Is the Euler product formula always divergent for 0

Is the Euler product formula always divergent for 0
2011-05-02T18:07:49Z

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?

Answer by GH for Is the Euler product formula always divergent for 0
2011-05-02T18:54:07Z

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 $\infty$.

Answer by Greg Martin for Is the Euler product formula always divergent for 0
2011-05-02T18:54:15Z

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 $s=1$.

Answer by David Speyer for Is the Euler product formula always divergent for 0
2011-05-03T15:38:55Z

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.