Does the following 'alternating' Euler product, with $p\_n$ the $n$-th prime number, converge for $\Re(s)>0$ ? 


$$\displaystyle \prod\_{n=1}^\infty \left( \dfrac{1}{1-\frac{1}{p\_{n}^{s}}} \right)^{(-1)^n}$$


Based on numerical evidence, I dare to conjecture that this is indeed the case (note that I could not find any zeros), but keen to find approaches towards a proof. Note that I also tried other triggers than $n$ to 'flip the factors', like for instance prime congruence to either $p\_n \pmod 6 =1$ or $5$ and $p\_n \pmod 4 =1$ or $3$. Even tried the Möbius function $\mu(n)$ as the exponent of $(-1)$, but did not observe any convergence in the domain $\Re(s)\le 1$. So, using $n$ to flip any other prime factor, seems a delicate choice to make convergence work or fail in this domain.


P.S.:


This question loosely builds on: [Equality of an alternating infinite product and an infinite sum](https://math.stackexchange.com/questions/700686/equality-between-an-infinite-product-and-an-infinite-series-how-can-i-reconcile) 



Does this 'alternating' Euler product converge for all $\Re(s) > 0$?
Taking logarithms, we arrive at the sum 


$$-\sum\_{n=1}^{\infty} (-1)^n \log(1-p\_n^{-s}).$$ For $Re(s)>1/k$, we can approximate the logarithm with its taylor series to $2k+1$ terms with an error of $o(p\_n^{-2})$ which converges absolutely. Thus, it suffices to show that for each $s>0$, the sum 


$$\sum\_{n=1}^{\infty} (-1)^n p\_n^{-s}$$ converges. To do this, we estimate


$$p\_{2n}^{-s}-p\_{2n+1}^{-s} = s^{-1}\int\_{p\_{2n}}^{p\_{2n+1}} x^{-1-s} dx = O\_s(G\_{2n}\cdot p\_{2n}^{-1-s})$$ where we define $G\_n= p\_{n+1}-p\_n$.


Now note that by the PNT we have $p\_n\sim n\log n$ and $\displaystyle\sum\_{n=X}^{2X-1} G\_n = p\_{2X}-p\_X \sim X\log X$.


\begin{align\*}
\sum\_{n=1}^{\infty} (-1)^n p\_n^{-s}&=\sum\_{n=1}^{\infty} (p\_{2n}^{-s}-p\_{2n+1}^{-s})\\
&\ll\_s \sum\_n G\_{2n}\cdot p\_{2n}^{-1-s}\\
&< \sum\_n G\_n \cdot p\_n^{-1-s}\\
&\ll \sum\_{m=0}^{\infty} \sum\_{n=2^m}^{2^{m+1}-1} G\_n (m2^m)^{-1-s}\\
&\ll \sum\_{m=0}^{\infty} (m2^m)^{-s}\\
\end{align\*}


Which converges for $s>0$.


