神鳥さんの級数問題(2)

神鳥さんの級数問題(2) https://mathlog.info/articles/366 を解いてみました

${\displaystyle \sum _{0<n\le k}\frac{(-1{)}^{n-1}(\genfrac{}{}{0ex}{}{k-1}{n-1})}{n^3{k}^2}}$

まず、${\displaystyle \sum _{n=1}^k\frac{(-1{)}^{n-1}(\genfrac{}{}{0ex}{}{k-1}{n-1})}{n^3}}$を求めます。

$$\begin{gathered} {\displaystyle \sum _{n=1}^k\frac{(-1{)}^{n-1}(\genfrac{}{}{0ex}{}{k-1}{n-1})}{n^3}} \\ ={\displaystyle \frac{1}{2}\sum _{n=1}^k(-1{)}^{n-1}(\genfrac{}{}{0ex}{}{k-1}{n-1}){\int }_0^1{x}^{n-1}{\log }^{2}xdx} \\ ={\displaystyle \frac{1}{2}{\int }_0^1(1-x{)}^{k-1}{\log }^{2}xdx} \\ ={\displaystyle \frac{1}{2}{{\left(\frac{d}{dx}\right)}^2\frac{\mathrm{\Gamma }(k)\mathrm{\Gamma }(x)}{\mathrm{\Gamma }(k+x)}|}_{x=1}} \\ ={\displaystyle \frac{1}{2k}((\psi (1)-\psi (1+k){)}^2+{\psi }^{\prime }(1)-{\psi }^{\prime }(1+k))} \\ ={\displaystyle \frac{1}{2k}({\left(\sum _{n=1}^k\frac{1}{n}\right)}^2+\sum _{0<n\le k}\frac{1}{n^2})} \end{gathered}$$

ここで、${\displaystyle {\left(\sum _{n=1}^k\frac{1}{n}\right)}^2}$はこのままだと扱いづらいので、${\displaystyle \sum _{0<n\le k}\frac{1}{n^2}+2\sum _{0<i<j\le k}\frac{1}{ij}}$に直します。

従って、
$$\begin{gathered} {\displaystyle \sum _{0<n\le k}\frac{(-1{)}^{n-1}(\genfrac{}{}{0ex}{}{k-1}{n-1})}{n^3{k}^2}} \\ ={\displaystyle \sum _{0<i<j\le k}(\frac{1}{ij{k}^3}+\frac{1}{j^2{k}^3})} \\ =\zeta (1,1,3)+\zeta (1,4)+\zeta (2,3)+\zeta (5) \end{gathered}$$

双対関係式より、$\zeta (1,1,3)=\zeta (1,4)$であり、神鳥さんの記事( https://mathlog.info/articles/206 )より、${\displaystyle \zeta (1,4)=2\zeta (5)-\zeta (2)\zeta (3),\zeta (2,3)=3\zeta (2)\zeta (3)-\frac{11}{2}\zeta (5)}$なので、

$$\begin{gathered} {\displaystyle \sum _{0<n\le k}\frac{(-1{)}^{n-1}(\genfrac{}{}{0ex}{}{k-1}{n-1})}{n^3{k}^2}} \\ {\displaystyle =\zeta (2)\zeta (3)-\frac{1}{2}\zeta (5)} \end{gathered}$$

解けましたね。