Primes and products

Well, here we will use Euler product formula.

$\displaystyle \zeta(s) = \sum_{n=1}^{\infty}{\frac{1}{{n}^{s}}} = \prod_{p}{\frac{1}{1- {p}^{-s}}}$

$\displaystyle \zeta(4) = \prod_{p}{\frac{1}{1 - {p}^{-4}}}$

$\displaystyle = \prod_{p}{\frac{1}{1 - {p}^{-2}}}\prod_{p}{\frac{1}{1 + {p}^{-2}}}$

$\displaystyle \zeta(4)= \zeta(2)\prod_{p}{\frac{1}{1 + {p}^{-2}}}$

$\displaystyle \prod_{p}{\frac{1}{1 + {p}^{-2}}} = \prod_{p}{\frac{{p}^{2}}{{p}^{2} + 1}} = \frac{\zeta(4)}{\zeta(2)}$

$\displaystyle \prod_{p}{\frac{1}{1 - {p}^{-2}}} = \prod_{p}{\frac{{p}^{2}}{{p}^{2} - 1}} =\zeta(2)$

$\displaystyle \frac{\displaystyle \prod_{p}{\frac{{p}^{2}}{{p}^{2} - 1}}}{\displaystyle \prod_{p}{\frac{{p}^{2}}{{p}^{2} + 1}}} = \displaystyle \prod_{p}{\frac{{p}^{2} +1}{{p}^{2} -1}} = \frac{{\zeta(2)}^{2}}{\zeta(4)}$

As one can easily simplify $S$ to the above form. We get the answer as -

$\displaystyle \frac{\displaystyle {(\frac{{\pi}^{2}}{6})}^{2}}{\frac{\displaystyle {\pi}^{4}}{90}} = \displaystyle \boxed{\frac{5}{2}}$

The answer is $\dfrac{5}{2}$

Edit : For those who want analytical solution it is : $\dfrac{(\zeta{(2)}^2)}{\zeta{(4)}}$