For $\Re(s)>1$, it is well known that $$\frac{1}{\zeta(s)} = \sum_{n=1}^{\infty} \frac{\mu(n)}{n^s}$$ where $\mu$ denotes the Mobius function and $\zeta$ is the Riemann zeta function. I have heard that if the series on the right-hand side has an analytic continuation REAL $s_{0} \in (1/2, 1)$, then $\zeta(s)\neq 0$ for every $s$ with$\Re(s)=s_0$. But since all nontrivial zeros of the zeta function are complex, why does the analytic continuation of the said series to real values of $s$ have anything to do with the complex zeros ?
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$\begingroup$ where did you hear that? $\endgroup$– ConradSep 18, 2019 at 15:36
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1$\begingroup$ See en.wikipedia.org/wiki/Dirichlet_series#Abscissa_of_convergence if $\sum_{n=1}^\infty\frac{a(n)}{n^{s_0}}$ converges then $$\sum_{n=1}^\infty\frac{a(n)}{n^s}= \lim_{N\to \infty} N^{s_0-s} (\sum_{n=1}^N a(n) n^{-s_0}) + \sum_{n=1}^{N-1} (\sum_{m=1}^n a(m)m^{-s_0}) (n^{s_0-s}-(n+1)^{s_0-s})$$ converges and is analytic for $\Re(s) > \Re(s_0)$ and where it is meromorphic it has no poles on $\Re(s_0)$. The converse is very specific to $\zeta(s)$ and follows from the same Tauberian theorems as in the proof of the PNT. $\endgroup$– reunsSep 19, 2019 at 5:37
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2 Answers
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The statement you have given is not quite correct, but it is close to being correct. $\frac{1}{\zeta(s)}$ is known to have an analytic continuation to the interval $(1/2,1)$, and indeed all the way up to $(-2,1)$ (up to the first trivial zero). What is true, however, is the following:
If the series $$\sum_{n=1}^\infty\frac{\mu(s)}{n^{s_0}}$$ converges for some $s_0\in(1/2,1)$, then $\zeta(s)$ has no zeros on the line $\mathrm{Re}(s)=s_0$.
This follows from the general fact which can be established just like the special case covered in this answer:
If an analytic continuation of a Dirichlet series has a pole on the line $\mathrm{Re}(s)=s_0$, then the Dirichlet series does not converge at the point $s_0$.
The brief idea of the proof is that if the series were to converge, we could estimate how large the value of the Dirichlet series is at points close to the line $\mathrm{Re}(s)=s_0$, and we can show that this value grows slower than it could if there was a pole.
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$\begingroup$ I think "pole" is a bit too specific in ths last phrase; the value grows more slowly than it could if the partial sums didn't converge. $\endgroup$ Dec 8, 2019 at 12:07
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$\begingroup$ @GregMartin What partial sums do you mean? The point of the argument is to deduce divergence of a sum from existence of a pole. Also I thought it is possible for a Dirichlet series to converge at a point and have a logarithmic singularity on the same abscissa - doesn't $\log\zeta(s)$ have that property? $\endgroup$– WojowuDec 8, 2019 at 13:38
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$\begingroup$ Sorry, I misread—I was thinking of the general situation of Dirichlet series convergence, but you are indeed postulating an actual pole of $1/\zeta(s)$. $\endgroup$ Dec 8, 2019 at 23:09
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Let $s_0>0$. The right statement is that the following are equivalent:
The sum $\sum_{n=1} \tfrac{\mu(n)}{n^s}$ converges for $s>s_0$.
$\zeta(s)$ has no zeroes with real part $>s_0$.
$1/\zeta(s)$ has an analytic continuation to the half space $\mathrm{Re}(s) > s_0$
$\sum_{n=1}^N \mu(n) = O(N^{s_0+\epsilon})$ for all $\epsilon >0$.
As you point out, $1/\zeta(s)$ certainly has an analytic continuation to the real interval $(0,\infty)$, since $\zeta$ doesn't vanish on the positive reals.
answered Sep 18, 2019 at 17:04
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$\begingroup$ Do you know any reference for a proof of your first condition from either the second or the third? $\endgroup$ Jun 8, 2023 at 1:12