Timeline for Does this 'alternating' Euler product converge for all $\Re(s) > 0$?

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Mar 9, 2014 at 17:40	comment	added	Agno		Thanks and understood. For the alternating prime product, I had of course hoped for convergence to known constants or to find some new zeros, but did not find any. However, for the alternating product over integers: $\prod_{n=1}^\infty \left( \dfrac{1}{1-\frac{1}{n^{s}}} \right)^{(-1)^n}$, there are a few interesting outcomes. E.g. $s=1$ gives $\frac{\pi}{2}$ and $s=2$ yields $\frac{\pi^2}{8}$. I guess there also exist more complicated closed forms for $s=3,4,5...$.
Mar 9, 2014 at 6:57	comment	added	jacob		Yup. In fact, you can prove $$\prod_{n=1}^{\infty}(1-a_n^{-s})^{(-1)^n}$$ converges for $Re(s)>0$ for any increasing sequence of positive integers $a_n$ this way. The PNT thing I used was overkill.
Mar 6, 2014 at 18:24	comment	added	Agno		Many thanks, Jacob. I honestly had not expected a proof being within reach at all. Does this also immediately prove the similar convergence of $\prod_{n=1}^\infty \left( \dfrac{1}{1-\frac{1}{n^{s}}} \right)^{(-1)^n}$, since the first two steps in your proof would be similar and lead to proving that $\sum_{n=1}^{\infty} (-1)^n n^{-s}$ (i.e. the Dirichlet Eta-function) converges for $\Re(s) > 0$ (which already has been proven)?
Mar 6, 2014 at 18:10	vote	accept	Agno
Mar 6, 2014 at 17:53	history	answered	jacob	CC BY-SA 3.0