It is well known that the values of the Riemann zeta function for even positive numbers are of the form:

$$\zeta(2k) = \rm rational * \pi ^{2k},$$

and more specifically $\zeta (2k)=(-1)^{{k+1}}{\frac {B_{{2k}}(2\pi )^{{2k}}}{2(2k)!}}\!$. It is not that far-fetched to consider that

$$\zeta(2k + 1) = \rm rational * \pi ^{2k + 1}.$$

Specifically for Apéry's constant (which is $\zeta(3)$), did someone prove something like that? The proof should be something like:

$\frac{\zeta(3)}{\pi^3}$ is rational / irrational / transcendental.

EDIT: Even if the question is still open (which I can see it is from the comments), is there any new development on this matter lately? Just curious.

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    $\begingroup$ No one has proved anything like this. $\endgroup$ Sep 16, 2017 at 18:02
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    $\begingroup$ This is a well-known question, see this MO-question. $\endgroup$ Sep 16, 2017 at 18:04
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    $\begingroup$ A significant amount of effort has gone into this, but the nature of Apery's constant is still largely mysterious. $\endgroup$
    – George C
    Sep 16, 2017 at 18:07
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    $\begingroup$ I am not sure whether the problem is open. But I would be rather surprised if $\large \frac{\zeta(3)}{\pi^3}$ turned out to be rational. My guess is that it is even transcendental (of course, only a guess). The continued fraction I calculated with PARI/GP with $20\ 000$ digits accuracy, contains $19501$ entries not exceeding $134656$. So, if the constant IS rational, numerator and denominator must be very large. $\endgroup$
    – Peter
    Sep 16, 2017 at 18:10
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    $\begingroup$ Cross-site duplicate $\endgroup$
    – Wojowu
    Sep 16, 2017 at 18:21

1 Answer 1


While Apéry proved in 1978 that $\zeta(3)$ is irrational, the irrationality of $\frac{\zeta(3)}{\pi^3}$ is still an open problem. There are some formulas expressing odd zeta values in terms of powers of $\pi$, the most known ones are due to Plouffe and Borwein & Bradley. Here are some examples:

$$ \begin{aligned} \zeta(3)&=\frac{7\pi^3}{180}-2\sum_{n=1}^\infty \frac{1}{n^3(e^{2\pi n}-1)},\\ \sum_{n=1}^\infty \frac{1}{n^3\,\binom {2n}n} &= -\frac{4}{3}\,\zeta(3)+\frac{\pi\sqrt{3}}{2\cdot 3^2}\,\left(\zeta(2, \tfrac{1}{3})-\zeta(2,\tfrac{2}{3}) \right). \end{aligned} $$

Moreover, in this Math.SE post we have:

$$ \frac{3}{2}\,\zeta(3) = \frac{\pi^3}{24}\sqrt{2}-2\sum_{k=1}^\infty \frac{1}{k^3(e^{\pi k\sqrt{2}}-1)}-\sum_{k=1}^\infty\frac{1}{k^3(e^{2\pi k\sqrt{2}}-1)}. $$

You can also check out this paper by Vepstas, which provides a nice generalization to some of these identities.


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