Does $\zeta(3)$ have a connection with $\pi$?

Solution 1:

The question is whether or not $\zeta(3)$ is connected with $\pi$. The answer is yes. Moreover, $\zeta(3) = \beta \pi^{\alpha}$ for some complex $\alpha, \beta$. Take $\alpha = 3$ and $\beta = 0.0387682...$. It is not known whether $\beta = 0.0387682...$ is algebraic or transcendental. It is known, however, that $\zeta(3)$ is irrational as shown by Apery.

Ramanujan conjectured and Grosswald proved that the following holds. 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 $n$, we take $\alpha = \beta = \pi$, \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, nothing is known about the algebraic nature of the infinite sum. This is a current topic of research. Indeed, it is conjectured that $\frac{\zeta(3)}{\pi^{3}}$ is transcendental.

Update: Recently, Takaaki Musha claims to have proved that $\frac{\zeta(2n+1)}{(2 \pi)^{2n+1}}$ is irrational for positive $n \geq 1$. However, some objection has since been raised (read comments below).

Solution 2:

Any complex number can be written as $\alpha \pi^\beta$ if you make no assumptions on $\alpha$ and $\beta$. You can even take $\beta = 0$.

It is however conjectured that $\zeta(2n+1)/\pi^{2n+1}$ is irrational for all $n \geq 1$. This conjecture is wide open.