# Sum them all up! (3)

**Calculus**Level 5

\[\displaystyle \large{ \sum_{\text{$\tau(n)$ is even}} \frac{1}{n^2} }\]

The sum runs over all positive integers \(n\) where \(\tau(n)\), the number of positive divisors of \(n\), is even. Find the value of the sum.