Does the Euler product formula diverge for any zero of the Riemann zeta function? Simple question (but not for me):
Does the Euler product formula diverge for any zero of the Riemann zeta function?
The reason why I ask this is that I heard we should not use the Euler product instead of the Riemann zeta function for Re(s)=<1 because it diverges on the critical strip, but I am not sure of that.
According to my numerical calculation, it seems that it converges for the (known) zeta zeros.
Additional Question 1: 
Is it clear that the Euler product for a nontrivial zeta zero is either divergent or convergent?
Additional Question 2:
If only one case is possible, which one is the right answer? Divergent or convergent?
 A: I am in a hurry now but let me tell what I think. I believe that in the critical strip and off the real axis $\prod_p (1-p^{-s})$ does not converge to any complex number (including zero). Using a similar idea as in my response to your related earlier question, this boils down to the fact that for any nonzero constants $c_1,\dots,c_n$ the sum $\sum_{m=1}^n c_m \sum_{p\in P}p^{-ms}$ oscillates wildly as $P\to\infty$. I dont's see this immediately, but I believe what happens is that the oscillation (or divergence) behavior of the inner sum depends heavily on $m$. More precisely, I believe that for $m=1$ you get a much wilder behavior than for the rest $m>1$. So altogether the above double sum inherits the behavior of $m=1$, namely the existence of very large partial sums for infinitely many $P$'s, and this prevents convergence or a tendency to pointing in special directions. I apologize if this is too vague, but certainly more than a comment.
EDIT 1: David Speyer showed that in the critical strip $\prod_p (1-p^{-s})$ does not converge to any nonzero complex number, see here. I believe that my approach above can also be made to work and yield more information. Perhaps the Riemann Hypothesis can be of great assistance here as $\sum_{p\in P}p^{-s}$ is very subtle. Note that for $\mathrm{Re}(s)=1$ and $s\neq 1$ the Euler product does converge to $\zeta(s)$, see Section 3.15 in Titchmarsh: The Theory of the Riemann Zeta-function.
EDIT 2: In my response to this question, I outline the proof that, assuming the Riemann Hypothesis, the partial products of $\prod_p (1-p^{-s})$ get arbitrary close to $0$ and $\infty$, at least for $\frac{1}{2}<\mathrm{Re}(s)<1$. I don't see any fundamental difficulty in extending this to $\mathrm{Re}(s)=\frac{1}{2}$.
A: I think that the following example might help you understand why we cannot write $\sum \frac{1}{n^s} = \prod_p(1-p^{-s})^{-1}$ for $0 \leq \Re(s) \leq 1$.  
Consider the formula for a geometric sequence, $\sum_n z^n = \frac{1}{1-z}$ which only holds for $|z| < 1$.  Notice that the function on right hand side of the equation makes sense for any $z \neq 1$; it is called an "analytic continuation" of the series
$\sum_n z^n$.  The important point is that the function $\frac{1}{1-z}$ is defined for all $z \neq 1$, but the series diverges when $|z| \geq 1$.  
