閉形式で書ける／書けそうな級数まとめ

${\bf Notation:}$ $\D\beta_r=\frac{\binom{2r}{r}}{2^{2r}},\quad G:{\rm Catalan's\, Constant}$

　
$\,1\,$ $\D \sum_{n=1}^\infty \frac{(-1)^{n-1}(4n-1)}{(2n)^3\beta_n^3}\sum_{m=0}^{n-1}(-1)^m\beta_m^3 = \frac{{\Gamma\L(\frac{1}{8}\R)}^2{\Gamma\L(\frac{3}{8}\R)}^2}{48\pi}$

$\,2\,$ $\D \sum_{n=1}^\infty \frac{(-1)^{n-1}}{(2n)^3\beta_n^3}\sum_{m=0}^{n-1}(-1)^m(4m+1)\beta_m^3 = \frac{{\Gamma\L(\frac{1}{8}\R)}^2{\Gamma\L(\frac{3}{8}\R)}^2}{96\pi}$

$\,3\,$ $\D \sum_{n=1}^\infty \frac{(-1)^{n-1}(4n-1)}{(2n)^3\beta_n^3}\sum_{m=0}^{n-1}(4m+1)\beta_m^4 = \frac{\pi}{2}$

$\,4\,$ $\D \sum_{n=1}^\infty \frac{(-1)^{n-1}(4n-1)}{(2n)^3\beta_n^3}\sum_{m=0}^{n-1}\beta_m^3 = \frac{\Gamma\L(\frac{1}{4}\R)^4}{32\pi}$

$\,5\,$ $\D \sum_{n=1}^\infty \frac{(-1)^{n-1}(4n-1)}{(2n)^3\beta_n^3}\sum_{m=0}^{n-1}(-1)^m(4m+1)\beta_m^5 = \frac{\Gamma\L(\frac{1}{4}\R)^4}{8\pi^2}$

$\,6\,$ $\D \sum_{n=0}^\infty \beta_n^3\L(2G+\sum_{m=1}^n \frac{1}{(2m)^2\beta_m^2}\R) = \frac{\Gamma\L(\frac{1}{4}\R)^4}{8\pi}$

$\,7\,$ $\D \sum_{n=1}^\infty \frac{1}{(2n)^3\beta_n^3}\L(\sum_{m=0}^{n-1} \beta_m^2\R)^2 = \frac{\Gamma\L(\frac{1}{4}\R)^4}{16}$

$\,8\,$ $\D \sum_{n=0}^\infty \frac{\beta_n^4}{4n+1}\L(\frac{7}{2}\zeta(3)+\sum_{m=1}^n \frac{4m-1}{(2m)^4\beta_m^4}\R) = \frac{11}{7680}\frac{\Gamma\L(\frac{1}{4}\R)^8}{\pi^2}$

$\,9\,$ $\D \sum_{n=1}^\infty \frac{1}{(4n-1)(2n)^4\beta_n^4}\sum_{m=0}^{n-1} (4m+1)\beta_m^4 = \frac{1}{7680}\frac{\Gamma\L(\frac{1}{4}\R)^8}{\pi^2}$

$10$ $\D \sum_{n=0}^\infty \frac{\beta_n^4}{4n+1} = \frac{\Gamma\L(\frac{1}{4}\R)^8}{80\pi^5}$

$11$ $\D \sum_{n=0}^\infty \frac{1}{(2n+1)^2}\L(\sum_{n< m}\frac{(-1)^{m-1}(4m-1)}{(2m)^3\beta_m^3}\R)\L(\sum_{m=0}^n (-1)^m(4m+1)\beta_m^3\R)+\sum_{n=1}^\infty \frac{1}{(2n)^2}\L(\sum_{n< m}\frac{(-1)^{m-1}(4m-1)}{(2m)^3\beta_m^3}\R)\L(\sum_{m=0}^{n-1} (-1)^m(4m+1)\beta_m^3\R) = \frac{7}{4}\zeta(3)$

$12$ $\D \sum_{n=0}^\infty \frac{1}{(2n+1)^2}\L(\sum_{n< m}\frac{(-1)^{m-1}(4m-1)}{(2m)^3\beta_m^3}\R)\L(\sum_{m=0}^n (4m+1)\beta_m^4\R)+\sum_{n=1}^\infty \frac{1}{(2n)^2}\L(\sum_{n< m}\frac{(-1)^{m-1}(4m-1)}{(2m)^3\beta_m^3}\R)\L(\sum_{m=0}^{n-1} (4m+1)\beta_m^4\R) = \frac{\pi^3}{16}$

$13$ $\D \sum_{n=0}^\infty \frac{1}{(2n+1)^3}\L(\frac{7}{2}\zeta(3)+\sum_{m=1}^n \frac{4m-1}{(2m)^4\beta_m^4}\R)\L(\sum_{m=0}^{n} (4m+1)\beta_m^4\R)+\sum_{n=1}^\infty \frac{1}{(2n)^3}\L(\frac{7}{2}\zeta(3)+\sum_{m=1}^n \frac{4m-1}{(2m)^4\beta_m^4}\R)\L(\sum_{m=0}^{n-1} (4m+1)\beta_m^4\R) = \frac{93}{16}\zeta(5)$

$14$ $\D \sum_{n=0}^\infty \frac{(4n+1)\L(\frac{1}{3}\R)_n\beta_n^5}{\L(\frac{7}{6}\R)_n}\L(\frac{7}{2}\zeta(3)+\sum_{m=1}^n \frac{4m-1}{(2m)^4\beta_m^4}\R) = \frac{\sqrt{3}}{2^7\sqrt[3]{2}}\frac{\Gamma\L(\frac{1}{3}\R)^{12}}{\pi^5}$

$15$ $\D \sum_{n=0}^\infty (4n+1)\beta_n^4\sum_{2n< m}\frac{(-1)^{m-1}}{m^2} = \frac{7\zeta(3)}{\pi^2}$

$16$ $\D \sum_{n=0}^\infty (-1)^n(4n+1)\beta_n^3\L(\sum_{n< m}\frac{(-1)^{m-1}(4m-1)}{(2m)^3\beta_m^3}\R)^2 = \frac{\pi}{2}\sum_{n=0}^\infty \frac{\beta_n}{2n+1}\sum_{m=0}^n \frac{\beta_m}{2m+1}$

$17$ $\D \sum_{n=0}^\infty (4n+1)\beta_n^4\L(\sum_{n< m}\frac{(-1)^{m-1}(4m-1)}{(2m)^3\beta_m^3}\R)^2 = \frac{7}{2}\zeta(3)$

$18$ $\D \sum_{n=0}^\infty (-1)^n(4n+1)\beta_n^5\L(\sum_{n< m}\frac{(-1)^{m-1}(4m-1)}{(2m)^3\beta_m^3}\R)^2 = \frac{\pi}{2}\sum_{n=0}^\infty \frac{1}{(2n+1)^2\beta_n}\sum_{m=0}^n \beta_m^3$

$18$ $\D \sum_{n=0}^\infty \L((4n+1)\beta_n^4\L(2G+\sum_{m=1}^n \frac{1}{(2m)^2\beta_m^2}\R)^2-\frac{4n+3}{(2n+1)^4\beta_n^4}\L(\sum_{m=0}^{n} \beta_m^2\R)^2 \R) = \frac{\pi^2\ln2}{4}$

$19$ $\D \sum_{n=0}^\infty (-1)^n(4n+1)\beta_n^3\L(\sum_{n< m}\frac{(-1)^{m-1}(4m-1)}{(2m)^3\beta_m^3}\R)\L(\sum_{n< m}\frac{(-1)^{m-1}(4m-1)}{(2m)^2(2m-1)\beta_m}\R) = \frac{7}{4}\zeta(3)$

$20$ $\D \sum_{n=0}^\infty (4n+1)\beta_n^4\L(\sum_{n< m}\frac{(-1)^{m-1}(4m-1)}{(2m)^3\beta_m^3}\R)\L(\sum_{n< m} \frac{(-1)^{m-1}(4m-1)}{(2m)^2(2m-1)\beta_m}\R) =\sum_{n=0}^\infty \frac{2}{(2n+1)^2}\sum_{k=0}^{2n} \frac{(-1)^k}{2k+1}$

$21$ $\D \sum_{n=0}^\infty (-1)^n(4n+1)\beta_n^3\L(\sum_{n< m}\frac{(-1)^{m-1}(4m-1)}{(2m)^3\beta_m^3}\R)\L(\sum_{n< m} \frac{(-1)^{m-1}(4m-1)}{(2m)^2(2m-1)\beta_m}\sum_{k=0}^{m-1}(-1)^k(4k+1)\beta_k^3\R) =\frac{\pi}{2}\sum_{n=0}^\infty \frac{\beta_n}{2n+1}\sum_{m=0}^n \beta_m^2$

$22$ $\D \sum_{n=0}^\infty (-1)^n(4n+1)\beta_n^3\L(\sum_{n< m}\frac{(-1)^{m-1}(4m-1)}{(2m)^3\beta_m^3}\R)\L(\sum_{n < m}\frac{(-1)^{m-1}(4m-1)}{(2m)^3\beta_m^3}\sum_{k=0}^{m-1}(4k+1)\beta_k^4\R) = \pi\ln2$

$23$ $\D \sum_{n=0}^\infty (-1)^n(4n+1)\beta_n^3\L(\sum_{n< m}\frac{(-1)^{m-1}(4m-1)}{(2m)^3\beta_m^3}\R)\L(\sum_{n\le m}\frac{(-1)^m\beta_m}{2m+1}-\sum_{n< m}\frac{(-1)^{m-1}\beta_m}{2m}\R) = \frac{\pi^2}{8}$

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

$25$ $\D \sum_{n=0}^\infty (4n+1)\beta_n^4\L(\sum_{n< m} \frac{(-1)^{m-1}(4m-1)}{(2m)^2(2m-1)\beta_m}\R)^2 = 2\pi G-\frac{7}{2}\zeta(3)$

$26$ $\D \sum_{n=1}^\infty \frac{(-1)^{n-1}(4n-1)}{(2n)^2(2n-1)\beta_n}\sum_{m=0}^{n-1}(4m+1)\beta_m^4 = \frac{2G}{\pi}$

$27$ $\D \sum_{n=1}^\infty \frac{(-1)^{n-1}(4n-1)}{(2n)^2(2n-1)\beta_n}\sum_{m=0}^{n-1}(-1)^m(4m+1)\beta_m^5 = \frac{4}{\pi}\sum_{n=0}^\infty \frac{\beta_n^2}{(4n+1)^2}$

$28$ $\D \sum_{n=0}^\infty (4n+1)\beta_n^4\L(\frac{7}{2}\zeta(3)+\sum_{m=1}^n \frac{4m-1}{(2m)^4\beta_m^4}\R)\L(\sum_{n< m}\frac{4m-1}{(2m)^3(2m-1)\beta_m^2}\R) = \frac{93}{8}\zeta(5)$

$29$ $\D \sum_{n=0}^\infty (-1)^n(4n+1)\beta_n^3\L(\sum_{n< m}\frac{(-1)^{m-1}(4m-1)}{(2m)^3\beta_m^3}\R)\L(\sum_{n< m}\frac{4m-1}{(2m)^3(2m-1)\beta_m^2}\sum_{k=0}^{m-1}(-1)^k(4k+1)\beta_k^3\R) = \frac{\pi^3}{8}$

$30$ $\D \sum_{n=0}^\infty (-1)^n(4n+1)\beta_n^5\L(\frac{\pi^2}{4}+\sum_{n< m}\frac{(-1)^{m-1}(4m-1)}{(2m)^3\beta_m^3}\R)\L(\sum_{n< m} \frac{(-1)^{m-1}(4m-1)}{(2m)^2(2m-1)\beta_m}\R) = \frac{\Gamma\L(\frac{1}{4}\R)^4}{32}$

$31$ $\D \sum_{n=1}^\infty \left(\frac{(-1)^{n-1}(4n-1)}{(2n)^3\beta_n^3}\sum_{k=0}^{n-1}(-1)^k(4k+1)\beta_k^3-\frac{1}{n}\right) =2\ln2$

$32$ $\D \sum_{n=1}^\infty \frac{(-1)^{n-1}(4n-1)}{(2n)^2(2n-1)\beta_n}\sum_{m=0}^{n-1}(-1)^m(4m+1)\beta_m^3 = \frac{\pi}{2}$