I was reading Chebychev's Theorem about the probability of a fraction to be irreducible and I came accross this evaluation $$\prod_{i: \text{ prime}}\left(1-\frac{1}{i^2}\right) = \frac{6}{{\pi}^2}$$ I wonder how can I prove this.
Any hint?
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As far as I can see, it is not something easy, at least for my level...and I feel a bit releaved know that I could not proove it. To be honest my first thought was the Wallis equality for pi, but of course it did not work. – dmtri Apr 03 '18 at 18:24
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Refer to Euler's product
$$\prod_{p} (1-p^{-s})^{-1} = \prod_{p} \Big(\sum_{n=0}^{\infty}p^{-ns}\Big) = \sum_{n=1}^{\infty} \frac{1}{n^{s}} = \zeta(s)$$
and to
Different methods to compute $\sum\limits_{k=1}^\infty \frac{1}{k^2}$ (Basel problem).
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1Of course, proving that $\zeta(2) = \pi^2/6$ is not exactly a trivial matter. – Michael Seifert Apr 03 '18 at 17:02
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1For that https://math.stackexchange.com/questions/8337/different-methods-to-compute-sum-limits-k-1-infty-frac1k2-basel-pro – user Apr 03 '18 at 17:03