Divergent Products over Prime Numbers

GENERA

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

\displaystyle \prod_{p\in\mathbb{P}} (1-\frac{1}{p^r}) = \frac{1}{\zeta(r)}

\displaystyle \prod_{p\in\mathbb{P}} (p^r-1) = \frac{(2\pi)^{2r}}{\zeta(r)}

\displaystyle \prod_{p\in\mathbb{P}} (1+\frac{1}{p^r}) = \frac{\zeta(r)}{\zeta(2r)}

\displaystyle \prod_{p\in\mathbb{P}} (p^r+1) = \frac{\zeta(r)}{\zeta(2r)} (2\pi)^{2r}

\displaystyle \prod_{p\in\mathbb{P}} (1+\frac{1}{p}+\frac{1}{p^2}+\cdots+\frac{1}{p^r}) = \frac{\zeta(1)}{\zeta(r+1)} = \frac{\gamma}{\zeta(r+1)}

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

\displaystyle \prod_{p\in\mathbb{P}} (1+p+p^2+\cdots+p^r) = \frac{\zeta(1)}{\zeta(r+1)} (2\pi)^{2r} = \frac{\gamma}{\zeta(r+1)} (2\pi)^{2r}

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

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