平方和の逆数和を求める

平方数の逆数和${\displaystyle \sum _{n\ge 1}{\displaystyle \frac{1}{1^2+\cdots +n^2}}}$の値を求めていきます.
$$\begin{array}{rcl}\sum _{n\ge 1}{\displaystyle \frac{1}{1^2+\cdots +n^2}}& =& 6\sum _{n\ge 1}{\displaystyle \frac{1}{n(n+1)(2n+1)}}\\ & =& 6\sum _{n\ge 1}({\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n+1}}-{\displaystyle \frac{4}{2n+1}})\\ & =& 6\sum _{n\ge 1}({\displaystyle \frac{1}{n}}-{\displaystyle \frac{2}{2n+1}}+{\displaystyle \frac{1}{n+1}}-{\displaystyle \frac{2}{2n+1}})\\ & =& 6(\sum _{n\ge 1}({\displaystyle \frac{1}{n}}-{\displaystyle \frac{1}{n+\frac{1}{2}}})+\sum _{n\ge 1}({\displaystyle \frac{1}{n+1}}-{\displaystyle \frac{1}{(n+1)-\frac{1}{2}}}))\\ & =& 6(\sum _{n\ge 1}({\displaystyle \frac{1}{n}}-{\displaystyle \frac{1}{n+\frac{1}{2}}})+\sum _{n\ge 1}({\displaystyle \frac{1}{n}}-{\displaystyle \frac{1}{n-\frac{1}{2}}}))+6\\ & =& 6(H_{\frac{1}{2}}+H_{-\frac{1}{2}})+6\\ & =& 6(\psi \left({\displaystyle \frac{1}{2}}\right)+{\displaystyle \frac{1}{\frac{1}{2}}}+\gamma +\psi \left({\displaystyle \frac{1}{2}}\right)+\gamma )+6\\ & =& 6(-\gamma -2\ln 2+2+\gamma -\gamma -2\ln 2+\gamma )+6\\ & =& 18-24\ln 2\end{array}$$
よって${\displaystyle \sum _{n\ge 1}{\displaystyle \frac{1}{1^2+\cdots +n^2}}=18-24\ln 2}$であることがわかりました.