Can be justified for integers $n\geq 1$ that

$$\zeta(2n+1)=\prod_{\text{p, prime}}\frac{1-\sigma(p^{2n})^{-1}}{1-p^{-2n}}?$$

Truly I don't know if I am wrong another time, when I use for an integer $m\geq 2$ that $$\zeta(m)=\prod_{\text{p, prime}}(1-p^{-m})^{-1}$$ to prove $$\frac{\zeta(m+1)}{\zeta(m)}=\prod_{\text{p, prime}}(1-\frac{1}{\sigma(p^m)}),$$ where $\sigma(l)$ is the sum of divisors function, and now using that Euler's product for Mobius function is $1/\zeta(m)$ we can conclude with the susbstitution $m=2n$.

Question. Please, can be justified or there is a mistake, that for integers $n\geq 1$ $$\zeta(2n+1)=\prod_{\text{p, prime}}\frac{1-\sigma(p^{2n})^{-1}}{1-p^{-2n}},$$ taking $\sigma(l)$ the sum of divisors function?

  • $\begingroup$ I am assuming that could be mistakes, if you know where please add a comment. Too I am assuming that the identity couldn't be useful since is difficult to compute the remainong factor involving the sum of divisors function. $\endgroup$ – user243301 Dec 9 '15 at 16:41

For any prime $p$, we have $$ \sigma(p^{2n}) = \frac{p^{2n+1}-1}{p-1} $$ and so $$ \frac{1-\sigma(p^{2n})^{-1}}{1-p^{-2n}} = \frac{1-\frac{p-1}{p^{2n+1}-1}}{1-p^{-2n}}=\frac{p^{2n+1}}{p^{2n+1}-1} = \left( 1 - \frac{1}{p^{2n+1}} \right)^{-1}. $$ Hence, $$\prod_{p, \text{prime}} \frac{1-\sigma(p^{2n})^{-1}}{1-p^{-2n}} = \prod_{p, \text{prime}} \left( 1 - \frac{1}{p^{2n+1}} \right)^{-1}=\zeta(2n+1). $$

  • $\begingroup$ Very thanks much @MatthewConroy I am learning that $\zeta(s)$ converges uniformly on compact subsets of $\Re s>1$, I believe that using M-Weirstrass test for infinite product. then Euler's products are defined. I am interesting in some factors of these products. I believe that too it is possible to prove for integeres $m\geq 2$, $\zeta(m)=\prod_p\frac{(p^{m-1})^2}{\sigma((p^{m-1})\phi((p^{m-1})}$, and $\zeta(m+1)/\zeta(m)=\prod_p\frac{p^2\phi(p^m)\sigma(p^{m-1})}{\phi(p^{m+1})\sigma(p^m)}$, where $\phi(l)$ is Euler's totient function. $\endgroup$ – user243301 Dec 9 '15 at 19:48
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    $\begingroup$ Since $\phi$ and $\sigma$ applied to prime powers have simple representations, I would first use those (as I did in my answer above) to write those expressions without $\phi$ and $\sigma$, and go from there. $\endgroup$ – Matthew Conroy Dec 9 '15 at 21:55
  • $\begingroup$ Very thanks much @MatthewConroy $\endgroup$ – user243301 Dec 9 '15 at 23:46

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