Product of a specific sequence of prime number squares

[Vishalrox]

Here is a good question for maths enthusiasts. I really find this sum very tough. Any help, advice or guidance for this question will be greatly appreciated.

find the product upto n terms

(1+(1/2^2)) . (1+(1/3^2)) . (1+(1/5^2)) . (1+(1/7^2)) . (1+(1/11^2))...upto n terms

Where the nth term is (1+(1/P^2)) or (1+(p^-2)) where p is a prime number.

I feel this question is really tough and couldn't even get the logic how to solve it.

 

[micromass]

You should read up on the Riemann-zeta function.

 

[RamaWolf]

In the infinite products below, the factors run over all primes $p_{k} \in \mathbb{P}$Theorem: $\prod \frac{p_{k}^{s}+1}{p_{k}^{s}}= \frac{\zeta(s)}{\zeta(2 s)}, with \zeta(s)$ the Riemann zeta functionProof: With the Euler product of the Riemann zeta function we have:

$\zeta(2 s) = \prod (1 - p_{k}^{-2 s})^{-1} = \prod \frac{p_{k}^{2 s}}{p_{k}^{2 s} - 1} = \prod \frac{p_{k}^{s}}{p_{k}^{s}-1} \frac{p_{k}^{s}}{p_{k}^{s}+1} = \zeta(s) \prod \frac{p_{k}^{s}}{p_{k}^{s}+1}$ Q.E.D.Now, with Im(s) = 0 and Re(s) = n we have for the Euler zeta function:

$\prod \frac{p_{k}^{n}+1}{p_{k}^{n}}= \frac{\zeta(n)}{\zeta(2 n)}$

and for the special case n = 2

$\prod \frac{p_{k}^{2}+1}{p_{k}^{2}}= \frac{\zeta(2)}{\zeta(4)} = \frac{15}{π^{2}}$

and that answers the posted question