Another summation

$\sum\limits_{\substack{\text{positive} \\ \text{even }n}}^{\infty} \dfrac{1}{n^2}$ Express the given summation as $\begin{array}{rl} \sum\limits_{k = 1}^{\infty} \dfrac{1}{\left(2k\right)^2} &= \sum\limits_{k = 1}^{\infty} \dfrac{1}{4} \cdot \dfrac{1}{k^2}\\ &= \dfrac{1}{4} \sum\limits_{k = 1}^{\infty} \dfrac{1}{k^2} \end{array}$ Since $\sum\limits_{k = 1}^{\infty} \dfrac{1}{k^2} = \dfrac{\pi^2}{6}$ this concludes that \begin{array}{rl} \sum\limits_{\substack{\text{positive} \\ \text{even }n}}^{\infty} \dfrac{1}{n^2} &= \dfrac{1}{4} \cdot \dfrac{\pi^2}{6}\\ &= \boxed{\dfrac{\pi^2}{24}} \end{array}