I know that $\sum_{n=1}^\infty n^{-2}=\pi^2/6$$\sum\limits_{n=1}^\infty n^{-2}=\pi^2/6$, but shouldn't the sum of rationals be rational? Is this akin to $\sum_{n=1}^\infty n=-1/12$$\sum\limits_{n=1}^\infty n=-1/12$? Or does that mean that, somehow, $\pi^2/6$ is rational?