級数botⅡの級数計算まとめ

320

級数botⅡ が投稿する数式の証明をまとめる。

No. 20

\begin{array}{r} ζ (3) = \frac{2 π}{7} \sum_{0 \leq m \leq n} \frac{{(\binom{2 m}{m})}^{2}}{2^{4 m} (2 n + 1)^{2}} \end{array}

証明

\begin{aligned} \frac{π}{2} \sum_{0 \leq m \leq n} \frac{{(\binom{2 m}{m})}^{2}}{2^{4 m} (2 n + 1)^{2}} & = \sum_{0 \leq n} \frac{1}{(2 n + 1)^{2}} \sum_{m = 0}^{n} \frac{(\binom{2 m}{m})}{2^{2 m}} \int_{0}^{1} \frac{x^{2 m}}{\sqrt{1 - x^{2}}} d x \\ = \sum_{0 \leq n} \frac{1}{(2 n + 1)^{2}} \int_{0}^{1} \frac{1}{1 - x^{2}} (1 - (2 n + 1) \frac{(\binom{2 n}{n})}{2^{2 n}} \int_{0}^{x} \frac{t^{2 n + 1}}{\sqrt{1 - t^{2}}} d t) d x \\ = \int_{0}^{1} \frac{1}{1 - x^{2}} (\frac{π^{2}}{8} - \int_{0}^{x} \frac{\sin^{- 1} t}{\sqrt{1 - t^{2}}} d t) d x \\ = \frac{1}{2} \int_{0}^{1} \frac{1}{1 - x^{2}} (\frac{π^{2}}{4} - (\sin^{- 1} x)^{2}) d x \\ = \frac{1}{2} \int_{0}^{\frac{π}{2}} \frac{\frac{π^{2}}{4} - x^{2}}{\cos x} d x \\ = \frac{7}{4} ζ (3) \end{aligned}

No. 22

\begin{array}{r} ζ (3) = 9 \sum_{0 < m < n} \frac{(\binom{2 m}{m})}{m n^{2} (\binom{2 n}{n})} \end{array}

証明

補題

\begin{aligned} 1. \sum_{0 < k} \frac{2}{k {(\binom{k + n}{k})}^{2}} = \frac{1}{n} - n (\binom{2 n}{n}) \sum_{n < m} \frac{3}{m^{2} (\binom{2 m}{m})} \\ 2. \sum_{0 < k} \frac{1}{k^{2} {(\binom{k + n}{k})}^{2}} = (\binom{2 n}{n}) \sum_{n < m} \frac{3}{m^{2} (\binom{2 m}{m})} \end{aligned}

\begin{aligned} S : & = \sum_{0 < m < n} \frac{(\binom{2 m}{m})}{m n^{2} (\binom{2 n}{n})} \\ = \frac{1}{3} \sum_{0 < k} \frac{1}{k^{2}} \sum_{0 < m} \frac{1}{m {(\binom{k + m}{k})}^{2}} \\ = \frac{1}{6} \sum_{0 < k} \frac{1}{k^{2}} (\frac{1}{k} - k (\binom{2 k}{k}) \sum_{k < m} \frac{3}{m^{2} (\binom{2 m}{m})}) \\ = \frac{ζ (3)}{6} - \frac{S}{2} \end{aligned}

よって

\begin{array}{r} ζ (3) = 9 S \end{array}

No. 34

\begin{array}{r} ζ (4) = 4 \sum_{0 < m < n} \frac{(\binom{2 m}{m})}{m^{2} n^{2} (\binom{2 n}{n})} + 12 \sum_{0 < k < m < n} \frac{(\binom{2 m}{m})}{k m n^{2} (\binom{2 n}{n})} \end{array}

証明

補題

\begin{aligned} 1. \sum_{0 < k} \frac{2}{k {(\binom{k + n}{k})}^{2}} = \frac{1}{n} - n (\binom{2 n}{n}) \sum_{n < m} \frac{3}{m^{2} (\binom{2 m}{m})} \\ 2. \sum_{0 < k} \frac{2}{k^{3} {(\binom{k + n}{k})}^{2}} = \sum_{n < k} \frac{1}{n^{3}} + 9 \sum_{n < k < m} \frac{(\binom{2 k}{k})}{k m^{2} (\binom{2 m}{m})} \end{aligned}

\begin{aligned} \sum_{0 < m, n} \frac{2}{m n^{3} {(\binom{m + n}{m})}^{2}} & = \sum_{0 < n} \frac{1}{n^{3}} (\frac{1}{n} - n (\binom{2 n}{n}) \sum_{n < m} \frac{3}{m^{2} (\binom{2 m}{m})}) \\ = ζ (4) - 3 \sum_{0 < m < n} \frac{(\binom{2 m}{m})}{m^{2} n^{2} (\binom{2 n}{n})} \\ \sum_{0 < m, n} \frac{2}{m n^{3} {(\binom{m + n}{m})}^{2}} & = \sum_{0 < m} \frac{1}{m} (\sum_{m < n} \frac{1}{n^{3}} + 9 \sum_{m < k < n} \frac{(\binom{2 k}{k})}{k n^{2} (\binom{2 n}{n})}) \\ = ζ (1, 3) + 9 \sum_{0 < k < m < n} \frac{(\binom{2 m}{m})}{k m n^{2} (\binom{2 n}{n})} \\ = \frac{ζ (4)}{4} + 9 \sum_{0 < k < m < n} \frac{(\binom{2 m}{m})}{k m n^{2} (\binom{2 n}{n})} \end{aligned}

よって

\begin{array}{r} ζ (4) = 4 \sum_{0 < m < n} \frac{(\binom{2 m}{m})}{m^{2} n^{2} (\binom{2 n}{n})} + 12 \sum_{0 < k < m < n} \frac{(\binom{2 m}{m})}{k m n^{2} (\binom{2 n}{n})} \end{array}

No. 41

\begin{array}{r} β (2) = \frac{1}{2} \sum_{0 \leq m \leq n} \frac{2^{n}}{(2 m + 1) (2 n + 1) (\binom{2 n}{n})} \end{array}

証明

\begin{aligned} \sum_{0 \leq m \leq n} \frac{2^{n}}{(2 m + 1) (2 n + 1) (\binom{2 n}{n})} & = \sum_{0 \leq m} \frac{1}{2 m + 1} \frac{2^{2 m + 1}}{(\binom{2 m}{m})} \int_{0}^{\frac{1}{\sqrt{2}}} \frac{x^{2 m}}{\sqrt{1 - x^{2}}} d x \\ = 2 \int_{0}^{\frac{1}{\sqrt{2}}} \frac{1}{\sqrt{1 - x^{2}}} \frac{\sin^{- 1} x}{x \sqrt{1 - x^{2}}} d x \\ = 2 \int_{0}^{\frac{π}{4}} \frac{x}{\sin x \cos x} d x \\ = \int_{0}^{\frac{π}{2}} \frac{x}{\sin x} d x \\ = 2 β (2) \end{aligned}

No. 42

\begin{array}{r} β (2) = \frac{1}{2} \sum_{0 < m < n} \frac{2^{m + n}}{m n (\binom{2 n}{n})} \end{array}

証明

\begin{aligned} \sum_{0 < m < n} \frac{2^{m + n}}{m n (\binom{2 n}{n})} & = \sum_{0 < m} \frac{2^{2 m}}{m (\binom{2 m}{m})} \int_{0}^{\frac{π}{2}} \frac{\cos x \sin^{2 m} x}{\sqrt{1 + \cos^{2} x}} d x \\ = 2 \int_{0}^{\frac{π}{2}} \frac{x \sin x}{\sqrt{1 + \cos^{2} x}} d x \\ = - 2 [x \sinh^{- 1} \cos x]_{0}^{\frac{π}{2}} + 2 \int_{0}^{\frac{π}{2}} \sinh^{- 1} \cos x d x \\ = 2 \sum_{0 \leq n} \frac{(- 1)^{n} (\binom{2 n}{n})}{2^{2 n} (2 n + 1)} \int_{0}^{\frac{π}{2}} \cos^{2 n + 1} x d x \\ = 2 \sum_{0 \leq n} \frac{(- 1)^{n}}{(2 n + 1)^{2}} \\ = 2 β (2) \end{aligned}

No. 50

\begin{array}{r} \sum_{0 \leq n} \frac{(\binom{2 n}{n})}{2^{4 n} (2 n + 1)^{3}} = \frac{7 π^{3}}{216} \end{array}

証明

\begin{aligned} \sum_{0 \leq n} \frac{(\binom{2 n}{n})}{2^{4 n} (2 n + 1)^{3}} & = \sum_{0 \leq n} \frac{(\binom{2 n}{n})}{2^{2 n}} \int_{0}^{\frac{1}{2}} x^{2 n} \ln^{2} \frac{1}{2 x} d x \\ = \int_{0}^{\frac{1}{2}} \frac{\ln^{2} \frac{1}{2 x}}{\sqrt{1 - x^{2}}} d x \\ = \int_{0}^{\frac{π}{6}} \ln^{2} \frac{1}{2 \sin x} d x \\ = \int_{0}^{\frac{π}{6}} \ln^{2} \frac{i}{e^{i x} - e^{- i x}} d x \\ = \int_{1}^{e^{\frac{π i}{6}}} \ln^{2} \frac{i x}{x^{2} - 1} \frac{d x}{i x} \\ = 2 i \int_{e^{\frac{π i}{3}}}^{1} \frac{1}{x} \ln^{2} \frac{- x}{(1 - x)^{2}} d x \end{aligned}

　 $\ln^{2}$ を展開し，各項は多重対数関数で書くことができる． ${L i}_{3} (e^{\frac{π i}{3}}) = \frac{ζ (3)}{3} + \frac{5 π^{3} i}{162}$ などの各特殊値を代入して $\frac{7 π^{3}}{216}$ となる．

投稿日：2022年3月21日

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級数botⅡの級数計算まとめ