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$$\prod_{k=1}^\infty \left(\frac{{p_k}^s}{{p_k}^s-1}\right) = \sum_{k=1}^\infty \frac 1 {k^s}$$

Let $s =3$ and $p$ range through all prime numbers.

$$\prod_{k=1}^\infty \left(\frac{{p_k}^3}{{p_k}^3-1}\right) = \sum_{k=1}^\infty \frac{1}{k^3}$$

$$\prod_{k=1}^\infty \left(\frac{{p_k}^3}{{p_k}^3-1}\right) \prod_{k=1}^\infty \left(\frac{{p_k}^3}{{p_k}^3+1}\right)=$$

$$\prod_{k=1}^{\infty}{\left(\frac{{p_k}^6}{{p_k}^6-1}\right)=}\sum_{k=1}^{\infty}{\frac{1}{k^6}=\frac{\pi^6}{945}}$$

$$\prod_{k=1}^\infty \left(\frac{{p_k}^3}{{p_k}^3-1}\right)\prod_{k=1}^\infty \left(\frac{{p_k}^3}{{p_k}^3+1}\right) = \frac{\pi^6}{945}$$

Which infinite product is a transcendental number?

$$\prod_{k=1}^{\infty}\left(\frac{{p_k}^3}{{p_k}^3-1}\right)or\ \prod_{k=1}^{\infty}\left(\frac{{p_k}^3}{{p_k}^3+1}\right)$$

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Most probably no one knows the answer. As an indication, note that $$ \prod_{k=1}^\infty \left(\frac{{p_k}^3}{{p_k}^3-1}\right) = \sum_{k=1}^\infty \frac{1}{k^3} =\zeta(3) =\text{Apéry's constant} $$

Wikipedia says that "it is still not known whether Apéry's constant is transcendental."

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  • $\begingroup$ What is $$\prod_{k=1}^{\infty}{\left(\frac{{p_k}^3}{{p_k}^3+1}\right)}$$ equal to? $\endgroup$ – AquinasLover1 Jun 29 at 13:39
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    $\begingroup$ It's precisely $\zeta(6)/\zeta(3)$ as you wrote in your question. Since $\zeta(3)$ has no known closed form, and $\zeta(6)$ does, there is no known closed form for $\zeta(6)/\zeta(3)$. $\endgroup$ – halrankard Jun 29 at 13:42
  • $\begingroup$ Is there anything wrong with the logic in the original post? $\endgroup$ – AquinasLover1 Jun 29 at 13:52
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    $\begingroup$ Maybe my comment on your other question was enough, but I'll put it here too in case others see. You've correctly written $\zeta(6)=\zeta(3)\beta$ where $\beta=\prod_{p}\frac{p^3}{p^3+1}$. We know $\zeta(6)$ transcendental, which says at least one of $\zeta(3)$ or $\beta$ is too. But we don't know exactly which for sure. We definitely don't know whether $\zeta(3)$ is transcendental and we probably don't know about $\beta$ too. $\endgroup$ – halrankard Jun 29 at 14:12
  • $\begingroup$ @halrankard, I think you could prove that both are transcendental if you were to modify Liouville's condition for Liouville numbers as so: $$0<|x-\frac{c_n}{d_n}|<\frac{1}{\prod_{k=1}^{n}d_k}$$ $\endgroup$ – AquinasLover1 Jun 30 at 1:11

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