Does this series converge to a rational multiple of $\pi^2$?

Consider the product over all prime numbers $$P=\prod_p\left(1+\frac{1}{p^2}\right)^{-1}=\prod_p\left(1-\frac{1}{p^2}+\frac{1}{(p^2)^2}+\dots\right).$$ If you think about how a term of this series will look like when you multiply it out, you will find that there is a term $1/n^2$ for every $n$ with an even number of prime factors, and a term $-1/n^2$ for every $n$ with an odd number of prime factors. Therefore, we find that $P=2S_2-\zeta(2)$. Since we know the value of $\zeta(2)$ and want to find the value of $S_2$, it's enough to find the value of $P$.

But observe $$P=\prod_p\left(1+\frac{1}{p^2}\right)^{-1}=\prod_p\frac{(1-1/p^4)^{-1}}{(1-1/p^2)^{-1}}=\frac{\prod_p(1-1/p^4)^{-1}}{\prod_p(1-1/p^2)^{-1}}=\frac{\zeta(4)}{\zeta(2)}=\frac{\pi^4/90}{\pi^2/6}=\frac{\pi^2}{15}.$$

Comparing this with the above, $$S_2=\frac{P+\zeta(2)}{2}=\frac{\pi^2/15+\pi^2/6}{2}=\frac{7\pi^2}{60}.$$