Does this 'alternating' Euler product converge for all $\Re(s) > 0$? 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 
 A: 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.
A: 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 $s>0$.
A: Combine two consecutive terms as follows:
$$
\left(1-\frac{1}{p_{2k}^s}\right)\left(1-\frac{1}{p_{2k+1}^s}\right)^{-1} = \frac{p_{2k+1}^s(p_{2k}^s-1)}{p_{2k}^s(p_{2k+1}^s-1)} = 1+\frac{p_{2k+1}^s-p_{2k}^s}{p_{2k}^s(p_{2k+1}^s-1)}
$$
Since $\sum\frac{1}{p_{2k}}=\infty$ the sum in product in question tends to $\infty$ as $s\rightarrow\frac{1}{2}$ from the right. Hence the product does not converge for any $s\in[0, \frac{1}{2}]$.
