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非常に興味深い級数一覧

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${\bf NOTATION}$


$\begin{align*}\displaystyle {\frak A}_n&=\frac{(-1)^n\binom{2n}{n}^3}{2^{6n}}\sum_{n< m}\frac{(-1)^{m-1}2^{6m}(4m-1)}{(2m)^3\binom{2m}{m}^3}\\ {\frak B}_n&=\frac{(-1)^n2^{6n}}{(2n+1)^3\binom{2n}{n}^3}\sum_{m=0}^n\frac{(-1)^m(4m+1)\binom{2m}{m}^3}{2^{6m}}\\ {\frak C}_n&=\frac{\binom{2n}{n}^4}{2^{4n}}\left(\frac{7}{2}\zeta(3)+\sum_{m=1}^n\frac{2^{8m}(4m-1)}{(2m)^4\binom{2m}{m}^4}\right)\\ {\frak D}_n&=\frac{2^{8n}}{(2n+1)^4\binom{2n}{n}^4}\sum_{m=0}^n\frac{(4m+1)\binom{2m}{m}^4}{2^{8m}}\\ {\frak E}_n&=\frac{\binom{2n}{n}^2}{2^{4n}}\left(2\beta(2)+\sum_{m=1}^n\frac{2^{4m}}{(2m)^2\binom{2m}{m}^2}\right)\\ {\frak F}_n&=\frac{2^{4n}}{(2n+1)^2\binom{2n}{n}^2}\sum_{m=0}^n \frac{\binom{2m}{m}^2}{2^{4m}}\\ V_n&=\frac{1}{2^{4n}}\sum_{m=0}^n \binom{2n-2m}{n-m}^2\binom{2m}{m}^2 \end{align*}$



${\bf PRODUCTS}$


$\begin{align*}\displaystyle \sum_{n=0}^\infty {\frak A}_n =2\sum_{n=0}^\infty {\frak B}_n =\frac{{\Gamma(\frac{1}{8})}^2{\Gamma(\frac{3}{8})}^2}{48\pi} \end{align*}$

$\begin{align*}\displaystyle \sum_{n=0}^\infty \frac{(-1)^n(4n+1)\binom{2n}{n}}{2^{2n}}{\frak A}_n=\frac{\pi}{2} \end{align*}$

$\begin{align*}\displaystyle 2\pi\sum_{n=0}^\infty (-1)^n{\frak A}_n =\frac{\pi^2}{2}\sum_{n=0}^\infty \frac{(4n+1)\binom{2n}{n}^2}{2^{4n}}{\frak A}_n =\frac{\pi}{2}\sum_{n=0}^\infty \frac{\binom{2n}{n}}{2^{2n}}{\frak E}_n =2\sum_{n=1}^\infty \frac{n\binom{2n}{n}}{2^{2n}}{\frak F}_{n-1}^2 =\sum_{n=0}^\infty \left(\frac{2{\frak E}_n}{4n+1}+\frac{2{\frak F}_n}{4n+3}\right) =\frac{{\Gamma(\frac{1}{4})}^4}{16} \end{align*}$

$\begin{align*}\displaystyle \pi\sum_{n=0}^\infty \frac{(-1)^n(4n+1)\binom{2n}{n}^3}{2^{6n}}{\frak A}_n =\sum_{n=0}^\infty \frac{(4n+1)\binom{2n}{n}^2}{2^{4n}}{\frak C}_n =\frac{\pi^2}{2}\sum_{n=0}^\infty \frac{\binom{2n}{n}^4}{2^{8n}} =\frac{4}{\pi^2}\sum_{n=0}^\infty \left({\frak C}_n\left(\frac{1}{4n+1}+\sum_{m=0}^{2n-1}\frac{2}{2m+1}\right)+{\frak D}_n\left(\frac{1}{4n+3}+\sum_{m=0}^{2n}\frac{2}{2m+1}\right)\right) \end{align*}$

$\begin{align*}\displaystyle \frac{\pi^4}{32}\sum_{n=0}^\infty \frac{(4n+1)\binom{2n}{n}^6}{2^{12n}} =\sum_{n=0}^\infty (4n+1){\frak A}_n^2 =\frac{2}{3}\sum_{n=0}^\infty {\frak C}_n =2\sum_{n=0}^\infty {\frak D}_{n} =\frac{\pi}{2}\sum_{n=0}^\infty \frac{\binom{2n}{n}^2}{2^{4n}}{\frak E}_n \end{align*}$

$\begin{align*}\displaystyle \sum_{n=0}^\infty \frac{{\frak C}_n}{4n+1} =11\sum_{n=0}^\infty \frac{{\frak D}_n}{4n+3} =\frac{11\pi^3}{96}\sum_{n=0}^\infty \frac{\binom{2n}{n}^4}{2^{8n}(4n+1)} =\frac{11}{7680}\frac{{\Gamma(\frac{1}{4})}^8}{\pi^2} \end{align*}$

$\begin{align*}\displaystyle \sum_{n=0}^\infty \Big((2n+1){\frak A}_n+(2n+2){\frak A}_{n+1}\Big){\frak B}_n =\frac{7}{4}\zeta(3) \end{align*}$

$\begin{align*}\displaystyle \sum_{n=0}^\infty \frac{(-1)^n(2n+1)\binom{2n}{n}}{2^{2n}}\Big((2n+1){\frak A}_n+(2n+2){\frak A}_{n+1}\Big){\frak D}_n =2\beta(3) \end{align*}$

$\begin{align*}\displaystyle \sum_{n=0}^\infty \Big((2n+1){\frak C}_n+(2n+2){\frak C}_{n+1}\Big){\frak D}_n =\frac{93}{16}\zeta(5) \end{align*}$

$\begin{align*}\displaystyle \frac{\pi^6}{64}\sum_{n=0}^\infty \frac{(4n+1)\binom{2n}{n}^8}{2^{16n}} =\sum_{n=0}^\infty \Big((4n+1){\frak C}_n^2-(4n+3){\frak D}_n^2\Big) \end{align*}$

$\begin{align*}\displaystyle \pi^2\sum_{n=0}^\infty \frac{(-1)^n\binom{2n}{n}^3}{2^{6n}}\L(2\sum_{m=0}^n \frac{(-1)^m\binom{2m}{m}}{2^{2m}}-\frac{(-1)^n\binom{2n}{n}}{2^{2n}}\R)=\frac{{\Gamma(\frac{1}{8})}^2{\Gamma(\frac{3}{8})}^2}{16\pi} +\sum_{n=0}^\infty \frac{(4n+1)\binom{4n}{2n}}{(2n+1)^3\binom{2n}{n}^2} \end{align*}$

$\begin{align*}\displaystyle \sum_{n=0}^\infty \frac{2^{4n}(4n+3){\frak B}_n}{(2n+1)^2\binom{2n}{n}^2}=\pi\sum_{n=0}^\infty \frac{\binom{2n}{n}^2}{2^{4n}(4n+1)^2}=\frac{\pi}{8}\sum_{n=0}^\infty \frac{\binom{2n}{n}^3}{2^{6n}}\L(\frac{\pi^2}{2}+\sum_{m=1}^n \frac{2^{2m}}{m^2\binom{2m}{m}}\R) \end{align*}$

$\begin{align*}\displaystyle \sum_{n=0}^\infty \frac{(4n+1)\binom{2n}{n}}{2^{2n}}\frac{{(\frac{1}{3})}_n}{{(\frac{7}{6})}_n}{\frak C}_n =\frac{\sqrt{3}\,{\Gamma(\frac{1}{3})}^{12}}{2^7\sqrt[3]{2}\,\pi^5} \end{align*}$

$\begin{align*}\displaystyle \frac{\pi^3}{4}\sum_{n=0}^\infty \frac{(4n+1)\binom{2n}{n}^4}{2^{8n}}V_{2n} =\sum_{n=0}^\infty \Big((4n+1){\frak C}_nV_{2n}+(4n+3){\frak D}_nV_{2n+1}\Big) =\frac{\Gamma\L(\frac{1}{4}\R)^8}{32\pi^4} \end{align*}$

$\begin{align*}\displaystyle \sum_{n=0}^\infty \L(\frac{(4n+1)\binom{2n}{n}^2}{2^{4n}}(2\ln2+H_{2n}){\frak C}_n+\frac{(4n+3)(2n+1)^2\binom{2n}{n}^2}{2^{4n}}{\frak D}_n^2\R) =\sum_{n=0}^\infty \frac{\binom{2n}{n}^4}{2^{8n}}(2\ln2+H_{2n}) \end{align*}$

$\begin{align*}\displaystyle \sum_{n=0}^\infty \frac{(4n+3)(2n+1)^2\binom{2n}{n}^2}{2^{4n}}{\frak D}_n^2=\frac{\pi^4}{16}\sum_{n=0}^\infty \frac{(4n+1)\binom{2n}{n}^6}{2^{12n}} \end{align*}$

$\begin{align*}\displaystyle \sum_{n=0}^\infty \frac{(4n+1)\binom{2n}{n}^4}{2^{8n}}\sum_{2n< m}\frac{(-1)^{m-1}}{m^2}=\frac{7\zeta(3)}{\pi^2} \end{align*}$

$\begin{align*}\displaystyle \sum_{n=0}^\infty \frac{2^{4n}(4n+1)}{\binom{2n}{n}^2}{\frak A}_n^2=\frac{7}{2}\zeta(3) \end{align*}$

$\begin{align*}\displaystyle \sum_{n=0}^\infty \L((4n+1){\frak E}_n^2-(4n+3){\frak F}_n^2\R)=\frac{\pi^2\ln2}{4} \end{align*}$



${\bf CONJECTURE}$


$\begin{align*}\displaystyle \sum_{n=0}^\infty (4n+1){\frak A}_n^2=\sum_{n=0}^\infty (4n+3){\frak B}_n^2 \end{align*}$

$\begin{align*}\displaystyle \sum_{n=0}^\infty (4n+1){\frak C}_n^2=5\sum_{n=0}^\infty (4n+3){\frak D}_n^2 \end{align*}$

$\begin{align*}\displaystyle \sum_{n=1}^\infty \frac{2n\binom{2n}{n}^4}{2^{8n}}{\frak D}_{n-1}+\sum_{n=0}^\infty \frac{(2n+1)\binom{2n}{n}^4}{2^{8n}}{\frak D}_{n}=\frac{\pi^2}{8} \end{align*}$

$\begin{align*}\displaystyle \sum_{n=1}^\infty \frac{2^{12n}(4n-1)}{(2n)^6\binom{2n}{n}^6}\left(\sum_{k=0}^{n-1}\frac{(4k+1)\binom{2k}{k}^4}{2^{8k}}\right)^2=\frac{\pi^4}{16}\sum_{n=0}^\infty \frac{(4n+1)\binom{2n}{n}^6}{2^{12n}} \end{align*}$

 

投稿日:202351

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