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][1]  


  [1]: http://math.stackexchange.com/questions/700686/equality-between-an-infinite-product-and-an-infinite-series-how-can-i-reconcile