Apery proved in 1976 that $\zeta(3)$ is irrational, and we know that for all integers $n$, $\zeta(2n)=\alpha \pi^{2n}$ for some $\alpha\in \mathbb{Q}$. Given these facts, it seems natural to ask whether we can have $\zeta(n)=\alpha \pi^n$ for all $n$ (I'm mainly interested in the case $n=3$). The proofs I've seen of the irrationality of $\zeta(3)$ don't seem to give this information.

My gut feeling is that the answer is no, but I can't find any reference proving this fact. I know that the answer hasn't been proven to be yes ($\zeta(3)$ isn't even known to be transcendental), but ruling out this possibility seems an easier problem.

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    $\begingroup$ You may find mathoverflow.net/questions/30659/… useful, with a relevant comment by Wadim Zudilin and answer by Emerton. It is conjectured algebraically independent of $\pi$, but it doesn't seem that even this weaker result is known. $\endgroup$ Apr 4, 2011 at 19:39
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    $\begingroup$ It's actually not natural to ask whether $\zeta(3)$ is a rational multiple of $\pi^3$! There are "natural" conjectures to make about special values of a huge class of zeta functions, including Riemann's, and once you have figured out what's going on then you realise that the (conjectural) story for Riemanns zeta should be quite different at odd and even integers. Evidence for this: look at the values of Riemann's zeta at negative integers! it always vanishes at $-2,-4,-6,\ldots$ and never vanishes at $-1,-3,-5,\ldots$. $\endgroup$ Apr 4, 2011 at 20:15
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    $\begingroup$ Just to expand on Kevin's answer in a way that might help you to do some relevant searches: the positive even integers and negative odd integers are "critical integers" for the motive of which $\zeta(s)$ is the associated motivic $L$-function (namely, $Spec(\mathbb{Q})$). These are the ones to which a conjecture of Deligne applies that "explains" why $\zeta(k)\pi^{-k}\in \mathbb{Q}$ for positive even $k$. $\endgroup$
    – Ramsey
    Apr 4, 2011 at 21:38
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    $\begingroup$ Perhaps one can ask if $\zeta(3)^a/\zeta(5)^b$ is rational for some pair of positive integers $(a,b)$... $\endgroup$ Aug 18, 2016 at 0:54
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    $\begingroup$ Intuitively speaking, all $\zeta(2n)$ are sums of squares, as is also the algebraic equation of the circle, $x^2+y^2=r^2,~$ whose constant $\pi$ is. $\endgroup$
    – Lucian
    Sep 1, 2019 at 18:51

5 Answers 5


I can't find references but I know it has been shown that if $\zeta(3)/\pi^3=a/b$ is rational then $a$ and $b$ are enormous.

EDIT: I found a reference, but not in a formal publication. At http://tech.groups.yahoo.com/group/primenumbers/message/22659?threaded=1&p=2 it says,

"Re: Numerology about the Apery Constant $\zeta(3)$

"I also attempted to use PSLQ to figure out whether $\zeta(3)/\pi^3$ was a low-degree low-height algebraic number. Result: If it has degree $\le10$ then its height is at least $10^{91}$."

This was posted by someone identifying himself as Warren Smith.

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    $\begingroup$ The link is broken. Wayback Machine says it has not archived that URL. =( $\endgroup$ Sep 2, 2022 at 2:03
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    $\begingroup$ Yes, sadly, Yahoo decided to delete Yahoo Groups some years ago. The arXiv preprint arXiv:0910.2684 by F. M. S. Lima reports on some computational experiments with $\zeta(3)$. Actually, if all one cares about is whether $\zeta(3)/\pi^3$ is rational (as opposed to algebraic), then nothing as fancy as PSLQ is needed; just try computing its continued fraction numerically. $\endgroup$ Dec 15, 2022 at 16:22

It is definitely not known if $\zeta(3)/\pi^3$ is rational or not. By the way, there is a paper of Felder and Willwacher where they prove that the weight of a certain graph appearing in Kontsevich's formality quasi-isomorphism is, up to a rational, $\zeta(3)/\pi^3$. The question whether Kontsevich's quasi-isomorphism is defined over $\mathbb{Q}$ or not, is still open. If the answer to this question would be "yes", then the associator defined by Alekseev and Torossian would have rational coefficients... and that would definitely be a great result!

Among the main recent advances concerning rationality of zeta values, there are the works of Tanguy Rivoal and Wadim Zudilin. One of the most advanced results is that there is at least one irrational in $\zeta(5)$, $\zeta(7)$, $\zeta(9)$, $\zeta(11)$.


If $\alpha, \beta > 0$ such that $\alpha \beta = \pi^{2}$, then for each non-negative integer $n$, \begin{align} \alpha^{-n} \left( \frac{\zeta(2n+1)}{2} + \sum_{k \geq 1} \frac{k^{-2n-1}}{e^{2 k \alpha} - 1} \right) & = (- \beta)^{-n} \left( \frac{\zeta(2n+1)}{2} + \sum_{k \geq 1} \frac{k^{-2n-1}}{e^{2 k \beta} - 1} \right) - \end{align} \begin{align} \qquad 2^{2n} \sum_{k = 0}^{n+1} (-1)^{k} \frac{B_{2k} \ B_{2n- 2k + 2}}{(2k)! \ (2n - 2k + 2)!} \alpha^{n - k + 1} \beta^{k}. \end{align} where $B_n$ is the $n^{\text{th}}$-Bernoulli number.

For odd positive integer $n$, \begin{align} \zeta(2n+1) = -2^{2n} \left( \sum_{k = 0}^{n+1} (-1)^{k} \frac{B_{2k} \ B_{2n- 2k + 2}}{(2k)! \ (2n - 2k + 2)!} \right) \pi^{2n+1} - 2 \sum_{k \geq 1} \frac{k^{-2n-1}}{e^{2 \pi k} - 1}. \end{align}

In particular, for $n = 1$, \begin{align} \zeta(3) = -4 \left( \sum_{k = 0}^{2} (-1)^{k} \frac{B_{2k} \ B_{2- 2k + 2}}{(2k)! \ (2 - 2k + 2)!} \right) \pi^{3} - 2 \sum_{k \geq 1} \frac{k^{-3}}{e^{2 \pi k} - 1}. \end{align}

Observe that the coefficient of $\pi^{3}$ is rational, however, it is my understanding that nothing is known about the algebraic nature of the infinite sum.


The same method that gives you those even cases also gives an answer in the odd cases. But (for both) it is the sum of $1/k^n$ over all nonzero integers $k$ ... so of course in the odd case you get $0$, and in the even case you divide by $2$ to get $\zeta(n)$.


In this article, Takaaki Musha proves that $\zeta(2n+1) \notin (2\pi )^{2n+1} \mathbb{Q}$. I haven't read it so I can't say more.

[Edit: published reference: Musha, Takaaki. Negation of the conjecture for odd zeta values. Int. J. Pure Appl. Math. 91, No. 1, 103-111 (2014). Not referenced by MathSciNet, and referenced by zbMATH (link).]

See this question.

  • $\begingroup$ As the linked question is a discussion about and around the linked article, it maybe worthwhile to look at the question first. (Or not, depending on how you feel about these things.) $\endgroup$ May 9, 2011 at 21:56
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    $\begingroup$ For what it's worth, the review in Math Reviews just quotes the summary of the paper. $\endgroup$ May 9, 2011 at 23:22

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