1 + a - c)_{n}} = \frac{Γ (1 + \frac{a}{2}) Γ (1 + a - b) Γ (1 + a - c) Γ (1 + \frac{a}{2} - b - c)}{Γ (1 + a) Γ (1 + \frac{a}{2} - b) Γ (1 + \frac{a}{2} - c) Γ (1 + a - b - c)} \end{array}$

まず, これを $a$ に関して偏微分すると,

$\begin{aligned} \sum_{n = 0}^{\infty} \frac{(a)_{n} (b)_{n} (c)_{n}}{n! (1 + a - b)_{n} (1 + a - c)_{n}} \sum_{k = 1}^{n} (\frac{1}{k + a - 1} - \frac{1}{k + a - b} - \frac{1}{k + a - c}) \\ = \frac{Γ (1 + \frac{a}{2}) Γ (1 + a - b) Γ (1 + a - c) Γ (1 + \frac{a}{2} - b - c)}{Γ (1 + a) Γ (1 + \frac{a}{2} - b) Γ (1 + \frac{a}{2} - c) Γ (1 + a - b - c)} (\frac{1}{2} (ψ (1 + \frac{a}{2}) - ψ (1 + \frac{a}{2} - b) - ψ (1 + \frac{a}{2} - c) + ψ (1 + \frac{a}{2} - b - c)) - ψ (1 + a) + ψ (1 + a - b) + ψ (1 + a - c) - ψ (1 + a - b - c)) \end{aligned}$

ここに $a = b = c = \frac{1}{2}$ を代入すると,

$\begin{aligned} \sum_{n = 0}^{\infty} \frac{{(\binom{2 n}{n})}^{3}}{2^{6 n}} \sum_{k = 1}^{n} (\frac{1}{k - \frac{1}{2}} - \frac{2}{k}) & = \frac{π}{Γ {(\frac{3}{4})}^{4}} (\frac{1}{2} (ψ (\frac{5}{4}) - 2 ψ (\frac{3}{4}) + ψ (\frac{1}{4})) - ψ (\frac{3}{2}) + 2 ψ (1) - ψ (\frac{1}{2})) \\ = \frac{π}{Γ {(\frac{3}{4})}^{4}} (4 \ln 2 - π) \end{aligned}$

また, Dixonの恒等式を $b$ に関して偏微分すると,

$\begin{aligned} \sum_{n = 0}^{\infty} \frac{(a)_{n} (b)_{n} (c)_{n}}{n! (1 + a - b)_{n} (1 + a - c)_{n}} \sum_{k = 1}^{n} (\frac{1}{k + b - 1} + \frac{1}{k + a - b}) \\ = \frac{Γ (1 + \frac{a}{2}) Γ (1 + a - b) Γ (1 + a - c) Γ (1 + \frac{a}{2} - b - c)}{Γ (1 + a) Γ (1 + \frac{a}{2} - b) Γ (1 + \frac{a}{2} - c) Γ (1 + a - b - c)} (ψ (1 + \frac{a}{2} - b) - ψ (1 + \frac{a}{2} - b - c) - ψ (1 + a - b) + ψ (1 + a - b - c)) \end{aligned}$

これに $a = b = c = \frac{1}{2}$ を代入すると,

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

よって,

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