Sum them all up! (3)

We note that $\sigma (n)$ is odd when $n$ is a perfect square. Therefore,

$\begin{aligned} \sum_{\sigma(n) \in \text{ even}} \frac 1{n^2} & = \small \color{#D61F06}{\cancel{\frac 1{1^2}}} + \frac 1{2^2} + \frac 1{3^2} + \color{#D61F06}{\cancel{\frac 1{4^2}}} + \frac 1{5^2} + \frac 1{6^2} + \frac 1{7^2} + \frac 1{8^2} + \color{#D61F06}{\cancel{\frac 1{9^2}}} + \frac 1{10^2} + ... \\ & = \sum_{n=1}^\infty \frac 1{n^2} - \sum_{n=1}^\infty \frac 1{(n^2)^2} \\ & = \sum_{n=1}^\infty \frac 1{n^2} - \sum_{n=1}^\infty \frac 1{n^4} \\ & = \zeta (2) - \zeta (4) \\ & = \frac {\pi^2}6 - \frac {\pi ^4}{90} \\ & \approx \boxed{0.563} \end{aligned}$