Is $\zeta(3)/\pi^3$ rational?

Question

Apery proved in his paper from 1979 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.

Answer 1

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.

Answer 2

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 $\zeta(5)$, $\zeta(7)$, $\zeta(9)$, $\zeta(11)$.

Answer 3

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.

Answer 4

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 $\zeta(n)$.

Answer 5

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

• Musha, Takaaki. Negation of the conjecture for odd zeta values. Int. J. Pure Appl. Math. 91, No. 1, 103-111 (2014) doi:10.12732/ijpam.v91i1.11, zbMATH entry.

See this question.