閉形式で書ける／書けそうな級数まとめ

540

$N o t a t i o n :$ $β_{r} = \frac{(\binom{2 r}{r})}{2^{2 r}}, G : C a t a l a n^{'} s C o n s t a n t$

　
$1$ $\sum_{n = 1}^{\infty} \frac{(- 1)^{n - 1} (4 n - 1)}{(2 n)^{3} β_{n}^{3}} \sum_{m = 0}^{n - 1} (- 1)^{m} β_{m}^{3} = \frac{{Γ (\frac{1}{8})}^{2} {Γ (\frac{3}{8})}^{2}}{48 π}$

$2$ $\sum_{n = 1}^{\infty} \frac{(- 1)^{n - 1}}{(2 n)^{3} β_{n}^{3}} \sum_{m = 0}^{n - 1} (- 1)^{m} (4 m + 1) β_{m}^{3} = \frac{{Γ (\frac{1}{8})}^{2} {Γ (\frac{3}{8})}^{2}}{96 π}$

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

$4$ $\sum_{n = 1}^{\infty} \frac{(- 1)^{n - 1} (4 n - 1)}{(2 n)^{3} β_{n}^{3}} \sum_{m = 0}^{n - 1} β_{m}^{3} = \frac{Γ {(\frac{1}{4})}^{4}}{32 π}$

$5$ $\sum_{n = 1}^{\infty} \frac{(- 1)^{n - 1} (4 n - 1)}{(2 n)^{3} β_{n}^{3}} \sum_{m = 0}^{n - 1} (- 1)^{m} (4 m + 1) β_{m}^{5} = \frac{Γ {(\frac{1}{4})}^{4}}{8 π^{2}}$

$6$ $\sum_{n = 0}^{\infty} β_{n}^{3} (2 G + \sum_{m = 1}^{n} \frac{1}{(2 m)^{2} β_{m}^{2}}) = \frac{Γ {(\frac{1}{4})}^{4}}{8 π}$

$7$ $\sum_{n = 1}^{\infty} \frac{1}{(2 n)^{3} β_{n}^{3}} {(\sum_{m = 0}^{n - 1} β_{m}^{2})}^{2} = \frac{Γ {(\frac{1}{4})}^{4}}{16}$

$8$ $\sum_{n = 0}^{\infty} \frac{β_{n}^{4}}{4 n + 1} (\frac{7}{2} ζ (3) + \sum_{m = 1}^{n} \frac{4 m - 1}{(2 m)^{4} β_{m}^{4}}) = \frac{11}{7680} \frac{Γ {(\frac{1}{4})}^{8}}{π^{2}}$

$9$ $\sum_{n = 1}^{\infty} \frac{1}{(4 n - 1) (2 n)^{4} β_{n}^{4}} \sum_{m = 0}^{n - 1} (4 m + 1) β_{m}^{4} = \frac{1}{7680} \frac{Γ {(\frac{1}{4})}^{8}}{π^{2}}$

$10$ $\sum_{n = 0}^{\infty} \frac{β_{n}^{4}}{4 n + 1} = \frac{Γ {(\frac{1}{4})}^{8}}{80 π^{5}}$

$11$ $\sum_{n = 0}^{\infty} \frac{1}{(2 n + 1)^{2}} (\sum_{n < m} \frac{(- 1)^{m - 1} (4 m - 1)}{(2 m)^{3} β_{m}^{3}}) (\sum_{m = 0}^{n} (- 1)^{m} (4 m + 1) β_{m}^{3}) + \sum_{n = 1}^{\infty} \frac{1}{(2 n)^{2}} (\sum_{n < m} \frac{(- 1)^{m - 1} (4 m - 1)}{(2 m)^{3} β_{m}^{3}}) (\sum_{m = 0}^{n - 1} (- 1)^{m} (4 m + 1) β_{m}^{3}) = \frac{7}{4} ζ (3)$

$12$ $\sum_{n = 0}^{\infty} \frac{1}{(2 n + 1)^{2}} (\sum_{n < m} \frac{(- 1)^{m - 1} (4 m - 1)}{(2 m)^{3} β_{m}^{3}}) (\sum_{m = 0}^{n} (4 m + 1) β_{m}^{4}) + \sum_{n = 1}^{\infty} \frac{1}{(2 n)^{2}} (\sum_{n < m} \frac{(- 1)^{m - 1} (4 m - 1)}{(2 m)^{3} β_{m}^{3}}) (\sum_{m = 0}^{n - 1} (4 m + 1) β_{m}^{4}) = \frac{π^{3}}{16}$

$13$ $\sum_{n = 0}^{\infty} \frac{1}{(2 n + 1)^{3}} (\frac{7}{2} ζ (3) + \sum_{m = 1}^{n} \frac{4 m - 1}{(2 m)^{4} β_{m}^{4}}) (\sum_{m = 0}^{n} (4 m + 1) β_{m}^{4}) + \sum_{n = 1}^{\infty} \frac{1}{(2 n)^{3}} (\frac{7}{2} ζ (3) + \sum_{m = 1}^{n} \frac{4 m - 1}{(2 m)^{4} β_{m}^{4}}) (\sum_{m = 0}^{n - 1} (4 m + 1) β_{m}^{4}) = \frac{93}{16} ζ (5)$

$14$ $\sum_{n = 0}^{\infty} \frac{(4 n + 1) {(\frac{1}{3})}_{n} β_{n}^{5}}{{(\frac{7}{6})}_{n}} (\frac{7}{2} ζ (3) + \sum_{m = 1}^{n} \frac{4 m - 1}{(2 m)^{4} β_{m}^{4}}) = \frac{\sqrt{3}}{2^{7} \sqrt[3]{2}} \frac{Γ {(\frac{1}{3})}^{12}}{π^{5}}$

$15$ $\sum_{n = 0}^{\infty} (4 n + 1) β_{n}^{4} \sum_{2 n < m} \frac{(- 1)^{m - 1}}{m^{2}} = \frac{7 ζ (3)}{π^{2}}$

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

$17$ $\sum_{n = 0}^{\infty} (4 n + 1) β_{n}^{4} {(\sum_{n < m} \frac{(- 1)^{m - 1} (4 m - 1)}{(2 m)^{3} β_{m}^{3}})}^{2} = \frac{7}{2} ζ (3)$

$18$ $\sum_{n = 0}^{\infty} (- 1)^{n} (4 n + 1) β_{n}^{5} {(\sum_{n < m} \frac{(- 1)^{m - 1} (4 m - 1)}{(2 m)^{3} β_{m}^{3}})}^{2} = \frac{π}{2} \sum_{n = 0}^{\infty} \frac{1}{(2 n + 1)^{2} β_{n}} \sum_{m = 0}^{n} β_{m}^{3}$

$18$ $\sum_{n = 0}^{\infty} ((4 n + 1) β_{n}^{4} {(2 G + \sum_{m = 1}^{n} \frac{1}{(2 m)^{2} β_{m}^{2}})}^{2} - \frac{4 n + 3}{(2 n + 1)^{4} β_{n}^{4}} {(\sum_{m = 0}^{n} β_{m}^{2})}^{2}) = \frac{π^{2} \ln 2}{4}$

$19$ $\sum_{n = 0}^{\infty} (- 1)^{n} (4 n + 1) β_{n}^{3} (\sum_{n < m} \frac{(- 1)^{m - 1} (4 m - 1)}{(2 m)^{3} β_{m}^{3}}) (\sum_{n < m} \frac{(- 1)^{m - 1} (4 m - 1)}{(2 m)^{2} (2 m - 1) β_{m}}) = \frac{7}{4} ζ (3)$

$20$ $\sum_{n = 0}^{\infty} (4 n + 1) β_{n}^{4} (\sum_{n < m} \frac{(- 1)^{m - 1} (4 m - 1)}{(2 m)^{3} β_{m}^{3}}) (\sum_{n < m} \frac{(- 1)^{m - 1} (4 m - 1)}{(2 m)^{2} (2 m - 1) β_{m}}) = \sum_{n = 0}^{\infty} \frac{2}{(2 n + 1)^{2}} \sum_{k = 0}^{2 n} \frac{(- 1)^{k}}{2 k + 1}$

$21$ $\sum_{n = 0}^{\infty} (- 1)^{n} (4 n + 1) β_{n}^{3} (\sum_{n < m} \frac{(- 1)^{m - 1} (4 m - 1)}{(2 m)^{3} β_{m}^{3}}) (\sum_{n < m} \frac{(- 1)^{m - 1} (4 m - 1)}{(2 m)^{2} (2 m - 1) β_{m}} \sum_{k = 0}^{m - 1} (- 1)^{k} (4 k + 1) β_{k}^{3}) = \frac{π}{2} \sum_{n = 0}^{\infty} \frac{β_{n}}{2 n + 1} \sum_{m = 0}^{n} β_{m}^{2}$

$22$ $\sum_{n = 0}^{\infty} (- 1)^{n} (4 n + 1) β_{n}^{3} (\sum_{n < m} \frac{(- 1)^{m - 1} (4 m - 1)}{(2 m)^{3} β_{m}^{3}}) (\sum_{n < m} \frac{(- 1)^{m - 1} (4 m - 1)}{(2 m)^{3} β_{m}^{3}} \sum_{k = 0}^{m - 1} (4 k + 1) β_{k}^{4}) = π \ln 2$

$23$ $\sum_{n = 0}^{\infty} (- 1)^{n} (4 n + 1) β_{n}^{3} (\sum_{n < m} \frac{(- 1)^{m - 1} (4 m - 1)}{(2 m)^{3} β_{m}^{3}}) (\sum_{n \leq m} \frac{(- 1)^{m} β_{m}}{2 m + 1} - \sum_{n < m} \frac{(- 1)^{m - 1} β_{m}}{2 m}) = \frac{π^{2}}{8}$

$24$ $\sum_{n = 1}^{\infty} \frac{4 n - 1}{(2 n)^{3} (2 n - 1) β_{n}^{2}} \sum_{m = 0}^{n - 1} (- 1)^{m} (4 m + 1) β_{m}^{3} = 2 G$

$25$ $\sum_{n = 0}^{\infty} (4 n + 1) β_{n}^{4} {(\sum_{n < m} \frac{(- 1)^{m - 1} (4 m - 1)}{(2 m)^{2} (2 m - 1) β_{m}})}^{2} = 2 π G - \frac{7}{2} ζ (3)$

$26$ $\sum_{n = 1}^{\infty} \frac{(- 1)^{n - 1} (4 n - 1)}{(2 n)^{2} (2 n - 1) β_{n}} \sum_{m = 0}^{n - 1} (4 m + 1) β_{m}^{4} = \frac{2 G}{π}$

$27$ $\sum_{n = 1}^{\infty} \frac{(- 1)^{n - 1} (4 n - 1)}{(2 n)^{2} (2 n - 1) β_{n}} \sum_{m = 0}^{n - 1} (- 1)^{m} (4 m + 1) β_{m}^{5} = \frac{4}{π} \sum_{n = 0}^{\infty} \frac{β_{n}^{2}}{(4 n + 1)^{2}}$

$28$ $\sum_{n = 0}^{\infty} (4 n + 1) β_{n}^{4} (\frac{7}{2} ζ (3) + \sum_{m = 1}^{n} \frac{4 m - 1}{(2 m)^{4} β_{m}^{4}}) (\sum_{n < m} \frac{4 m - 1}{(2 m)^{3} (2 m - 1) β_{m}^{2}}) = \frac{93}{8} ζ (5)$

$29$ $\sum_{n = 0}^{\infty} (- 1)^{n} (4 n + 1) β_{n}^{3} (\sum_{n < m} \frac{(- 1)^{m - 1} (4 m - 1)}{(2 m)^{3} β_{m}^{3}}) (\sum_{n < m} \frac{4 m - 1}{(2 m)^{3} (2 m - 1) β_{m}^{2}} \sum_{k = 0}^{m - 1} (- 1)^{k} (4 k + 1) β_{k}^{3}) = \frac{π^{3}}{8}$

$30$ $\sum_{n = 0}^{\infty} (- 1)^{n} (4 n + 1) β_{n}^{5} (\frac{π^{2}}{4} + \sum_{n < m} \frac{(- 1)^{m - 1} (4 m - 1)}{(2 m)^{3} β_{m}^{3}}) (\sum_{n < m} \frac{(- 1)^{m - 1} (4 m - 1)}{(2 m)^{2} (2 m - 1) β_{m}}) = \frac{Γ {(\frac{1}{4})}^{4}}{32}$

$31$ $\sum_{n = 1}^{\infty} (\frac{(- 1)^{n - 1} (4 n - 1)}{(2 n)^{3} β_{n}^{3}} \sum_{k = 0}^{n - 1} (- 1)^{k} (4 k + 1) β_{k}^{3} - \frac{1}{n}) = 2 \ln 2$

$32$ $\sum_{n = 1}^{\infty} \frac{(- 1)^{n - 1} (4 n - 1)}{(2 n)^{2} (2 n - 1) β_{n}} \sum_{m = 0}^{n - 1} (- 1)^{m} (4 m + 1) β_{m}^{3} = \frac{π}{2}$

投稿日：2023年8月9日

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閉形式で書ける／書けそうな級数まとめ