Divergent Products over Prime Numbers

GENERA

$\displaystyle \prod_{p\in\mathbb{P}} \, p^r = (2\pi)^{2r}$

$\displaystyle \prod_{p\in\mathbb{P}} \, \left(1\!-\!\dfrac{1}{p^r}\right) = \dfrac{1}{\zeta(r)}$

$\displaystyle \prod_{p\in\mathbb{P}} \, (p^r\!-\!1) = \dfrac{(2\pi)^{2r}}{\zeta(r)}$

$\displaystyle \prod_{p\in\mathbb{P}} \, \left(1\!+\!\frac{1}{p^r}\right) = \dfrac{\zeta(r)}{\zeta(2r)}$

$\displaystyle \prod_{p\in\mathbb{P}} \, (1\!+\!p^r) = (2\pi)^{2r} \, \dfrac{\zeta(r)}{\zeta(2r)}$

$\displaystyle \prod_{p\in\mathbb{P}} \, \left(1\!+\!\frac{1}{p}\!+\!\dfrac{1}{p^2}\!+\!\cdots\!+\!\frac{1}{p^r}\right) = \dfrac{\zeta(1)}{\zeta(r\!+\!1)} = \dfrac{1}{\zeta(r\!+\!1)}$

$\displaystyle \prod_{p\in\mathbb{P}} \, \sum_{m=0}^{r-1} \, \dfrac{1}{p^m} = \frac{\zeta(1)}{\zeta(r)} = \dfrac{1}{\zeta(r)}$

$\displaystyle \prod_{p\in\mathbb{P}} \, (1\!+\!p\!+\!p^2\!+\!\cdots\!+\!p^r) = \dfrac{\zeta(1)}{\zeta(r\!+\!1)} \, (2\pi)^{2r} = \dfrac{(2\pi)^{2r}}{\zeta(r\!+\!1)}$

$\displaystyle \prod_{p\in\mathbb{P}} \, \sum_{m=0}^{r-1} \, p^m = \dfrac{\zeta(1)}{\zeta(r)} \, (2\pi)^{2r-2} = \dfrac{(2\pi)^{2r-2}}{\zeta(r)}$

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