It is known that the Euler product formula converges for $\Re(s)>1$
(and there it represents the Riemann zeta function).


**My question:** Is the Euler product always divergent for
$0 < \Re(s) < 1$ ?


I thought that the absolute value of the Euler product formula is positively divergent under the above condition. Is it apparent? 



Is the Euler product formula always divergent for 0<Re(s)<1?
Let 
$$t\_P = \sum\_{p < P} \log \left| \frac{1}{1-p^{-s}} \right|$$
with $s=\sigma+it$, $\sigma \in (0,1)$ and $t$ a nonzero real.
The point of this answer is to show that the $t\_P$ jump around a great deal. Specifically, for any $M$ and $N$, there are $P$ and $Q$ with $N < P < Q$ such that $t\_Q - t\_P > M$, and other $P'$ and $Q'$ with $N < P' < Q'$ such that $t\_{Q'} - t\_{P'} < -M$


Thus $t\_P$ cannot approach any finite limit. It could still approach $\pm \infty$; think of $\sum (-1)^n (3+(-1)^n)^n$, which has arbitrarily large increases and decreases, but does climb to $\infty$. However, this result still means you should be very suspicious of any numerical data which seems to indicate that $t\_P$ has a definite trend: There is always enough future oscillation remaining to wipe out any gains you have made towards $\pm \infty$.


Obviously, this implies the analogous statements about $\prod \left| \frac{1}{1-p^{-s}} \right|$: It cannot approach a finite limit, and you should not trust numerical evidence that it is going to $0$ or $\infty$. And, of course, life is only more complicated if you keep track of the argument of the Euler product as well as its magnitude.




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So, a proof. We will treat $\sigma$ and $t$ as completely fixed, so constants in $O$'s can depend on them.


Choose a small positive real $\delta$. This will be a once and for all choice, but I will record dependences on it explicitly, because I need to see that I can take a small enough choice to make everything work.


Let $(P,Q)$ be of the form 
$$(e^{(2 \pi k-\delta)/t}, e^{(2 \pi k+\delta)/t})$$
for some positive integer $k$. By choosing $k$ large, we can arrange that $P$ and $Q$ are larger than any required $N$.


For any prime $p$ in this range,
$$|1-p^{-s}| = |1-p^{-\sigma} e^{i \theta}|$$
for some $\theta \in (2 \pi k - \delta, 2 \pi k + \delta)$. So this is
$$1-p^{-\sigma}(1 + O(\delta^2))$$
and
$$ \log \left| \frac{1}{1-p^{-s}} \right| = p^{-\sigma} (1+O(\delta^2))(1+O(p^{-\sigma}))$$
If $(P,Q)$ is large enough, the first error term dominates and 
$$t\_Q - t\_P \geq \sum\_{e^{2 \pi k - \delta}/t < p < e^{2 \pi k + \delta}/t} p^{-\sigma}(1+O(\delta^2)) = \# \{p: e^{(2 \pi k - \delta)/t} < p < e^{(2 \pi k + \delta)/t} \} e^{-2 \pi k \sigma/t} (1+O(\delta)).$$
(The error term has changed because the new dominant error is approximating $e^{\delta \sigma/t}$ as $1+O(\delta)$.


By the prime number theorem, the number of primes in this range is 
$$\left( e^{(2 \pi k + \delta)/t} - e^{(2 \pi k - \delta)/t} \right) \frac{1}{2 \pi k/t} (1 + O(1/k)) = \frac{2 \delta e^{2 \pi k/t}}{(2 \pi k/t)} (1+O(\delta)+O(1/k)).$$ 


In short, we have bounded $t\_Q - t\_P$ below by 
$$\frac{\delta t e^{2 \pi k(1-\sigma)/t}}{2 \pi k}(1+O(\delta) + O(1/k)).$$
Assuming our initial choice of $\delta$ was small enough, and using $\sigma<1$, this goes to $\infty$.


Now, repeat the argument with $(P,Q) = (e^{((2k+1)\pi -\delta)/t}, e^{((2k+1)\pi +\delta)/t})$ to show that $t\_Q - t\_P$ can be arbitrarily negative as well.




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I don't have a gut instinct for whether this sum goes to $- \infty$, goes to $\infty$, or oscillates indefinitely. However, it should be clear that this sum is very far from being the $\zeta$ function.


