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大学数学基礎解説

非常に興味深い級数一覧

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N O T A T I O N

\begin{aligned} A_{n} & = \frac{(- 1)^{n} {(\binom{2 n}{n})}^{3}}{2^{6 n}} \sum_{n < m} \frac{(- 1)^{m - 1} 2^{6 m} (4 m - 1)}{(2 m)^{3} {(\binom{2 m}{m})}^{3}} \\ B_{n} & = \frac{(- 1)^{n} 2^{6 n}}{(2 n + 1)^{3} {(\binom{2 n}{n})}^{3}} \sum_{m = 0}^{n} \frac{(- 1)^{m} (4 m + 1) {(\binom{2 m}{m})}^{3}}{2^{6 m}} \\ C_{n} & = \frac{{(\binom{2 n}{n})}^{4}}{2^{4 n}} (\frac{7}{2} ζ (3) + \sum_{m = 1}^{n} \frac{2^{8 m} (4 m - 1)}{(2 m)^{4} {(\binom{2 m}{m})}^{4}}) \\ D_{n} & = \frac{2^{8 n}}{(2 n + 1)^{4} {(\binom{2 n}{n})}^{4}} \sum_{m = 0}^{n} \frac{(4 m + 1) {(\binom{2 m}{m})}^{4}}{2^{8 m}} \\ E_{n} & = \frac{{(\binom{2 n}{n})}^{2}}{2^{4 n}} (2 β (2) + \sum_{m = 1}^{n} \frac{2^{4 m}}{(2 m)^{2} {(\binom{2 m}{m})}^{2}}) \\ F_{n} & = \frac{2^{4 n}}{(2 n + 1)^{2} {(\binom{2 n}{n})}^{2}} \sum_{m = 0}^{n} \frac{{(\binom{2 m}{m})}^{2}}{2^{4 m}} \\ V_{n} & = \frac{1}{2^{4 n}} \sum_{m = 0}^{n} {(\binom{2 n - 2 m}{n - m})}^{2} {(\binom{2 m}{m})}^{2} \end{aligned}

P R O D U C T S

\begin{array}{r} \sum_{n = 0}^{\infty} A_{n} = 2 \sum_{n = 0}^{\infty} B_{n} = \frac{{Γ (\frac{1}{8})}^{2} {Γ (\frac{3}{8})}^{2}}{48 π} \end{array}

\begin{array}{r} \sum_{n = 0}^{\infty} \frac{(- 1)^{n} (4 n + 1) (\binom{2 n}{n})}{2^{2 n}} A_{n} = \frac{π}{2} \end{array}

\begin{array}{r} 2 π \sum_{n = 0}^{\infty} (- 1)^{n} A_{n} = \frac{π^{2}}{2} \sum_{n = 0}^{\infty} \frac{(4 n + 1) {(\binom{2 n}{n})}^{2}}{2^{4 n}} A_{n} = \frac{π}{2} \sum_{n = 0}^{\infty} \frac{(\binom{2 n}{n})}{2^{2 n}} E_{n} = 2 \sum_{n = 1}^{\infty} \frac{n (\binom{2 n}{n})}{2^{2 n}} F_{n - 1}^{2} = \sum_{n = 0}^{\infty} (\frac{2 E_{n}}{4 n + 1} + \frac{2 F_{n}}{4 n + 3}) = \frac{{Γ (\frac{1}{4})}^{4}}{16} \end{array}

\begin{array}{r} π \sum_{n = 0}^{\infty} \frac{(- 1)^{n} (4 n + 1) {(\binom{2 n}{n})}^{3}}{2^{6 n}} A_{n} = \sum_{n = 0}^{\infty} \frac{(4 n + 1) {(\binom{2 n}{n})}^{2}}{2^{4 n}} C_{n} = \frac{π^{2}}{2} \sum_{n = 0}^{\infty} \frac{{(\binom{2 n}{n})}^{4}}{2^{8 n}} = \frac{4}{π^{2}} \sum_{n = 0}^{\infty} (C_{n} (\frac{1}{4 n + 1} + \sum_{m = 0}^{2 n - 1} \frac{2}{2 m + 1}) + D_{n} (\frac{1}{4 n + 3} + \sum_{m = 0}^{2 n} \frac{2}{2 m + 1})) \end{array}

\begin{array}{r} \frac{π^{4}}{32} \sum_{n = 0}^{\infty} \frac{(4 n + 1) {(\binom{2 n}{n})}^{6}}{2^{12 n}} = \sum_{n = 0}^{\infty} (4 n + 1) A_{n}^{2} = \frac{2}{3} \sum_{n = 0}^{\infty} C_{n} = 2 \sum_{n = 0}^{\infty} D_{n} = \frac{π}{2} \sum_{n = 0}^{\infty} \frac{{(\binom{2 n}{n})}^{2}}{2^{4 n}} E_{n} \end{array}

\begin{array}{r} \sum_{n = 0}^{\infty} \frac{C_{n}}{4 n + 1} = 11 \sum_{n = 0}^{\infty} \frac{D_{n}}{4 n + 3} = \frac{11 π^{3}}{96} \sum_{n = 0}^{\infty} \frac{{(\binom{2 n}{n})}^{4}}{2^{8 n} (4 n + 1)} = \frac{11}{7680} \frac{{Γ (\frac{1}{4})}^{8}}{π^{2}} \end{array}

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

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

\begin{array}{r} \sum_{n = 0}^{\infty} ((2 n + 1) C_{n} + (2 n + 2) C_{n + 1}) D_{n} = \frac{93}{16} ζ (5) \end{array}

\begin{array}{r} \frac{π^{6}}{64} \sum_{n = 0}^{\infty} \frac{(4 n + 1) {(\binom{2 n}{n})}^{8}}{2^{16 n}} = \sum_{n = 0}^{\infty} ((4 n + 1) C_{n}^{2} - (4 n + 3) D_{n}^{2}) \end{array}

\begin{array}{r} π^{2} \sum_{n = 0}^{\infty} \frac{(- 1)^{n} {(\binom{2 n}{n})}^{3}}{2^{6 n}} (2 \sum_{m = 0}^{n} \frac{(- 1)^{m} (\binom{2 m}{m})}{2^{2 m}} - \frac{(- 1)^{n} (\binom{2 n}{n})}{2^{2 n}}) = \frac{{Γ (\frac{1}{8})}^{2} {Γ (\frac{3}{8})}^{2}}{16 π} + \sum_{n = 0}^{\infty} \frac{(4 n + 1) (\binom{4 n}{2 n})}{(2 n + 1)^{3} {(\binom{2 n}{n})}^{2}} \end{array}

\begin{array}{r} \sum_{n = 0}^{\infty} \frac{2^{4 n} (4 n + 3) B_{n}}{(2 n + 1)^{2} {(\binom{2 n}{n})}^{2}} = π \sum_{n = 0}^{\infty} \frac{{(\binom{2 n}{n})}^{2}}{2^{4 n} (4 n + 1)^{2}} = \frac{π}{8} \sum_{n = 0}^{\infty} \frac{{(\binom{2 n}{n})}^{3}}{2^{6 n}} (\frac{π^{2}}{2} + \sum_{m = 1}^{n} \frac{2^{2 m}}{m^{2} (\binom{2 m}{m})}) \end{array}

\begin{array}{r} \sum_{n = 0}^{\infty} \frac{(4 n + 1) (\binom{2 n}{n})}{2^{2 n}} \frac{{(\frac{1}{3})}_{n}}{{(\frac{7}{6})}_{n}} C_{n} = \frac{\sqrt{3} {Γ (\frac{1}{3})}^{12}}{2^{7} \sqrt[3]{2} π^{5}} \end{array}

\begin{array}{r} \frac{π^{3}}{4} \sum_{n = 0}^{\infty} \frac{(4 n + 1) {(\binom{2 n}{n})}^{4}}{2^{8 n}} V_{2 n} = \sum_{n = 0}^{\infty} ((4 n + 1) C_{n} V_{2 n} + (4 n + 3) D_{n} V_{2 n + 1}) = \frac{Γ {(\frac{1}{4})}^{8}}{32 π^{4}} \end{array}

\begin{array}{r} \sum_{n = 0}^{\infty} (\frac{(4 n + 1) {(\binom{2 n}{n})}^{2}}{2^{4 n}} (2 \ln 2 + H_{2 n}) C_{n} + \frac{(4 n + 3) (2 n + 1)^{2} {(\binom{2 n}{n})}^{2}}{2^{4 n}} D_{n}^{2}) = \sum_{n = 0}^{\infty} \frac{{(\binom{2 n}{n})}^{4}}{2^{8 n}} (2 \ln 2 + H_{2 n}) \end{array}

\begin{array}{r} \sum_{n = 0}^{\infty} \frac{(4 n + 3) (2 n + 1)^{2} {(\binom{2 n}{n})}^{2}}{2^{4 n}} D_{n}^{2} = \frac{π^{4}}{16} \sum_{n = 0}^{\infty} \frac{(4 n + 1) {(\binom{2 n}{n})}^{6}}{2^{12 n}} \end{array}

\begin{array}{r} \sum_{n = 0}^{\infty} \frac{(4 n + 1) {(\binom{2 n}{n})}^{4}}{2^{8 n}} \sum_{2 n < m} \frac{(- 1)^{m - 1}}{m^{2}} = \frac{7 ζ (3)}{π^{2}} \end{array}

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

\begin{array}{r} \sum_{n = 0}^{\infty} ((4 n + 1) E_{n}^{2} - (4 n + 3) F_{n}^{2}) = \frac{π^{2} \ln 2}{4} \end{array}

C O N J E C T U R E

\begin{array}{r} \sum_{n = 0}^{\infty} (4 n + 1) A_{n}^{2} = \sum_{n = 0}^{\infty} (4 n + 3) B_{n}^{2} \end{array}

\begin{array}{r} \sum_{n = 0}^{\infty} (4 n + 1) C_{n}^{2} = 5 \sum_{n = 0}^{\infty} (4 n + 3) D_{n}^{2} \end{array}

\begin{array}{r} \sum_{n = 1}^{\infty} \frac{2 n {(\binom{2 n}{n})}^{4}}{2^{8 n}} D_{n - 1} + \sum_{n = 0}^{\infty} \frac{(2 n + 1) {(\binom{2 n}{n})}^{4}}{2^{8 n}} D_{n} = \frac{π^{2}}{8} \end{array}

　

投稿日：2023年5月1日

更新日：2024年10月8日

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非常に興味深い級数一覧