平方和の逆数和を求める

平方数の逆数和$\displaystyle\sum_{n\geq1}\dfrac{1}{1^2+\cdots+n^2}$の値を求めていきます.
$$ \begin{eqnarray*} \sum_{n\geq1}\dfrac{1}{1^2+\cdots+n^2}&=&6\sum_{n\geq1}\dfrac{1}{n(n+1)(2n+1)}\\ &=&6\sum_{n\geq1}\left(\dfrac{1}{n}+\dfrac{1}{n+1}-\dfrac{4}{2n+1}\right)\\ &=&6\sum_{n\geq1}\left(\dfrac{1}{n}-\dfrac{2}{2n+1}+\dfrac{1}{n+1}-\dfrac{2}{2n+1}\right)\\ &=&6\left(\sum_{n\geq1}\left(\dfrac{1}{n}-\dfrac{1}{n+\frac{1}{2}}\right)+\sum_{n\geq1}\left(\dfrac{1}{n+1}-\dfrac{1}{(n+1)-\frac{1}{2}}\right)\right)\\ &=&6\left(\sum_{n\geq1}\left(\dfrac{1}{n}-\dfrac{1}{n+\frac{1}{2}}\right)+\sum_{n\geq1}\left(\dfrac{1}{n}-\dfrac{1}{n-\frac{1}{2}}\right)\right)+6\\ &=&6\left(H_{\frac{1}{2}}+H_{-\frac{1}{2}}\right)+6\\ &=&6\left(\psi\left(\dfrac{1}{2}\right)+\dfrac{1}{\frac{1}{2}}+\gamma+\psi\left(\dfrac{1}{2}\right)+\gamma\right)+6\\ &=&6(-\gamma-2\ln2+2+\gamma-\gamma-2\ln2+\gamma)+6\\ &=&18-24\ln2 \end{eqnarray*} $$
よって$\displaystyle\sum_{n\geq1}\dfrac{1}{1^2+\cdots+n^2}=18-24\ln2$であることがわかりました.