# Is Apéry's constant a rational multiple of $\pi ^ 3$?

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.

• No one has proved anything like this. Sep 16, 2017 at 18:02
• This is a well-known question, see this MO-question. Sep 16, 2017 at 18:04
• A significant amount of effort has gone into this, but the nature of Apery's constant is still largely mysterious. Sep 16, 2017 at 18:07
• 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. Sep 16, 2017 at 18:10
• Cross-site duplicate Sep 16, 2017 at 18:21

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