高冪二項係数付き級数の関係式

$\displaystyle \beta_n\coloneqq \frac{\binom{2n}{n}}{2^{2n}}$
気が向き次第順次追加します

\begin{align} &\frac{\pi^2}{4}\sum_{0\leq n}\beta_n^5\left(2\beta(2)+\sum_{m=1}^n\frac{1}{(2m)^2\beta_m^2}\right)+\frac{2}{\pi}\sum_{0< n}\frac{1}{(2n)^5\beta_n^5}\left(\sum_{m=0}^{n-1}\beta_m^2\right)^2\\ &=\sum_{0< n}\frac{1}{(2n)^2\beta_n}\sum_{m=0}^{n-1}\beta_m^5\left(2\beta(2)+\sum_{k=1}^m\frac{1}{(2k)^2\beta_k^2}\right)+\sum_{0\leq n}\frac{1}{(2n+1)^2\beta_n}\sum_{m=0}^{n}\beta_m^5\left(2\beta(2)+\sum_{k=1}^m\frac{1}{(2k)^2\beta_k^2}\right) \end{align}

\begin{align} &\frac{\pi^2}{4}\sum_{0\leq n}\beta_n^5\left(2\beta(2)+\sum_{m=1}^n\frac{1}{(2m)^2\beta_m^2}\right) +\sum_{0\leq n}\beta_n^5\left(2\beta(2)+\sum_{m=1}^n\frac{1}{(2m)^2\beta_m^2}\right)^2\\ =&\frac{\pi}{2}\sum_{0\leq n}\beta_n^5\left(\frac{7}{2}\zeta(3)+\sum_{m=1}^n\frac{4m-1}{(2m)^4\beta_m^4}\right)+\sum_{0< n}\frac{1}{(2n)^5\beta_n^5}\left(\sum_{m=0}^{n-1}\beta_m^2\right)\sum_{m=0}^{n-1}\beta_m^4(4m+1) \end{align}

\begin{align} \pi\sum_{0< n}\frac{1}{(2n)^4\beta_n^4}\left(\sum_{m=0}^{n-1}\beta_m^2\right)^3+&\sum_{0< n}\frac{1}{(2n)^4\beta_n^4}\left(\sum_{m=0}^{n-1}\beta_m^2\right)\sum_{m=0}^{n-1}\beta_m^2\left(\ln2+\sum_{k=0}^{m-1}\frac{1}{2k+1}\right)\\ =\pi&\sum_{0< n}\frac{1}{(2n)^4\beta_n^4}\left(\sum_{m=0}^{n-1}(4m+1)\beta_m^4\right)\sum_{m=0}^{n-1}\beta_m^2\left(2\ln2+\sum_{k=1}^m\frac{1}{2k}\right) \end{align}