TITLE: Does this 'alternating' Euler product converge for all $\Re(s) > 0$?
QUESTION [5 upvotes]: 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

REPLY [11 votes]: As Jacob mentioned, the question comes down to the conditional convergence of $\sum p^{-s}(-1)^n$ (for essentially the reason that $\prod(1-x)$ converges and is nonzero if $\sum x$ converges and $\sum x^2$ absolutely converges.
Now, if you look in Titchmarsh's Theory of Functions, there is a theorem that the region of conditional convergence of a Dirichlet series is a half plane (along with part of its boundary).  This is proven by showing that convergence at some $s$ gives convergence on every wedge with vertex at $s$ and otherwise strictly to the right.
Thus since the series converges for every real $x>0$ (alternating series), it also converges in the right half plane.