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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?

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    $\begingroup$ In view of the fact that on one of your recent questions there was some confusion on terminology, might I ask you to be a bit more detailed and precise regarding what you want to know. Something like this: Let s be a zero [perhaps explanation, whether you want to include the trivial ones or want to assume RH, though this might actually not be relevant] of the R-zeta funtion. Is the Euler product [Formula of the product] necessarily divergent [plus explanation regarding you convention regarding 'convergence' ot zero] $\endgroup$
    – user9072
    Commented May 3, 2011 at 10:21
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    $\begingroup$ It seems you are not overly interested in my suggestions, which is fine. But, in case you just misunderstood what I asked for (and this also applies to your more recent question): to give a motivation for your question (as you now did) is also a good idea. Yet, I think it is still more important that you are precise. Say, what does 'use the Euler product instead of the Riemmann Zeta function' mean? (And didn't GH and others yesterday give you an argument that the product diverges?) $\endgroup$
    – user9072
    Commented May 3, 2011 at 13:15
  • $\begingroup$ This question is actually different from the one I asked before. In my former question, I wanted to know about a chance to have a convergent value on the critical strip for a complex number s, and I could find an example where it does not diverge among the answers. So, I want to ask only for the nontrivial zeros of the Riemann zeta function. It is expected that there are only two possible cases: 1. it diverges. 2. it converges. $\endgroup$
    – Seongsoo Choi
    Commented May 3, 2011 at 15:17
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    $\begingroup$ In addition, I heard from some mathematician that I cannot write the Euler product formula as the Riemann zeta function for 0<=Re(s)<=1. I want to know the exact reason why it is impossible rather than a somewhat vague answer that, "it just diverges there." $\endgroup$
    – Seongsoo Choi
    Commented May 3, 2011 at 15:22
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    $\begingroup$ Seongsoo Choi, I do not want to further insist on communicating my personal preference regarding how a question should be asked as this is addmittedly subjective. Yet, if you actually want to know what is in your last comment, I'd seem simplest to me to ask about this directly. But, needless to say, you are free to ask what and how you like. It was merely my intention to give some feeedback how you could make it easier for me (and perhaps others) to answer your question. I won't do so again (in the sense that you should not worry about me to much). $\endgroup$
    – user9072
    Commented May 3, 2011 at 16:20

2 Answers 2

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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 $\mathrm{Re}(s)=\frac{1}{2}$.

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    $\begingroup$ When Re(s) =1/2, the terms for m >2 are absolutely convergent. So you only have to deal with the first two. $\endgroup$
    – David E Speyer
    Commented May 3, 2011 at 14:14
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    $\begingroup$ Work of Montgomery on large values of $\zeta$ on the half-line uses this in essence, I suspect. springerlink.com/content/h62kj118v1687613 "Assume RH, $1/2<\sigma<1$ fixed, for $T\le t\le 2T$ we have $\log\zeta(s)=\sum_{p\le T} p^{-s}+O(1)$, calling this $F(t)$. As the numbers $p^{-it}$ have a random appearance, we consider $G(\theta)=\sum_{p\le T} p^{-\sigma}e^{2\pi i\theta_p}$ where the $\theta_p$ are independent..We can prove that with great uniformity the distribution function of $F(t)$, for $T\le t\le 2T$ is very close to that of $G(\theta)$ for $\theta\in(R/Z)^{\pi(T)}$." $\endgroup$
    – Junkie
    Commented May 3, 2011 at 14:46
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    $\begingroup$ @David: I wanted to make a statement for any point in the critical strip. You are right that for $1/3<\mathrm{Re}(s)<1$ we only need $n=2$ (and for $1/2<\mathrm{Re}(s)<1$ we only need $n=1$). $\endgroup$
    – GH from MO
    Commented May 3, 2011 at 17:58
  • $\begingroup$ Dear GH, did David Speyer not only prove that the product does noit converge absolutely? It might still converge conditionally. $\endgroup$
    – Marc Palm
    Commented May 4, 2011 at 5:01
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    $\begingroup$ Dear pm, David Speyer proved that in the critical strip $\left|\prod_p(1-p^{-s})^{-1}\right|$ does not converge to a positive real number. In particular, $\prod_p(1-p^{-s})^{-1}$ does not converge to a nonzero complex number. $\endgroup$
    – GH from MO
    Commented May 4, 2011 at 9:53
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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 $|z| \geq 1$.

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