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This question already has an answer here:

Is it known if $\frac{\zeta(3)}{\pi^3}\in\mathbb{Q}$?

It is obvious that $\frac{\zeta(2n)}{\pi^{2n}}\in\mathbb{Q}$, but since there is no closed form for the odd values, are we left to be unable to determine if the solution could have such a form?

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marked as duplicate by Did sequences-and-series May 5 '16 at 11:59

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    $\begingroup$ mathoverflow.net/questions/60595/is-zeta3-pi3-rational $\endgroup$ – Wojowu May 5 '16 at 11:46
  • $\begingroup$ @Wojowu Thanks. $\endgroup$ – Simply Beautiful Art May 5 '16 at 11:47
  • $\begingroup$ I'm not sure it is that "obvious" that $\frac{\zeta(2n)}{\pi^{2n}}\in\mathbb{Q}$... $\endgroup$ – fretty May 5 '16 at 12:14
  • $\begingroup$ @fretty Did you look at the general solution to $\zeta(2n)$? Every component with the exception of the $\pi^{2n}$ is rational, and multiplied and divided together, ${\mathbb{Q}\over\mathbb{Q}}\in\mathbb{Q}$ $\endgroup$ – Simply Beautiful Art May 5 '16 at 20:54
  • $\begingroup$ But the formula itself is hardly obvious... $\endgroup$ – fretty May 5 '16 at 22:29
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Well $\zeta(3)$ is what known as Apery's constant, and it is proved to be irrational. But whether $\frac{\zeta(3)}{\pi^3}$ is rational or not is not known. For little more info you can check this link

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