級数一覧　𝟏

主にぼくが計算した級数を列挙する。ぼくが自力で導出できなかったものにはその出処を示す。

定義・記法

　　　$\BA &\zeta(k_1,\cdots,\overline{k_s},\cdots,k_a) =\sum_{0< n_1<\cdots< n_a}\frac{(-1)^{k_s}}{n_1^{k_1}\cdots n_a^{k_a}}\\ &\zeta^{\star}(k_1,\cdots,k_a) =\sum_{0< n_1\le \cdots\le n_a}\frac{1}{n_1^{k_1}\cdots n_a^{k_a}}\\ &\zeta_N(k_1,\cdots,k_a)=\sum_{0< n_1<\cdots< n_a\le N}\frac{1}{n_1^{k_1}\cdots n_a^{k_a}}\\ &{\zeta}^{\star}_N(k_1,\cdots,k_a)=\sum_{0< n_1\le \cdots\le n_a\le N}\frac{1}{n_1^{k_1}\cdots n_a^{k_a}}\\ &t(k_1,\cdots,\overline{k_s},\cdots,k_a) =\sum_{0\le n_1<\cdots< n_a}\frac{(-1)^{k_s}}{(2n_1+1)^{k_1}\cdots(2n_a+1)^{k_a}}\\ &\beta(s)=\sum_{0\le n}\frac{(-1)^n}{(2n+1)^s}\\ &\eta(s)=(1-2^{1-s})\zeta(s) \\ &{\rm Li}_{k_1,\cdots,k_a}(z)=\sum_{0< n_1<\cdots< n_a}\frac{z^{k_a}}{n_1^{k_1}\cdots n_a^{k_a}} \\ &\chi_{\nu}(z)=\sum_{0\le n}\frac{z^{2n+1}}{(2n+1)^{\nu}}\\ &\psi(1+z)=-\gamma+\sum_{0< n} \frac{z}{n(n+z)}\\ &\psi'(z)=\sum_{0\le n} \frac{1}{(n+z)^2}\\ &(z)_n=\frac{\Gamma(n+z)}{\Gamma(z)}\\ &\underbrace{k,\cdots ,k}_{n}={\{k\}}_n\\ &H_n=\sum_{m=1}^n\frac{1}{m}\\ &{_p}F_q\L[\begin{matrix}a_1,\cdots,a_p\\b_1,\cdots,b_q \end{matrix};z \R] =\sum_{0\le n}\frac{(a_1,\cdots,a_p)_n}{(b_1,\cdots,b_q)_n}\frac{z^n}{n!} \EA$

あくまでぼくが知っている範囲なので、実は計算できるというのがあれば教えてほしい。
また、間違った結果を書いている危険性があると提言するのも吝かで無い感じも僅かながら存在するので、あれば教えてほしい。
新たな結果を得たら追加していく。

$ \D　　　\sum_{0< n}\frac{\zeta_{n-1}(\{1\}_{a-1}){\zeta}^{\star}_n(\{1\}_{b})}{n^3} =\zeta^{\star}(b+1,a+1)-\sum_{j=1}^a\binom{j+b}{b}\zeta(a-j+1,j+b+1) $　　　　(訂正後)$ $

$ \D　　　{\rm Li}_{\{1\}_{a-1},b+1}(1-x) =\sum_{j=0}^{b-1}(-1)^j\zeta(\{1\}_{a-1},b-j+1){\rm Li}_{\{1\}_{j}}(x) +(-1)^b\sum_{0\le n_1\le \cdots\le n_b\le a}{\rm Li}_{\{1\}_{n_1}}(1-x)\,{\rm Li}_{a-n_b+1,n_b-n_{b-1}+1,\cdots,n_2-n_1+1}(x) $

$ \D　　　 \sum_{0< n}\left(\frac{1}{n^{a+1}}-\sum_{j=0}^a\frac{x^n}{n^{a-j+1}} {\rm Li}_{{\{1\}}_{j}}(1-x) \right) \sum_{k=0}^{b}(-1)^k\zeta_{n-1}(\{1\}_{b-k}){\rm Li}_{{\{1\}}_k}(x) ={\rm Li}_{{\{1\}}_{a-1},b+2}(1-x) $　　　　(出処：コメント欄)$ $

$ \D　　　\sum_{0< n_1<\cdots< n_a}\frac{1}{n_1\cdots n_{a-1}n_a^3}\frac{n_a!}{(\alpha)_{n_a}} =\sum_{0\le m< n}\frac{1}{(m+\alpha)n^{a+1}}-\sum_{j=1}^a \sum_{0< m\le n}\frac{1}{m^{a-j+1}(n+\alpha)^{j+1}} $

$ \D　　　\sum_{k=0}^{m-1}{\zeta}^{\star}_n({\{1\}}_k){\zeta}^{\star}_n(m-k)=m{\zeta}^{\star}_n({\{1\}}_{m}) $

　　　$\BA \D \sum_{m=1}^\infty\frac{(-1)^{m-1}}{m^{n+1}}+\sum_{j=1}^n \frac{1}{j!}\sum_{m=1}^\infty \frac{(-1)^{m-1}{\zeta}_{m-1}({\{1\}}_{j-1})}{m^{n-j+1}} =\sum_{j=0}^n\frac{\ln^j2}{j!}{\rm Li}_{n-j+1}\L(\frac{1}{2}\R) \EA$

　　　$\BA \D \sum_{0< m_1<\cdots< m_a\le n_1<\cdots< n_b}\frac{2^{2m_a+2n_1}}{(2m_1)^2\cdots(2m_a)^2\binom{2m_a}{m_a}\binom{2n_1}{n_1}(2n_1+1)^2\cdots(2n_b+1)^2} =\frac{1}{(2a-1)!(2b)!}\int_0^\frac{\pi}{2}\frac{x^{2a-1}\L(\frac{\pi}{2}-x\R)^{2b}}{\sin x}\,dx \EA$

　　　$\BA \D \sum_{0< l< m< n}\frac{2^{2l}}{l^2\binom{2l}{l}}\frac{\binom{2m}{m}}{2^{2m}m}\frac{2^{2n}}{n^2\binom{2n}{n}} =\frac{93}{16}\zeta(5)-\frac{21}{8}\zeta(2)\zeta(3) \EA$

$ \D　　　\sum_{0< n}(-1)^{n-1}\L(-1+(2n)^2\ln\frac{(2n)^2}{(2n)^2-1} \R) =-\frac{1}{2}+\frac{8\beta(2)}{\pi}-\frac{14\zeta(3)}{\pi^2} $

$ \D　　　{_2}F_1\L[\begin{matrix}\frac{1}{2},\frac{1}{2}\\\frac{1}{2}+\alpha \end{matrix};\frac{1}{2} \R] =\frac{1}{\sqrt{2}}\frac{\Gamma\L(\frac{\alpha}{2} \R)\Gamma\L(\frac{1}{2}+\alpha \R)}{\Gamma\L(\frac{1+\alpha}{2} \R)\Gamma(\alpha)} $

$ \D　　　\sum_{0\le n}\L(\frac{(\alpha)_n}{(1-\alpha)_n} \R)^2 =\frac{1}{2}+\frac{\sqrt{\pi}}{2}\frac{\Gamma^2(1-\alpha)\Gamma\L(\frac{1}{2}-2\alpha \R)}{\Gamma^2\L(\frac{1}{2}-\alpha \R)\Gamma(1-2\alpha)} $

$ \D　　　\beta(2)=\frac{1}{\sqrt{2}}\sum_{0\le m\le n}\frac{\binom{2m}{m}}{2^{2m}(2m+1)}\frac{1}{2^n(2n+1)} $

$ \D　　　\sum_{0<2m+1< n}\frac{(-1)^{n}}{(2m+1)(2n+1)}=\frac{\pi^2}{24} $

$ \D　　　\sum_{0\le 2m< n}\frac{\binom{2m}{m}\binom{4m}{2m}}{2^{6m}(2m+1)}\frac{2^{2n}}{n^2\binom{2n}{n}} =8\beta(2)-\pi\ln2 $

$ \D　　　\sum_{0\le m< n}\frac{2^{2m}}{(2m+1)^2\binom{2m}{m}}\frac{\binom{2n}{n}}{2^{2n}n} =4\beta(2)\ln2+8\sum_{0\le m< n}\frac{1}{(2m+1)(4n+1)^2}-8\sum_{0\le m\le n}\frac{1}{(2m+1)(4n+3)^2} $

$ \D　　　\sum_{0< m< n}\frac{2^{m+n}}{m^2n\binom{2n}{n}} =\sum_{0\le 2m< n}\frac{\binom{2m}{m}\binom{4m+2}{2m+1}}{2^{6m}(2m+1)}\frac{2^{2n}}{n^2\binom{2n}{n}} -\frac{\pi^3}{12}-\pi\ln^2 2 $

$ \D　　　\sum_{0\le n}\frac{\sqrt{2}}{2^n(2n+1)^2}+\frac{\pi}{8}\sum_{0\le n}\frac{\binom{2n}{n}\binom{4n+2}{2n+1}}{2^{6n}(2n+1)}=\frac{\pi^2}{4} $

$ \D　　　\sum_{0\le m< n}\frac{(-1)^{n-1}}{(2m+1)n}=\frac{\pi^2}{16} $

$ \D　　　\sum_{0< n}\frac{2^{2n}}{n^2\binom{2n}{n}}\L(\frac{2x}{1+x^2} \R)^{2n} =8\sum_{0\le m< n}\frac{(-1)^{n-1}x^{2n}}{(2m+1)n} $

$ \D　　　 \sum_{0\le m\le n}\frac{2^{2n}}{(2m+1)(2n+1)^2\binom{2n}{n}}-2\sum_{0< m\le n}\frac{(-1)^{n-1}}{m(2n+1)^2} =\frac{5\pi^3}{48}-2\beta(2)\ln2 $

$ \D　　　\sum_{0\le m\le n}\frac{\binom{2m}{m}^2}{2^{4m-2n}(2m+1)(2n+1)^2\binom{2n}{n}}=\frac{\pi^3}{16} $

$ \D　　　\sum_{0\le n}\frac{2^{4n}}{(2n+1)^4\binom{2n}{n}^2}=2t(4)-8t(1,3)+2\pi\,t(1,\overline{2}) $

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

$\BA \D　　　\frac{\pi^3}{8}-\pi\ln^2 2-\sum_{0< n}\frac{\binom{2n}{n}H_n}{2^{2n}(2n+1)^2} &=4\beta(2)\ln2-4\sum_{0< n}\frac{(-1)^{n-1}H_n}{(2n+1)^2}\\ &=\sum_{0\le n}\frac{2^{2n+1}}{(2n+1)^3\binom{2n}{n}}\\ &=\pi\sum_{0\le 2m< n}\frac{\binom{2m}{m}^2(-1)^{n-1}}{2^{4m}(2m+1)n}\\ &=8\sum_{0\le m< n}\frac{(-1)^m-(-1)^n}{(2m+1)(2n+1)^2} \EA$

$ \D　　　\sum_{0< n}\frac{(-1)^{n-1}2^{2n}}{n^2\binom{2n}{n}}\sum_{m=0}^{n-1}\frac{\binom{2m}{m}}{2^{2m}}\L(\frac{\pi^2}{2}-\sum_{l=1}^m\frac{2^{2l}}{l^2\binom{2l}{l}}\R) =\sum_{0< n}\frac{(-1)^{n-1}2^{2n}}{n^2\binom{2n}{n}}\L(2H_n+\frac{1-(-1)^n}{n} \R) $

$ \D　　　8\sum_{0< m< n}\frac{\binom{2m}{m}}{2^{2m}m^2}\frac{2^n}{n\binom{2n}{n}} =\pi\sum_{0< n}\frac{\binom{2n}{n}^2}{2^{4n}n^2}+8\sum_{0< m\le n}\frac{1}{m^2}\frac{(-1)^{n-1}}{2n+1} $

$ \D　　　\sum_{0\le m< n}\frac{1}{(2m+1)^a n^2}\L(\frac{2^{2n}}{\binom{2n}{n}}-1 \R) =\sum_{0\le m< n}\L((-1)^{a-1}\frac{2\ln2+H_m}{(2m+1)^a}+2\sum_{j=0}^{a-2}\frac{(-1)^jt(a-j)}{(2m+1)^{j+1}} \R)\frac{\binom{2n}{n}}{2^{2n}n} $

$ \D　　　\sum_{0< m< n}\frac{1}{m^an^2}\L(\frac{2^{2n}}{\binom{2n}{n}}-1 \R) =\sum_{0< m\le n}\L((-1)^{a-1}\frac{H_m}{m^{a}}+\sum_{j=0}^{a-2}\frac{(-1)^j\zeta(a-j)}{m^{j+1}} \R)\frac{\binom{2n}{n}}{2^{2n}n} $

$\BA \D　　　\sum_{0< m< n}\frac{2^{2m}}{m^2\binom{2m}{m}}\frac{1}{(2n-1)^2} &=2\pi\sum_{0\le 2l<2m< n}\frac{1}{(2l+1)^2}\frac{\binom{2m}{m}^2}{2^{4m}}\frac{(-1)^{n-1}}{n}\\ &=\sum_{0\le m< n}\L(\frac{\pi^2}{2m+1}-\frac{8}{(2m+1)^3} \R)\frac{(-1)^{n-1}}{n}\\ &=\frac{\pi^4}{16}-\sum_{0\le m< n}\frac{1}{(2m+1)^2n^2} -\sum_{0\le m< n}\L(\frac{\pi^2}{4}\frac{1}{2m+1}-\frac{2\ln2+H_m}{(2m+1)^2} \R)\frac{\binom{2n}{n}}{2^{2n}n}　　　　　~ \EA$

$ \D　　　\sum_{0< m< n}\frac{2^{2m}}{m^2\binom{2m}{m}}\frac{1}{(2n-1)^2} =8\beta(2)^2-\frac{\pi^4}{16} $　　　　( 出処 )$ $

$ \D　　　\sum_{m=0}^n \frac{(-1)^{m}}{(2m+1)^{a+1}}\binom{n}{m} =\frac{2^{2n}}{(2n+1)\binom{2n}{n}}\sum_{0\le k_1\le\cdots\le k_a\le n}\frac{1}{(2k_1+1)\cdots(2k_a+1)} $　　　　(出処：コメント欄)$ $

$ \D　　　\sum_{0< n}\frac{\binom{m+n}{m}}{n\binom{2m+2n}{2m}}=2\sum_{2m\le k}\frac{(-1)^{k}}{k} $

$ \D　　　\sum_{0< n}\frac{2^{2n}}{n^2\binom{2n}{n}}\frac{\binom{m+n}{m}}{\binom{2m+2n}{2m}}=\sum_{m\le n}\frac{1}{\left(n+\frac{1}{2}\right)^2} $

$ \D　　　2\sum_{0< m\le n}\frac{1}{m(2n+1)^2} =\sum_{0<2m\le n}\frac{2^{2m}}{m^2\binom{2m}{m}}\frac{(-1)^n}{n} =\frac{7}{2}\zeta(3)-\frac{1}{2}\pi^2\ln2 $　　　　　(最右辺：コメント欄)$ $

$ \D　　　\sum_{0< n}\L(\frac{1}{n}-2\sum_{n< m}\frac{(-1)^{m+n+1}}{m} \R)=\ln2 $

$\BA \D　　　\sum_{0< l< m< n}\frac{1}{lmn}\sum_{k=0}^{n-1}\L(\frac{(-1)^k\pi}{2k+1}-\frac{2}{(2k+1)^2} \R) =&\frac{\pi}{4}\sum_{0\le n}\frac{2^{2n}}{\binom{2n}{n}}\L(\frac{2}{(2n+1)^4}-\frac{2\ln2}{(2n+1)^3}+\frac{\ln^2 2}{(2n+1)^2}\R)\\ &-\frac{1}{8}\sum_{0< n}\frac{2^{4n}}{n^3\binom{2n}{n}^2}\L(\L(\sum_{m=1}^{2n-1}\frac{(-1)^m}{m}\R)^2+\sum_{2n\le m}\frac{(-1)^m}{m^2}\R)\\ &-\frac{7}{16}\pi^2\zeta(3)+\frac{31}{4}\zeta(5)-\pi\beta(4)　　　　　~ \EA$

$ \D　　　\sum_{0< n}\frac{2^{4n}}{n^3\binom{2n}{n}^2}\left(\left(\sum_{2n\le m}\frac{(-1)^m}{m} \right)^2+\sum_{2n\le m}\frac{(-1)^m}{m^2} \right)=2\sum_{0\le m< n}\frac{\binom{2n}{n}}{(2m+1)^2n^32^{2n}} $

$ \D　　　\sum_{0\le n}\frac{2^{2n}}{\binom{2n}{n}}\L(\frac{2}{(2n+1)^4}-\frac{2\ln2}{(2n+1)^3}+\frac{\ln^2 2}{(2n+1)^2}\R) =\frac{\pi}{2}\sum_{0\le n}\frac{\binom{2n}{n}^2}{2^{4n}(2n+1)} \L(\L(\sum_{m=1}^{2n}\frac{(-1)^m}{m} \R)^2+\sum_{2n< m}\frac{(-1)^{m-1}}{m^2}\R) $

$ \D　　　\sum_{0\le m< n}\frac{\binom{2m}{m}}{2^{2m}(2m+1)}\frac{2^{2n}}{(2n+1)^2\binom{2n}{n}}=t(3) $

$ \D　　　\sum_{0< m\le n}\frac{2^{2m+2n}}{m^2(2n+1)^2\binom{2m}{m}\binom{2n}{n}}=7\pi\zeta(3)-24\beta(4) $

$ \D　　　\sum_{0\le n_1\le \cdots\le n_a} \frac{(2x)^{2n_a}\sqrt{1-x^2}}{(2n_1+1)^2\cdots(2n_{a-1}+1)^2(2n_a+1)\binom{2n_a}{n_a}} =\sum_{0\le n_1\le \cdots\le n_a} \frac{\binom{2n_1}{n_1}^2(2x)^{2n_a}}{2^{4n_1}(2n_1+1)(2n_2+1)^2\cdots(2n_a+1)^2\binom{2n_a}{n_a}} $

$\BA \D　　　\sum_{0\le m< n}\frac{(-1)^m}{2m+1}\frac{\binom{2n}{n}}{2^{2n}(2n+1)} &=\frac{1}{2}\sum_{0\le n}\frac{(-1)^n\binom{2n}{n}}{2^{2n}(2n+1)^2}\\ &=\frac{\pi^2}{24}+\frac{1}{\sqrt{2}}\sum_{0\le m\le n}\frac{1}{(2m+1)(2n+1)2^{n}}-\frac{1}{4}\sum_{0< m\le n}\frac{1}{mn2^n} \EA$

$ \D　　　\sum_{0< n}\frac{2^{4n}}{n^4\binom{2n}{n}^2} =\sum_{0\le m< n}\L(\frac{\pi(-1)^m}{2m+1}-\frac{2}{(2m+1)^2}\R)\frac{8}{n^2} -\sum_{0\le m< n}\frac{8}{(2m+1)n^3} -\sum_{0< m\le n}\frac{16}{m}\L(\frac{\pi(-1)^n}{(2n+1)^2}-\frac{2}{(2n+1)^3}\R) $

$ \D　　　\sum_{0\le n}\frac{(-1)^n\binom{2n}{n}}{2^{2n}(2n+1)^2} =2\sqrt{2}\sum_{0\le m< n}\frac{(-1)^{n-1}}{2m+1}\L(\frac{1}{4n-1}+\frac{1}{4n+1}\R) $

$ \D　　　\sum_{0\le n}\frac{\binom{2n}{n}}{2^{2n}(2n+1)^2}\L(\frac{(-1)^n\pi}{2}-\frac{1}{2n+1} \R) =\pi\sum_{0< n}\frac{\binom{4n}{2n}}{2^{2n+3}n^2\binom{2n}{n}} $

$ \D　　　\sqrt{2\pi}\sum_{0\le n}\frac{2^n{\Gamma}^2\L(\frac{n}{2}+\frac{3}{4}\R)}{(2n+1)^2n!} =\frac{\pi^3}{8}+\frac{\pi}{8}\sum_{0< n}\frac{2^{2n}\binom{2n}{n}}{n^2\binom{4n}{2n}} $

$ \D　　　\sum_{0< n}\frac{2^{2n}\binom{2n}{n}}{n^2\binom{4n}{2n}}+\pi\sum_{0\le n}\frac{\binom{2n}{n}\binom{4n+2}{2n+1}}{2^{6n}(2n+1)}=\pi^2 $

$ \D　　　\sum_{0< n}\frac{1}{2^{2n}n^2}\L(\frac{\binom{4n}{2n}}{\binom{2n}{n}}-(-1)^n\binom{2n}{n} \R) =\zeta(2) $

$ \D　　　\sum_{0\le m< n}\frac{\binom{2n}{n}}{(2m+1)(2n+1)^2 2^{2n}}=\frac{\pi\ln^22}{2}-\frac{\pi^3}{48} $

$ \D　　　\sum_{0\le m\le n}\frac{2^{2n}}{(2m+1)(2n+1)^2\binom{2n}{n}}-\frac{\pi}{4}\sum_{0< n}\frac{\binom{2n}{n}^2H_n}{2^{4n}(2n+1)} =4\beta(2)\ln2 $

$ \D　　　\frac{\pi}{2}\sum_{0< n}\frac{H_n}{2^{4n}n}\binom{2n}{n}^2 =16\beta(2)\ln2-\frac{\pi^3}{3}-16t(1,\V{2}) $

$\BA \D　　　\sum_{0\le n}\frac{2^{6n}}{(2n+1)^3\binom{2n}{n}\binom{4n+2}{2n+1}} -\frac{\pi}{32}\sum_{0< n}\frac{2^{2n}\binom{2n}{n}}{n^2\binom{4n}{2n}} &=\frac{\pi}{8}\sum_{0\le n}\frac{\binom{2n}{n}^2}{2^{4n}(2n+1)^2}\\ &=\frac{\pi^3}{16}-\sqrt{2}\sum_{0\le m\le n}\frac{\binom{2m}{m}}{2^{2m}(2m+1)}\frac{1}{2^n(2n+1)^2}\\ &=\frac{1}{4}\sum_{0\le m< n}\frac{(-1)^m}{2m+1}\frac{1}{n^2}+\frac{1}{2}\sum_{0< n}\frac{(-1)^{n-1}H_n}{(2n+1)^2}\\ &=2\sum_{0\le m< n}\frac{(-1)^m}{(2m+1)(2n+1)^2}\\ &=\beta(3)+2\sum_{0\le m< n}\frac{(-1)^n}{(2m+1)(2n+1)^2}\\ &=\frac{1}{4}\sum_{0< m\le 2n+1}\frac{(-1)^{m-1}}{m}\frac{2^{2n}}{(2n+1)^2\binom{2n}{n}} \EA$

$\BA \D　　　\sum_{0\le m< n}\frac{\binom{2m}{m}}{2^{2m}(2m+1)}\frac{(-1)^{n-1}}{n} &=\sum_{0\le n}\frac{1}{(2n+1)^2}\frac{\binom{4n+2}{2n+1}}{2^{2n+2}\binom{2n}{n}}\\ &=\frac{\pi}{2}\ln\L(-1+\sqrt{2} \R)+4\beta(2)-4\sum_{0\le n}\frac{(-1)^n\L(-1+\sqrt{2} \R)^{2n+1}}{(2n+1)^2} \EA$

$ \D　　　\frac{\pi}{8}\sum_{0\le n}\frac{\binom{2n}{n}\binom{4n+2}{2n+1}}{2^{6n}(2n+1)^2} =\pi\beta(2)-\sum_{0\le n}\frac{(-1)^n\binom{2n}{n}}{2^{2n}(2n+1)^3}-2\sum_{0\le m< n}\frac{(-1)^m}{(2m+1)^2}\frac{\binom{2n}{n}}{2^{2n}(2n+1)} $

$ \D　　　\frac{\pi}{8}\sum_{0< n}\frac{\binom{4n}{2n}}{2^{2n}n^3\binom{2n}{n}} =\frac{\pi}{6}\ln^3 2+\frac{\pi^3}{24}\ln2-\frac{3}{2}\pi\zeta(3)-\frac{\pi^2}{8}\sum_{0\le n}\frac{\binom{2n}{n}\binom{4n+2}{2n+1}}{2^{6n}(2n+1)^2}+4\sum_{0\le m< n}\frac{1}{(2m+1)^3}\frac{\binom{2n}{n}}{2^{2n}(2n+1)} $

$ \D　　　\frac{\pi}{2}\sum_{0\le m< n}\frac{\binom{2m}{m}}{2^{2m}(2m+1)}\frac{\binom{4n}{2n}}{2^{4n}n} =\frac{\pi^2}{2}\ln\L(1+\sqrt{2} \R)-\sum_{0\le m< n}\frac{(-1)^m\binom{2m}{m}}{2^{2m}(2m+1)}\frac{1-(-1)^n}{n^2} $

$ \D　　　\sum_{0\le m< n}\frac{\binom{4m+2}{2m+1}}{2^{4m}(2m+1)}\frac{2^{2n}}{n^2\binom{2n}{n}} =8\sum_{0\le m< n}\frac{(-1)^m\binom{2m}{m}}{2^{2m}(2m+1)}\frac{1}{n^2} $

$ \D　　　 \sum_{j=0}^{a-1}\frac{(-1)^{a-j-1}\pi^{2j}}{(2j+1)!}\sum_{0\le m< n}\frac{\binom{2m}{m}}{2^{2m}(2m+1)}\frac{(-1)^{n-1}}{n^{2a-2j-1}} =\sum_{0< n_1<\cdots< n_a}\frac{1}{n_1^2\cdots n_{a-1}^2(2n_a-1)^2}\frac{\binom{4n_a-2}{2n_a-1}}{2^{2n_a}\binom{2n_a-2}{n_a-1}} $

$ \D　　　\pi\sum_{0\le m< n}\frac{1}{(2m+1)^2}\frac{\binom{2n}{n}\binom{4n}{2n}}{2^{6n}n} -\sum_{0< m\le n}\frac{1-(-1)^m}{m^3}\frac{\binom{2n}{n}}{2^{2n}(2n+1)} =\frac{3}{4}\pi^3\ln2-2\pi^2\beta(2)-\frac{7}{8}\pi\zeta(3) $

$\BA \D　　　\frac{\Gamma^2\L(\frac{1}{4} \R)}{\sqrt{2\pi}} \sum_{0< n}\frac{2^{2n}}{n^2\binom{2n}{n}}\frac{\L(\frac{1}{4} \R)_n}{\L(\frac{3}{4} \R)_n} =\frac{\pi}{\sqrt{2}}\sum_{0\le m< n}\frac{(-1)^m\binom{2m}{m}^2}{2^{4m}n^2} &-\frac{\pi\sqrt{\pi}}{\Gamma^2\L(\frac{1}{4} \R)}\sum_{0\le n}\frac{2^{4n+3}}{(2n+1)^2\binom{4n+2}{2n+1}}\frac{\L(\frac{3}{4} \R)_n}{\L(\frac{5}{4} \R)_n}\\ &+\frac{\Gamma^2\L(\frac{1}{4} \R)}{\sqrt{\pi}}\sum_{0< n}\frac{2^{4n-4}}{n^2\binom{4n}{2n}}\frac{\L(\frac{1}{4} \R)_n}{\L(\frac{3}{4} \R)_n} \EA$

$\BA \D　　　\frac{\sqrt{2}}{64}\L(\psi'\L(\frac{1}{8}\R)+\psi'\L(\frac{3}{8}\R)-\psi'\L(\frac{5}{8}\R)-\psi'\L(\frac{7}{8}\R) \R) &=\beta(2)+\sum_{0\le n}\frac{2^{2n}\binom{2n}{n}}{(2n+1)^2\binom{4n+2}{2n+1}}\\ &=\beta(2)+\frac{\sqrt{2}}{16}\sum_{0\le m< n}\frac{\binom{2m}{m}\binom{4m}{2m}}{2^{6m}}\frac{2^{2n}}{n^2\binom{2n}{n}}\\ &=\frac{4t(3)}{\pi}+\frac{\sqrt{2}}{16}\sum_{0\le m< n}\frac{\binom{2m}{m}\binom{4m}{2m}}{2^{6m}}\frac{1}{n^2}\\ &=\frac{4t(3)}{\pi}+\frac{1}{16\pi}\sum_{0< n}\frac{2^{6n}}{n^3\binom{2n}{n}\binom{4n}{2n}} \EA$

$ \D　　　\sum_{0< n}\frac{(-1)^{n-1}}{n^{2r}}\frac{\left(\alpha\right)_n}{\left(1-\alpha\right)_n} +\frac{1}{2}\sum_{0< n_1\le\cdots\le n_r}\frac{2^{2n_1}}{\binom{2n_1}{n_1}}\frac{\L(\frac{1}{2}-\alpha \R)_{n_1}}{(1-\alpha)_{n_1}}\frac{1}{n_1^2\cdots n_r^2} =\eta(2r) $

$ \D　　　\sum_{0< n}\frac{(\alpha)_n}{n^2(1-\alpha)_n}=\zeta(2) -\frac{2\sqrt{\pi}\,\Gamma(1-\alpha)}{\Gamma\L(\frac{1}{2}-\alpha\R)}\sum_{0\le n}\frac{\L(\frac{1}{2}+\alpha\R)_n}{(2n+1)^2\L(\frac{1}{2} \R)_n} +\frac{1}{2}\sum_{0< n}\frac{2^{2n}}{n^2\binom{2n}{n}}\frac{\L(\frac{1}{2}-\alpha \R)_{n}}{(1-\alpha)_{n}} $

$ \D　　　\sum_{0\le m< n}\frac{1}{(2m+1)n}\frac{\L(\frac{1}{4} \R)_n}{\L(\frac{3}{4} \R)_n} =\frac{\pi^2}{16}+\beta(2) $

$\BA \D　　　\frac{1}{2}\sum_{0< n}\frac{\L(\frac{1}{2}\R)_n}{n^2\left(\frac{3}{4} \right)_n} &=\sum_{0< n}\frac{2^n}{n^2\binom{2n}{n}}-\sum_{0< m< n}\frac{2^n}{mn\binom{2n}{n}}\\ &=\frac{\pi\ln2}{2}+\frac{\pi^2}{8}-2\beta(2) \EA$

$\BA \D　　　\frac{1}{2}\sum_{0< n}\frac{\L(\frac{1}{2}\R)_n}{n^3\left(\frac{3}{4} \right)_n} =&\sum_{0< n}\frac{2^n}{n^3\binom{2n}{n}}-\sum_{0< m< n}\frac{2^n}{m^2n\binom{2n}{n}}-\sum_{0< m< n}\frac{2^n}{mn^2\binom{2n}{n}}+\sum_{0< l< m< n}\frac{2^n}{lmn\binom{2n}{n}}\\ =&32\sum_{0< l< m\le \frac{n}{2}}\frac{(-1)^n}{(4l-1)(4m-1)(2n+1)} \EA$

$\BA \D　　　\frac{1}{2}\sum_{0< n}\frac{\L(\frac{1}{2}\R)_n}{n^4\left(\frac{3}{4} \right)_n} =&\sum_{0< n}\frac{2^n}{n^4\binom{2n}{n}} -\sum_{0< m< n}\frac{2^n}{m^3n\binom{2n}{n}} -\sum_{0< m< n}\frac{2^n}{m^2n^2\binom{2n}{n}} -\sum_{0< m< n}\frac{2^n}{mn^3\binom{2n}{n}}\\ &+\sum_{0< l< m< n}\frac{2^n}{l^2mn\binom{2n}{n}} +\sum_{0< l< m< n}\frac{2^n}{lm^2n\binom{2n}{n}} +\sum_{0< l< m< n}\frac{2^n}{lmn^2\binom{2n}{n}} -\sum_{0< k< l< m< n}\frac{2^n}{klmn\binom{2n}{n}} \EA$

$ \D　　　\frac{1}{2}\sum_{0< n}\frac{\L(\frac{1}{2}\R)_n}{n^r\left(\frac{3}{4} \right)_n} =\sum_{j=1}^{r} (-1)^{j-1} \sum_{\substack{0< n_1<\cdots< n_j\\k_1+\cdots+k_j=r,\, k_i\in\mathbb{Z}_{>0}}}\frac{2^{n_j}}{n_1^{k_1}\cdots n_j^{k_j} \binom{2n_j}{n_j}}　　　　　　　($推測$) $

$ \D　　　\sum_{0\le m< n}\frac{\L(\frac{1}{4} \R)_m}{m!}\frac{2^{2m}}{(2m+1)\binom{2m}{m}}\frac{\binom{2n}{n}}{2^{2n}(2n+1)} =\pi\L(1-\frac{4\sqrt{2\pi}}{\Gamma^2\L(\frac{1}{4}\R)}\R) $

$ \D　　　\sum_{0\le m\le n}\frac{1}{2m+1}\L(1+\frac{\binom{2m}{m}}{2^{2m}}\R)\frac{2^{2n+2}}{(2n+1)^2\binom{2n}{n}}-\sum_{0\le m< n}\frac{1}{2m+1}\L(1+\frac{\binom{2m}{m}}{2^{2m}}\R)\frac{2^{2n}}{n^2\binom{2n}{n}} =8\beta(2)\ln2 $

$ \D　　　2\beta(2)^2 =\sum_{0\le m< n}\L(\frac{1}{(2m+1)^2}\frac{1-(-1)^n}{n^2}+\frac{1}{2m+1}\frac{1-(-1)^n}{n^3}-\frac{\pi^2}{2}\frac{1}{2m+1}\frac{(-1)^{n-1}}{n} \R) $

$ \D　　　6\sum_{0\lt n} \frac{2^{4n}}{(2n+1)^3\binom{2n}{n}^2}\sum_{m=0}^{n-1}\frac{1}{2m+1}\sum_{k=1}^n\frac{1}{k^2} =\sum_{0\le m\lt n}\left(\frac{\pi^3(-1)^m}{(2m+1)^2}-\frac{24\pi(-1)^m}{(2m+1)^4}+\frac{48}{(2m+1)^5} \right)\frac{(-1)^{n-1}}{n} $

$ \D　　　\sum_{0< m\le n}\frac{(-1)^{n-1}}{m^2(2n+1)}+4\sqrt{2}\sum_{0\le n}\frac{\binom{2n}{n}}{2^{3n}(2n+1)^3} =\frac{\pi^3}{6}+\frac{5}{4}\pi\ln^2 2-2\beta(2)\ln2+\frac{\pi}{8}\sum_{0< n}\frac{\binom{2n}{n}^2}{2^{4n}n^2} $

$ \D　　　\sum_{0< m< n}\frac{1}{mn}\frac{2^n\binom{2m}{m}}{2^m\binom{2n}{n}} =\frac{\pi\ln2}{2} $

$ \D　　　\sqrt{\pi}\sum_{0< n}\frac{1}{2^n n^2}\frac{n!}{\Gamma^2\L(\frac{5}{4}+\frac{n}{2} \R)} =2\sum_{0< m+1< n}\frac{\binom{2m}{m}^2}{2^{5m}(m+1)}\frac{1}{n^2} $

$ \D　　　\frac{\binom{2m}{m}}{2^{2m}}\sum_{m< k}\frac{2^{2k}}{k(k+n)\binom{2k}{k}} +\frac{\binom{2n}{n}}{2^{2n}}\sum_{n< k}\frac{2^{2k}}{k(k+m)\binom{2k}{k}} =\pi^2\frac{\binom{2m}{m}\binom{2n}{n}}{2^{2m+2n}} $

$ \D　　　\frac{2^{2m}}{m\binom{2m}{m}}\sum_{k=0}^{m-1}\frac{\binom{2k}{k}}{2^{2k}(k+n)} +\frac{2^{2n}}{n\binom{2n}{n}}\sum_{k=0}^{n-1}\frac{\binom{2k}{k}}{2^{2k}(k+m)} =\frac{2^{2m+2n}}{mn\binom{2m}{m}\binom{2n}{n}} $

$ \D　　　\sum_{0< m\le n}\frac{2^{2m}}{m^2\binom{2m}{m}}\frac{\binom{2n}{n}}{2^{2n}n^2} =\pi^2\ln^2 2-\sum_{0< m\le n}\frac{2^{2n}}{mn^3\binom{2n}{n}} $

$\BA \D　　　\frac{4\beta(2)^2}{\pi}+\sum_{0\le m< n}\frac{2^{4m}}{(2m+1)^2\binom{2m}{m}^2}\frac{\binom{2n}{n}^2}{2^{4n}(2n+1)} &=2\beta(2)\ln2+\frac{1}{2}\sum_{0\le n}\frac{2^{2n}H_n}{(2n+1)^2\binom{2n}{n}}\\ &=2\beta(3)+\sum_{0< m\le n}\frac{2^{2m}}{m^2\binom{2m}{m}}\frac{\binom{2n}{n}}{2^{2n}(2n+1)} \EA$

$ \D　　　\sum_{0< m\le n}\frac{2^{2m}}{m^2\binom{2m}{m}}\frac{\binom{2n}{n}}{2^{2n}(2n+1)^2} +2\sum_{0\le n}\frac{2^{2n}\left(H_n+2\ln2 \right)}{(2n+1)^3\binom{2n}{n}} =8t(2)(\beta(2)-2\pi\ln2) $

$ \D　　　\frac{2^{2n}}{\binom{2n}{n}}\sum_{m< k}\frac{2^{2k}}{k^2\binom{2k}{k}}\frac{k!n!}{(k+n)!} =\frac{2^{2m}}{\binom{2m}{m}}\sum_{n< k}\frac{2^{2k}}{k^2\binom{2k}{k}}\frac{k!m!}{(k+m)!} $

$ \D　　　\sum_{0\le m< n}\frac{2^{2n}}{(2m+1)n^3\binom{2n}{n}} =-\frac{\pi^4}{24}-32t(\overline{3},\overline{1}) $

$ \D　　　\sum_{0\le m< n}\frac{\binom{2m}{m}}{2^{2m}(2m+1)}\frac{2^{2n}}{(2n+1)^3\binom{2n}{n}} =-4t(\overline{1},\overline{3}) $

$ \D　　　\sum_{0<2m\le n}\frac{1}{(2m-1)^2}\frac{(-1)^n}{n}=\frac{1}{8}\sum_{0< m< n}\frac{\binom{2m}{m}}{2^{2m}m}\frac{2^{2n}}{n^2\binom{2n}{n}}=\frac{\pi^2\ln2}{8}-\frac{7}{16}\zeta(3) $

$ \D　　　16\sum_{0\le m< n}\frac{\binom{2n}{n}}{(2m+1)^2(2n+1)^2 2^{2n}} =2\pi\sum_{0<2m< n}\frac{(-1)^{n-1}}{m^2 n} =\pi\sum_{0< m< n}\frac{\binom{2m}{m}}{2^{2m}mn^2} =\frac{\pi^3\ln2}{3}-\frac{3\pi\zeta(3)}{2} $

$ \D　　　\sum_{0\le m< n}\frac{(-1)^{m+n-1}}{(2m+1)^2n} =2t(\V{3})+2t(2,\V{1})+2t(1,\V{2})-t(\V{2})\ln2 $

$ \D　　　\sum_{0<2m+1< n}\frac{2^{4m}}{(2m+1)^2\binom{2m}{m}^2}\frac{(-1)^{n-1}}{n} =-2t(\V{3})-4t(2,\V{1})-4t(1,\V{2})+2t(\V{2})\ln2 $

$\BA \D　　　\pi\beta(2)\ln2 =&-2t(4)+2\pi t(\V{1},2)-4\pi t(1,\V{2})+8t(1,3)\\ &+\sum_{0\le m< n}\L(\frac{\pi}{2}\frac{(-1)^m}{2m+1}-\frac{1}{(2m+1)^2} \R)\frac{1}{n^2} -\sum_{0< m\le n}\frac{4}{m}\L(\frac{\pi}{2}\frac{(-1)^n}{(2n+1)^2}-\frac{1}{(2n+1)^3}\R) \EA$

$ \D　　　\sum_{0\le n}\frac{\L(\frac{1}{2}+\alpha \R)_n}{\L(n+\frac{1}{2} \R)^2\L(\frac{1}{2} \R)_n} =\sum_{0< n}\frac{2^{2n}}{n(n-\alpha)\binom{2n}{n}} $

$ \D　　　\sum_{0\le n}\frac{2^{2n}}{(2n+1)^2\binom{2n}{n}}=2\beta(2) $

$\BA \D　　　\sum_{0\le n}\frac{2^{2n}}{(2n+1)^3\binom{2n}{n}} &=2\beta(2)\ln2-2\sum_{0< m\le n}\frac{(-1)^{n-1}}{m(2n+1)^2}\\ &=4t(\overline{1},2)-4t(1,\overline{2})\\ &=2t(\V{3})+4t(2,\V{1}) \EA$

$\BA \D　　　\sum_{0\le n}\frac{2^{2n}}{(2n+1)^4\binom{2n}{n}} &=\frac{\ln2}{2}\sum_{0\le n}\frac{2^{2n}}{(2n+1)^3\binom{2n}{n}} +\frac{7}{8}\pi\zeta(3)-\frac{1}{8}\pi^3\ln2 -2\sum_{0< l\le m\le n}\frac{(-1)^{n}}{l(2m+1)^2(2n+1)} -2\sum_{0< l\le m< n}\frac{(-1)^{n}}{l(2m+1)(2n+1)^2}\\ &=4t(2)t(\V{2})-2t(\V{4})-8t(\V{2},2)+8t(1,1,\V{2})-8t(1,\V{1},2)\\ &=2t(\V{4})+4t(2,\V{2})+4t(3,\V{1})+8t(2,1,\V{1}) \EA$

$ \D　　　\sum_{0\leq n}\frac{2^{2n}}{(2n+1)^a\binom{2n}n}=\sum_{\substack{{\rm wt}({\boldsymbol k})=a\\k_1\geq 2}}2^jt(k_1,\dots,k_{j-1},\overline{k_j}) $　　　　　(訂正後。出処：コメント欄)$ $

$ \D　　　\sum_{0< n}\frac{2^{2n}}{n^2\binom{2n}{n}}=4t(2) $

$ \D　　　\sum_{0< n}\frac{2^{2n}}{n^3\binom{2n}{n}} =-4t(3)-6\zeta(2)\zeta(\V{1}) $

$ \D　　　\sum_{0< n}\frac{2^{2n}}{n^4\binom{2n}{n}} =\zeta(2,2)+4\zeta(\V{2},2)+4\zeta(\V{1},3)-4\zeta(1,\V{3})+12\zeta(2)\zeta(1,\V{1}) $

$ \D　　　\sum_{0< m< n}\frac{2^{2n}}{mn^2\binom{2n}{n}} =7\zeta(3) $

$ \D　　　\sum_{0< m< n}\frac{2^{2n}}{m^2n^2\binom{2n}{n}} =4t(4) $

$ \D　　　\sum_{0< m< n}\frac{2^{2n}}{mn^3\binom{2n}{n}} =14\zeta(2,2)-4\zeta(\V{2},2)-4\zeta(\V{1},3)+4\zeta(1,\V{3})+4\zeta(3,\V{1})-4\zeta(\V{3},\V{1}) $

$ \D　　　\sum_{n=1}^\infty \frac{\binom{2n}{n}^2}{2^{4n}n^2}\left(8\beta(2)+\sum_{m=1}^n\frac{2^{4m}}{m^2\binom{2m}{m}^2} \right) =2\pi\sum_{n=0}^\infty \frac{\binom{2n}{n}^2}{2^{4n}}\left(\frac{\zeta(2)}{2n+1}-\frac{4}{(2n+1)^2}\left(\sum_{m=0}^n\frac{1}{2m+1}-\ln2\right)\right) $

$ \D　　　\sum_{n=0}^\infty \frac{\binom{2n}{n}^2}{2^{4n}(2n+1)^2}\left(8\beta(2)+\sum_{m=1}^n\frac{2^{4m}}{m^2\binom{2m}{m}^2} \right) =2\pi t(3)+4t(2)\left(\pi\ln2-2\beta(2) \right)-\frac{\pi}{2}\sum_{n=1}^\infty \frac{\binom{2n}{n}^2}{2^{4n}n^2}\sum_{m=0}^{n-1}\frac{1}{2m+1} $

$\BA \D　　　\sum_{n=1}^\infty \frac{\binom{2n}{n}}{2^{2n}n}\sum_{k=0}^{n-1}\frac{1}{2k+1}\sum_{l=0}^{n-1}\frac{1}{(2l+1)^2} &=2t(4)+t(2)\sum_{n=1}^\infty \frac{\binom{2n}{n}^2H_n}{2^{4n}n}\\ &=\frac{\pi}{8}\sum_{n=1}^\infty\frac{1}{n^2}\sum_{m=0}^{n-1}\frac{\binom{2m}{m}}{2^{2m}(2m+1)}+\frac{\pi}{8}\sum_{n=1}^\infty \frac{\binom{2n}{n}}{2^{2n}(2n+1)}\sum_{m=1}^{n}\frac{2^{2m}}{m^2\binom{2m}{m}}\\ &=\pi\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^2}\sum_{m=0}^{n-1}\frac{(-1)^m}{2m+1}\\ &=\frac{\pi}{8}\sum_{n=1}^\infty\frac{2^{4n}}{n^3\binom{2n}{n}^2}\sum_{m=0}^{n-1}\frac{\binom{2m}{m}^2}{2^{4m}} \EA$

$ \D　　　\frac{\pi x}{2}\sum_{n=0}^\infty \frac{\binom{2n}{n}^2}{2^{4n}(n+x)} =\sum_{n=0}^\infty \frac{1}{2n+1}\frac{\left(\frac{1}{2}+x \right)_n}{\left(1+x \right)_n} $

$ \D　　　{_3}F_2\left[\begin{matrix}\frac{1}{2},\frac{1}{2},\alpha\\ \frac{3}{2},\beta \end{matrix};1 \right] =\frac{\Gamma(\beta)\Gamma\left(\frac{1}{2}-\alpha+\beta \right)}{\Gamma(\beta-\alpha)\Gamma\left(\frac{1}{2}+\alpha \right)} {_3}F_2\left[\begin{matrix}1,1,\alpha\\ \frac{3}{2},\frac{1}{2}+\beta \end{matrix};1 \right] $

$\BA \D　　　\sum_{n=0}^\infty \frac{\binom{2n}{n}^3}{2^{6n}}\left(3\zeta(2)+\sum_{m=1}^n\frac{2^{2m}}{m^2\binom{2m}{m}} \right) &=\frac{1}{\pi}\sum_{n=1}^\infty \frac{2^{6n}}{n^3\binom{2n}{n}^3}\sum_{m=0}^{n-1}\frac{\binom{2m}{m}^2}{2^{4m}}\\ &=4\sum_{n=0}^\infty\frac{2^{2n}}{(2n+1)^2\binom{2n}{n}}\sum_{m=0}^n\frac{\binom{2m}{m}^3}{2^{6m}} \EA$

$\BA \D　　　\sum_{n=1}^{\infty}\frac{\binom{2n}{n}^3}{2^{6n}n}\left(3\zeta(2)+\sum_{m=1}^n\frac{2^{2m}}{m^2\binom{2m}{m}} \right) &=18\zeta(2)\ln2-4t(3)-4\pi\beta(2)-\frac{1}{\pi}\sum_{n=1}^\infty \frac{2^{6n}}{n^4\binom{2n}{n}^3}\sum_{m=0}^{n-1}\frac{\binom{2m}{m}^2}{2^{4m}} \EA$

$ \D　　　8\beta(2)=\sum_{n=1}^\infty \frac{2^{2n}}{n^2\binom{2n}{n}}\sum_{m=0}^{n-1}\frac{\binom{2m}{m}^2}{2^{4m}} $

$\BA \D　　　2\zeta(2)\beta(2)+\sum_{n=1}^\infty \frac{1}{n^2}\sum_{m=0}^{n-1}\frac{2^{4m}}{(2m+1)^2\binom{2m}{m}^2} &=\sum_{n=0}^\infty\frac{2^{2n+2}}{(2n+1)^2\binom{2n}{n}}\sum_{m=0}^n\left(\frac{\pi}{2}\frac{\binom{2m}{m}}{2^{2m}(2m+1)}-\frac{1}{(2m+1)^2} \right)\\ &=2\sum_{n=1}^\infty\frac{\binom{2n}{n}\left(2\ln2+H_n \right)}{2^{2n}(2n+1)}\sum_{m=0}^{n-1}\frac{1}{(2m+1)^2}\\ &=4\pi t(3)-8\beta(4) \EA$

$ \D　　　\frac{\pi}{2}\frac{1}{\sqrt{1-x^2}}\sum_{n=0}^\infty \frac{\binom{2n}{n}^2}{2^{4n}}x^{2n} =\sum_{n=0}^\infty \frac{2^{2n}x^{2n}}{(2n+1)\binom{2n}{n}}\sum_{m=0}^n\frac{\binom{2m}{m}^2}{2^{4m}x^{2m}} $

$\BA \D　　　\chi_2\L(\frac{1}{\sqrt{2}} \R) &=\frac{5}{6}t(2)-\frac{1}{8}\sum_{n=1}^\infty \frac{\binom{4n}{2n}}{2^{2n}n^2\binom{2n}{n}}\\ &=t(2)-\frac{\pi}{16}\sum_{n=0}^\infty \frac{\binom{2n}{n}\binom{4n+2}{2n+1}}{2^{6n}(2n+1)} \EA$

$ \D　　　\sum_{n=1}^\infty \frac{2^{2n}}{n^2\binom{2n}{n}}\sum_{m=0}^{n-1}\frac{\binom{2m}{m}^2}{2^{4m}}=2\pi\ln2+2\sum_{n=0}^\infty\frac{\binom{2n}{n}H_n}{2^{2n}(2n+1)} $

$ \D　　　\frac{\pi}{2}\frac{\tanh^{-1}x}{\sqrt{1-x^2}}+\frac{2}{1+x}\sum_{n=0}^\infty \frac{\left(-\frac{1-x}{1+x} \right)^n}{(2n+1)^2} =\sum_{n=0}^\infty \frac{\binom{2n}{n}}{2^{2n}}\,x^{2n}\left(2\beta(2)+\frac{1}{4}\sum_{m=1}^n\frac{2^{4m}}{m^2\binom{2m}{m}^2} \right) $

$ \D　　　2\sum_{n=1}^\infty \frac{(\alpha)_n}{n^2(2\alpha)_n}=\zeta(2)+\psi'(2\alpha)-\left(\psi(\alpha)-\psi(2\alpha)\right)^2 $

$ \D　　　\sum_{n=0}^\infty \frac{\binom{2n}{n}^3}{2^{6n}}\sum_{m=1}^n\frac{2^{4m}}{m^2\binom{2m}{m}^2} =\left(\pi^2-8\beta(2) \right)\sum_{n=0}^\infty \frac{\binom{2n}{n}^3}{2^{6n}} $

$ \D　　　\zeta(2)+\sum_{n=1}^\infty \frac{\binom{2n}{n}^2H_n}{2^{4n}n} =\frac{1}{\pi}\sum_{n=1}^\infty \frac{2^{4n}}{n^3\binom{2n}{n}^2}\sum_{m=0}^{n-1}\frac{\binom{2m}{m}^2}{2^{4m}} $

$ \D　　　\zeta(3)+\sum_{n=1}^\infty \frac{\binom{2n}{n}^2}{2^{4n}n}\left(\zeta(2)-\frac{H_n}{n}\right) =\frac{1}{\pi}\sum_{n=1}^\infty \frac{2^{4n}}{n^4\binom{2n}{n}^2}\sum_{m=0}^{n-1}\frac{\binom{2m}{m}^2}{2^{4m}} $

$ \D　　　\sum_{n=1}^\infty \frac{2^{2n}}{n^3\binom{2n}{n}}\sum_{m=0}^{n-1}\frac{\binom{2m}{m}^2}{2^{4m}} =4\pi\sum_{n=0}^\infty\frac{\binom{2n}{n}^2}{2^{4n}(2n+1)^2}-16\beta(2)\ln2 $

$ \D　　　\sum_{m=1}^\infty\frac{\binom{2m}{m}}{2^{2m+1}m}\sum_{l=0}^{m-1}\frac{2^{4l}}{(2l+1)^3\binom{2l}{l}^2} -\sum_{m=0}^{n-1}\frac{2^{4m}}{(2m+1)^2\binom{2m}{m}^2}\left(\frac{\pi^2}{24}-\frac{1}{2m+1}\left(-\ln2+\sum_{k=0}^{m}\frac{1}{2k+1} \right) \right) =\frac{2^{4n}}{\binom{2n}{n}^2}\sum_{m=1}^\infty\frac{2^{4m}}{(2m)^3\binom{2m}{m}^2}\sum_{l=0}^{m-1}\frac{\binom{2l}{l}^2}{2^{4l}(2l+2n+1)} $

　　　$ \D \sum_{n=1}^\infty \left(\frac{2x}{1+x^2}\right)^{2n}\frac{\binom{2n}{n}}{2^{2n}n}\sum_{m=1}^n\frac{2^{2m}}{m^2\binom{2m}{m}} =4\chi_3(x^2) $

　　　$ \D \sum_{0\le m< n}\frac{2^{2n}}{n^2\binom{2n}{n}}\frac{\binom{2m}{m}^2}{2^{4m}}\left((-1)^m+\frac{1}{2^{m+\frac{1}{2}}}\right)=\sqrt{\frac{\pi}{2}}\frac{\Gamma^2\left(\frac{1}{4} \right)}{2} $

$ \D　　　\sum_{0< m\le n}\frac{2^{2m}}{m^2\binom{2m}{m}}\frac{(-1)^{n-1}\binom{2n}{n}^2}{2^{4n}} =\sum_{0\le m< n}\frac{\binom{2m}{m}^2}{2^{4m}n^2}\left((-1)^m-\frac{1}{2^{m+\frac{1}{2}}} \right) $

　　　$ \D 2\sum_{0< m\le n}\frac{2^{4m}}{m^2\binom{2m}{m}^2}\frac{\binom{2n}{n}}{2^{2n}}(-1)^{m+n} =\sum_{0\le m< n}\frac{\binom{2m}{m}^2}{2^{5m}}\frac{2^{2n}}{n^2\binom{2n}{n}} $

　　　$ \D \sum_{0< m< n}\frac{1}{m^3n^2\binom{2n}{n}}=\frac{58}{9}\zeta(5)-\frac{4}{9}\zeta(2)\zeta(3)-2\pi\sum_{0< n}\frac{1}{n^4}\sin\frac{\pi n}{3} $

　　　$ \D \sum_{0< m< n}\frac{2^{2m}}{m^2\binom{2m}{m}}\frac{1}{n^2}=\sum_{0< m< n}\frac{\binom{2m}{m}}{2^{2m}m^2}\frac{2^{2n}}{n^2\binom{2n}{n}} $

　　　$ \D \sum_{j=0}^{r-1}\sum_{0< m< n}\frac{(-1)^{m-1}}{m^{j+1}n^{r-j}}\L(\frac{(-1)^n}{m}-\frac{1}{n}\R) =\sum_{0< m< n}\frac{(-1)^{n-1}}{mn}\L(\frac{1}{m^r}-\frac{1}{n^r}\R) $

　　　$ \D \sum_{m=0}^{n-1}\frac{(32m+17)2^{6m}}{(4m+1)\binom{2m}{m}\binom{4m}{2m}}=\frac{(4n-1)2^{6n}}{\binom{2n}{n}\binom{4n}{2n}}+1　 $

\begin{align*} 　　　　　&\frac{\Gamma\left(\frac{1}{2}\right)\Gamma(1+a)\Gamma^2(b)}{\Gamma(a+b)\Gamma\left(\frac{1}{2}+b\right)}\frac{(1+a)_n}{(a+b)_n}\sum_{0\le m}\frac{\binom{2m}{m}(b)_m}{\left(\frac{1}{2}+b\right)_m2^{2m}(a+b+m+n)}\\ =&\frac{\left(\frac{1}{2}+a\right)_n}{\left(\frac{1}{2}+a+b\right)_n}\left(\sum_{0\le m}\frac{(1-b)_m}{m!\left(\frac{1}{2}+a+m\right)}\left(\gamma+\psi(1+a)+2\ln2+\sum_{k=0}^{m-1}\frac{1}{k+1+a}\right)+\frac{2}{1+2a}\frac{\Gamma(1+a)\Gamma(b)}{\Gamma(1+a+b)}\sum_{m=0}^{n-1}\frac{\left(\frac{1}{2}+a+b\right)_m(1+a)_m}{\left(\frac{3}{2}+a\right)_m(1+a+b)_m}\right) \end{align*}
　　　　　　（ただし$a>-\frac{1}{2}$）

$ \D　　　 \sum_{m=0}^\infty \frac{\binom{2m}{m}}{2^{2m}}\frac{\L(\frac{1}{2}+a\R)_m}{(1+a)_m(a+m+n)} =\frac{\Gamma(a)}{\Gamma\L(\frac{1}{2}\R)\Gamma\L(\frac{1}{2}+a\R)}\frac{2^{2n}}{n\binom{2n}{n}}\frac{(a)_n}{\L(\frac{1}{2}+a\R)_n} \sum_{m=0}^{n-1}\frac{\binom{2m}{m}}{2^{2m}}\frac{\L(\frac{1}{2}+a\R)_m}{(1+a)_m} $

$\underline{　\hspace{100cm}}$

$ \D　　　Z_C(s)=\sum_{n=0}^\infty \frac{\binom{2n}{n}}{2^{2n}\left(n+\frac{1}{2}\right)^s} \, ,　　Z_{C^2}(s)=\sum_{n=0}^\infty \frac{\binom{2n}{n}^2}{2^{4n}\left(n+\frac{1}{2}\right)^s} \, ,　　T_1(s)=\sum_{n=1}^\infty \frac{2^{2n}}{n^s\binom{2n}{n}}\sum_{m=0}^{n-1}\frac{\binom{2m}{m}^2}{2^{4m}} $

と定義するとき

$ \D　　　\sum_{j=0}^n Z_C(j+1)T_1(n-j+2)=\pi^2Z_{C^2}(n+1) $

$ \D　　　 $

$ \D　　　 $