# Is the Euler product formula always divergent for 0<Re(s)<1?

It is known that the Euler product formula converges for Re(s)>1. (which represents the Riemann zeta function.)

My question: Is the Euler product formula 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?

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I see this has gathered 3 votes to close. Could someone explain why? The question doesn't seem obvious to me, although I suspect that it is standard to analytic number theorists. –  David Speyer May 2 '11 at 18:40
It all depends on whether you call infinite products with a limit of zero convergent or divergent. It is quite standard in the literature to call them divergent. However, if you regard them as convergent, then the question may become nontrivial. –  Michael Renardy May 2 '11 at 19:00

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.

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.

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.

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Your answer is very helpful. Thank you. I have another question: If s=σ-i(π/lnp), (1-p^-s)=(1+p^-σ)>1. In this case, it seems that the Euler product converges to zero or some value, which means that the product may not always diverge, right or any problem? –  Seongsoo Choi May 3 '11 at 16:36
I don't understand what the formula $s = \sigma - i \pi/\log p$ means. We are taking the product as $p$ varies. Are you imposing that this is true for one particular prime? That's not going to have much effect on the product as a whole. –  David Speyer May 3 '11 at 16:39
You are right. It's my mistake. I couldn't take account of the fact that p is variable here, but I really agree about its oscillating characteristic. –  Seongsoo Choi May 3 '11 at 17:31
David, I am glad you clarified this! –  GH from MO May 3 '11 at 18:17
Thanks! And thanks for your answer, which is very nice and to the point! I'm glad this pushed me to finally work this out. I've always known there were subtleties about evaluating things in the critical strip, but I didn't know what they were. –  David Speyer May 3 '11 at 18:21

Here is a quick argument that $\prod_p(1-p^{-s})^{-1}$ is divergent for $\frac{1}{2}<\mathrm{Re}(s)<1$. Assume it is convergent (meaning it has a nonzero limit), then $\sum_p -\log(1-p^{-s})$ is also convergent. Using $-\log(1-p^{-s})=p^{-s}+O(p^{-2s})$ and $\mathrm{Re}(s)>\frac{1}{2}$ we see that $\sum_p p^{-s}$ is convergent. By a standard result (e.g. Montgomery-Vaughan: Multiplicative Number Theory I, Page 11, Theorem 1.1.) this would imply that $\sum_p p^{-s}$ is convergent for some real number $s<1$ which is false.

EDIT: This is a partial response to David Speyer's comment/question whether $\prod_p(1-p^{-s})^{-1}$ diverges to zero, diverges to $\infty$, or oscillates. On the real axis the product clearly diverges to zero by Mertens' theorem, so David really asks about the behavior of $$\mathrm{Re}\sum_{p\leq N}\frac{1}{p^{\sigma+it}} = \sum_{p\leq N}\frac{\cos(t\log p)}{p^\sigma}$$ for $\frac{1}{2}<\sigma<1$ and $t\neq 0$: whether it diverges to $\pm\infty$ or it oscillates as $N\to\infty$. Let us assume the Riemann Hypothesis, then $$\psi(x):=\sum_{n\leq x}\Lambda(n)=x+O(x^{1/2}\log^2 x).$$ Up to $O(1)$ error, the sum in question equals $$\sum_{n\leq N}\frac{\cos(t\log n)}{n^\sigma\log n}\Lambda(N)=\int_{2-}^N\frac{\cos(t\log x)}{x^\sigma\log x}d\psi(x).$$ Using all the hypotheses it follows by two integrations by parts (in between we approximate $\psi(x)$ by $x$) that $$\sum_{p\leq N}\frac{\cos(t\log p)}{p^\sigma} = \int_2^N\frac{\cos(t\log x)}{x^\sigma\log x}dx + O(1).$$ The right hand side is purely analytic, it has no reference to primes. It should be straightforward to prove that the $\limsup$ and $\liminf$ of the right hand side is $+\infty$ and $-\infty$, respectively, which would show that the Euler product oscillates: the absolute value of the partial products get arbitrary close to $0$ and $\infty$.

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Out of curiosity, does it diverge to $0$, diverge to $\infty$, or oscillate? I'm betting on oscillation, but I don't know. –  David Speyer May 3 '11 at 14:21
David, I believe oscillation follows from RH, see "EDIT" in my response. –  GH from MO May 3 '11 at 19:49
Thanks! That should resolve everything then. –  David Speyer May 3 '11 at 21:05
Yes, this $cos(tlnp)$ term is really important I think. I guess this term might be interconnected with the law of prime numbers. –  Seongsoo Choi May 4 '11 at 4:07

By definition, a product $\prod (1-a_n)^{-1}$ converges or diverges depending on whether $\sum a_n$ converges or diverges EDIT: THIS IS NOT CORRECT - SEE COMMENTS BELOW - GM. So the divergence of the Euler product $\prod_p (1-p^{-s})^{-1}$ inside the critical strip $0 < \Re s < 1$ is equivalent to the divergence of $$F(s) = \sum_p \frac1{p^s}.$$ Now $F(s)$ is a regular old Dirichlet series - its coefficients are the indicator function of the primes - and so it converges in some right half-plane by the general theory of Dirichlet series (see for example Montgomery and Vaughan's Multiplicative Number Theory I. Classical theory, chapter 1). But it definitely diverges at $s=1$ by Mertens's formula $$\sum_{p\le x} \frac1p \sim \log\log x,$$ and so it cannot converge (even conditionally) anywhere to the left of $s=1$.

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@Greg: Your definition (of convergence of a product) is only valid for positive real numbers $a_n$. In fact what you say is a theorem that works in this context. In general the situation is more subtle (I think), see my response. –  GH from MO May 2 '11 at 18:57
For an example of what GH is saying, $\sum (-1)^n n^{-1/3}$ is convergent, but $\sum \log(1-(-1)^n n^{-1/3})$ diverges to $-\infty$. –  David Speyer May 3 '11 at 14:18
Thanks both of you - I see what you mean and I agree. I think it might be possible that my incorrect proof could be turned into a correct proof. It would require quantitative statements about the divergence of $\sum_{p\le x} p^{-s}$ and the behavior of $\sum_{p\le x} \big( \log(1-p^{-s})^{-1} - p^{-s} \big)$: the former might well diverge provably faster than the latter. The theorem I referenced does have a quantitative version; but I haven't checked the details. –  Greg Martin May 5 '11 at 6:02