Primes again 3

If $S$ is squarefree, it means that it does not have any repeated prime factors. This means that any number that is the product of unique primes is a squarefree number.

So $\sum_{S \text{ is squarefree, }S>0} \frac{1}{S^{n}}=1+\sum _{ a }{ \frac { 1 }{ ({ p }_{ a })^{n} } } +\sum _{ a>b }{ \frac { 1 }{ ({ p }_{ a }{ p }_{ b } )^{n}} } +\sum _{ a>b>c }{ \frac { 1 }{ ({ p }_{ a }{ p }_{ b }{ p }_{ c })^{n} } } ...$

Where $p_{k}$ is the $\text{k}^{\text{th}}$ prime. This is equivalent to the expansion of $\prod _{ \text{p is prime} }{ 1+{ p }^{ -n } }$

Hence, $\sum_{S \text{ is squarefree, }S>0} \frac{1}{S^{2}}=\prod _{ \text{p is prime} }{ 1+{ p }^{ -2 } }=\frac { \prod _{\text{p is prime} }{ 1-{ p }^{ -4 } } }{ \prod _{ \text{p is prime} }{ 1-{ p }^{ -2 } } }$ Through Euler's product, $\prod _{\text{p is prime} }{ 1+{ p }^{ -n } } =\frac { 1 }{ \zeta (n) }$ .

So the answer is $\frac { \zeta (2) }{ \zeta (4) } =\frac{15}{\pi^{2}}$

The sum converges due to $\sum \frac{1}{S^{2}}<\zeta (2)$ of which $\zeta{(2)}$ is known to converge.