camilus
- 146
- 0
Should \prod_\mathbb{P} \left( \sum_{\mathbb{Z} \ge 0} p^{-s n} \right) ^{-1} have that ^(-1) after it? Or am I missing something..? Are you rewriting 1/(1-p^-s) using geometric series?lurflurf said:We know
$$\prod_\mathbb{P} (1-p^{-s})^{-1}=\prod_\mathbb{P} \left( \sum_{\mathbb{Z} \ge 0} p^{-s n} \right) ^{-1} = \sum_{\mathbb{Z}>0} n^{-s}$$
with the minus sign it is a little more complicated
$$\prod_\mathbb{P} (1+p^{-s})^{-1}=\prod_\mathbb{P} \left( \sum_{\mathbb{Z} \ge 0} (-p)^{-s n} \right) ^{-1} = \sum_{\mathbb{Z}>0} (-1)^{\sum ord_p(k)}n^{-s}$$
Where the ord_p(k) makes sure we get the right sign (it counts the minuses), clearly
$$\left( \prod_\mathbb{P} (1-p^{-s})^{-1} \right) \left( \prod_\mathbb{P} (1+p^{-s})^{-1} \right) =\zeta(2n)$$
but instead of using that your link makes a simple estimate