等式を示す

みさき氏の級数を示します！

以下、$\displaystyle C_r=2^{-2r}\binom{2r}{r}=\frac{(2r-1)!!}{(2r)!!}$とする。

\begin{align} \sum_{n=1}^\infty\frac{H_n^{(2)}}{n4^n}\binom{2n}{n}&=\sum_{n=1}^\infty \frac{C_n}{n}\sum_{m=1}^n\frac{1}{m^2} \\ &=\sum_{0< m\leq n}\frac{C_n}{m^2n} \\ &=\sum_{0< m\leq n}\int_0^1t^{m-1}dt\frac{C_n}{mn} \\ &=\int_0^1\frac{dt}{t}\sum_{0< m\leq n}t^m\frac{C_n}{mn} \\ &=\int_0^1\frac{dt}{t}\sum_{0< m\leq n}\int_0^tx^{m-1}dx\frac{C_n}{n} \\ &=\int_0^1\frac{dt}{t}\int_0^tdx\sum_{0\leq m< n}x^m\frac{C_n}{n} \\ &=\int_0^1\frac{dt}{t}\int_0^tdx\sum_{0< n}\frac{1-x^n}{1-x}\frac{C_n}{n} \\ &=\int_0^1\frac{dt}{t}\int_0^t\frac{dx}{1-x}\sum_{0< n}(1-x^n)\frac{C_n}{n} \\ &=\int_0^1\frac{dt}{t}\int_0^t\frac{dx}{1-x}\int_x^1\frac{(1-u)^{-1/2}-1}{u}du \\ &=\int_{0< x< t<1,0< x< u<1}\frac{dt}{t}\frac{dx}{1-x}\frac{(1-u)^{-1/2}-1}{u}du \\ &=\int_{0< x< t< u<1}\frac{dt}{t}\frac{dx}{1-x}\frac{(1-u)^{-1/2}-1}{u}du+\int_{0< x< u < t<1}\frac{dt}{t}\frac{dx}{1-x}\frac{(1-u)^{-1/2}-1}{u}du \\ &=\int_0^1\frac{(1-u)^{-1/2}-1}{u}du\int_0^u\frac{dt}{t}\int_0^t\frac{dx}{1-x}+\int_0^1\frac{dt}{t}\int_0^t\frac{(1-u)^{-1/2}-1}{u}du\int_0^u\frac{dx}{1-x} \\ &=-4\int_0^1\frac{du}{-1-u}\int_u^1\left(\frac{1}{1+t}+\frac{1}{-1-t}\right)dt\int_t^1\frac{dx}{x}-4\int_0^1\left(\frac{1}{1+t}+\frac{1}{-1-t}\right)dt\int_t^1\frac{du}{-1-u}\int_u^1\frac{dx}{x} \\ &=-4\int_0^1a(b_0+b_1)b_1+ab_1(b_0+b_1) \\ &=-4\int_0^1z_{2,0}z_{1,1}+z_{2,1}z_{1,1}+z_{2,1}z_{1,0}+z_{2,1}z_{1,1} \\ &=-4(\zeta(\ol1,2)+\zeta(\ol1,\ol2)+2\zeta(1,\ol2)) \\ &=-4\left(\zeta(3)-\frac{\pi^2\ln2}{4}+\frac{\pi^2\ln2}{4}-\frac{13}{8}\zeta(3)+\frac14\zeta(3)\right) \\ &=\frac32\zeta(3) \end{align}

式の分量が多かった…