二項係数の3乗が入った二重級数まとめ

\begin{align*} \beta_n:=\frac{\binom{2n}{n}}{2^{2n}},\quad A:=\sum_{n=0}^{\infty}\beta_n^3,\quad B&:=\sum_{n=1}^{\infty}\frac 1{n^3\beta_n^3}, \quad C:=\sum_{n=0}^{\infty}\frac{(-1)^n}{\left(n+\frac 12\right)^2} \end{align*}
とする.

分母が整数なもの

\begin{align*} \sum_{n=0}^{\infty}\beta_n^3&=A\\ \sum_{n=0}^{\infty}\beta_n^3\sum_{k=1}^n\frac 1k&=2A\left(\frac{\pi}3-\ln2\right)\\ \sum_{n=0}^{\infty}\beta_n^3\sum_{k=1}^n\frac 1{k^2}&=A\left(C-\frac{\pi^2}3\right)\\ \sum_{n=0}^{\infty}\beta_n^3\sum_{k=1}^n\frac 1{k^3}&=\text{?}\\ \sum_{n=0}^{\infty}\beta_n^3\sum_{k=1}^n\frac 1{k^2\beta_k}&=\frac{A\pi^2}2-\frac{B}{\pi}\\ \sum_{n=0}^{\infty}\beta_n^3\sum_{k=1}^n\frac 1{k^3\beta_k}&=\text{?}\\ \sum_{n=0}^{\infty}\beta_n^3\sum_{k=1}^n\frac 1{k^2\beta_k^2}&=(\pi^2-2C)A\\ \sum_{n=0}^{\infty}\beta_n^3\sum_{k=1}^n\frac 1{k^3\beta_k^2}&=(\pi^3-14\zeta(3))A-B\\ \sum_{n=0}^{\infty}\beta_n^3\sum_{k=1}^n\frac 1{k^3\beta_k^3}&=A\left(B-\frac{A\pi^3}3\right)\\ \end{align*}

分母が半整数なもの

\begin{align*} \sum_{n=0}^{\infty}\beta_n^3\sum_{k=0}^{n-1}\frac 1{k+\frac 12}&=\frac{A\pi}3\\ \sum_{n=0}^{\infty}\beta_n^3\sum_{k=0}^{n-1}\frac 1{\left(k+\frac 12\right)^2}&=\left(\frac{\pi^2}2-C\right)A\\ \sum_{n=0}^{\infty}\beta_n^3\sum_{k=0}^{n-1}\frac 1{\left(k+\frac 12\right)^3}&=\text{?}\\ \sum_{n=0}^{\infty}\beta_n^3\sum_{k=0}^{n-1}\frac 1{\left(k+\frac 12\right)^2\beta_k}&=(2C-\pi^2)A+\frac{B}{\pi}\\ \sum_{n=0}^{\infty}\beta_n^3\sum_{k=0}^{n-1}\frac 1{\left(k+\frac 12\right)^3\beta_k}&=\text{?}\\ \sum_{n=0}^{\infty}\beta_n^3\sum_{k=0}^{n-1}\frac 1{\left(k+\frac 12\right)^2\beta_k^2}&=(\pi^2-2C)A\\ \sum_{n=0}^{\infty}\beta_n^3\sum_{k=0}^{n-1}\frac 1{\left(k+\frac 12\right)^3\beta_k^2}&=(28\zeta(3)-\pi^3)A\\ \sum_{n=0}^{\infty}\beta_n^3\sum_{k=0}^{n-1}\frac 1{\left(k+\frac 12\right)^3\beta_k^3}&=A\left(B-\frac{A\pi^3}3\right)\\ \end{align*}

$\left(k+\frac 14\right)\beta_k^3$が入っているもの

\begin{align*} \sum_{n=1}^{\infty}\frac 1{n^2}\sum_{k=0}^{n-1}\left(k+\frac 14\right)\beta_k^3&=A-\frac 1{A}\left(\frac 12+\frac{C}{\pi^2}\right)\\ \sum_{n=1}^{\infty}\frac 1{n^3}\sum_{k=0}^{n-1}\left(k+\frac 14\right)\beta_k^3&=\text{?}\\ \sum_{n=1}^{\infty}\left(\frac 1{n^2\beta_n}\sum_{k=0}^{n-1}\left(k+\frac 14\right)\beta_k^3-\frac{2}{\pi n}\right)&=\text{?}\\ \sum_{n=1}^{\infty}\frac 1{n^3\beta_n}\sum_{k=0}^{n-1}\left(k+\frac 14\right)\beta_k^3&=\text{?}\\ \sum_{n=1}^{\infty}\frac 1{n^3\beta_n^2}\sum_{k=0}^{n-1}\left(k+\frac 14\right)\beta_k^3&=\text{?}\\ \sum_{n=1}^{\infty}\left(\frac 1{n^3\beta_n^3}\sum_{k=0}^{n-1}\left(k+\frac 14\right)\beta_k^3-\frac 2n\right)&=4\ln2-\pi\\ \sum_{n=0}^{\infty}\frac 1{\left(n+\frac 12\right)^2}\sum_{k=0}^n\left(k+\frac 14\right)\beta_k^3&=\frac{C}{\pi^2A}+A\\ \sum_{n=0}^{\infty}\frac 1{\left(n+\frac 12\right)^3}\sum_{k=0}^n\left(k+\frac 14\right)\beta_k^3&=\text{?}\\ \sum_{n=0}^{\infty}\left(\frac 1{\left(n+\frac 12\right)^2\beta_n}\sum_{k=0}^n\left(k+\frac 14\right)\beta_k^3-\frac {2}{\pi(n+1)}\right)&=\text{?}\\ \sum_{n=0}^{\infty}\frac 1{\left(n+\frac 12\right)^3\beta_n}\sum_{k=0}^n\left(k+\frac 14\right)\beta_k^3&=\text{?}\\ \sum_{n=0}^{\infty}\frac 1{\left(n+\frac 12\right)^3\beta_n^2}\sum_{k=0}^n\left(k+\frac 14\right)\beta_k^3&=\frac 1{A}\left(\frac{2C}{\pi}+\pi\right)\\ \sum_{n=0}^{\infty}\left(\frac 1{\left(n+\frac 12\right)^3\beta_n^3}\sum_{k=0}^{n}\left(k+\frac 14\right)\beta_k^3-\frac 2{n+1}\right)&=4\ln2-\pi\\ \end{align*}