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Question is pretty much the title. It is pretty easy to show that $\zeta(2n)$ is irrational for all $n$ once you know that $\zeta(2n)$ is a rational multiple of $\pi^{2n}$ (and then also use the fact that $\pi$ is transcendental or some other related result). My question is, does anyone know of a proof that $\zeta(2)$ is irrational without evaluating it as $\frac{\pi^2}{6}$ and using the fact that $\pi^2$ is irrational? Perhaps one that uses its series definition or integral representations?

Not homework, just curious.

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  • $\begingroup$ See mathoverflow.net/questions/30659/… (referred to in mathoverflow.net/questions/202858/…) $\endgroup$ – Kyle Miller Feb 6 '16 at 2:49
  • $\begingroup$ $\displaystyle \frac{1}{\zeta(2)} = \prod_p (1-p^{-2}) = \prod_p \frac{p^2-1}{p^2} = \prod_p \frac{(p+1)(p-1)}{p^2} = \prod_p \frac{4 \frac{p+1}{2}\frac{p-1}{2}}{p^2}$ which cannot be rational because ... $\endgroup$ – reuns Feb 6 '16 at 3:12
  • $\begingroup$ because the $\prod_p 4$ will never simplify with the denominators $\endgroup$ – reuns Feb 6 '16 at 3:22
  • $\begingroup$ @user1952009 Woah that was remarkably simple, and I think I understand it. If you would like to make that an answer and add details, then I will accept it. $\endgroup$ – ASKASK Feb 6 '16 at 3:25
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    $\begingroup$ @user1952009, $(4/5)(20/17)(68/65)(260/257)...=1$, so I'm afraid your argument doesn't work. $\endgroup$ – Gerry Myerson Feb 6 '16 at 3:46
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Beukers proved $\zeta(2)$ irrational without using $\zeta(2)=\pi^2/6$. An exposition of his proof is here.

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