45
$\begingroup$

Let $\mu(n)$ denote the Mobius function with the well-known Dirichlet series representation $$ \frac{1}{\zeta(s)} = \sum_{n=1}^{\infty} \frac{\mu(n)}{n^{s}}. $$ Basic theorems about Dirichlet series imply that if the Dirichlet series on the right converges for some $s = \sigma + it$, then it converges for all $s$ with real part $> \sigma$. Hence, $\sum_{n=1}^{\infty} \frac{\mu(n)}{\sqrt{n}}$ converging is a sufficient condition for the Riemann hypothesis.

One way to approach this question is to use partial summation. Let $M(x) = \sum_{n \leq x} \mu(n)$. Then $$ \sum_{n \leq x} \frac{\mu(n)}{\sqrt{n}} = \frac{M(x)}{\sqrt{x}} + \frac{1}{2} \int_{1}^{x} \frac{M(t)}{t^{3/2}} \, dt. $$ Odlyzko and te Riele proved that $\liminf_{x \to \infty} \frac{M(x)}{\sqrt{x}} < -1.009$ and $\limsup_{x \to \infty} \frac{M(x)}{\sqrt{x}} > 1.06$. Much earlier, Ingham had showed that $M(x)/\sqrt{x}$ was unbounded assuming the linear independence of the imaginary parts of the zeroes of $\zeta(s)$.

In addition, Gonek has an unpublished conjecture (mentioned in Ng's paper "The distribution of the summatory function of the Mobius function") that $$ -\infty < \liminf_{x \to \infty} \frac{M(x)}{\sqrt{x} (\log \log \log x)^{5/4}} < 0 <\limsup_{x \to \infty} \frac{M(x)}{\sqrt{x} (\log \log \log x)^{5/4}} < \infty. $$

Using these results and conjectures to address the original question seems to be challenging, because of possible cancellation between $\frac{M(x)}{\sqrt{x}}$ and $\int_{1}^{x} \frac{M(t)}{t^{3/2}} \, dt$. My questions are the following:

  1. Are known results about $M(x)$ enough to determine if $\sum_{n=1}^{\infty} \frac{\mu(n)}{\sqrt{n}}$ diverges?

  2. If not, does Gonek's conjecture (or any other plausible conjectures) imply that $\sum_{n=1}^{\infty} \frac{\mu(n)}{\sqrt{n}}$ diverges?

$\endgroup$

1 Answer 1

56
$\begingroup$

One can show that $\sum_{n=1}^{\infty} \mu(n)/\sqrt{n}$ diverges. Suppose to the contrary that it converges, which as you note implies RH. Put $M_0(x)=\sum_{n\le x} \mu(n)/\sqrt{n}$, and our assumption is that $M_0(x)=C+o(1)$ as $x\to \infty$.

Note that for any $s=\sigma+it$ with $\sigma>1/2$ we have $$ \int_0^{\infty} sM_0(e^x)e^{-sx} dx = \sum_{n=1}^{\infty} \frac{\mu(n)}{\sqrt{n}} \int_{\log n}^{\infty} se^{-sx} dx = \frac{1}{\zeta(s+1/2)}. \tag{1} $$ Since $1/\zeta(s+1/2)$ is analytic (by RH) in $\sigma >0$, the identity above also holds in this larger domain. But from our hypothesis we note that the LHS above is $$ \int_0^{\infty} s(C+o(1)) e^{-sx} dx = C + o(|s|/\sigma). $$ Now take $s=\sigma+i\gamma$, where $\gamma =14.1\ldots $ is the ordinate of the first zero of $\zeta(s)$. Then the RHS of (1) is $\sim C_0/\sigma$ for a constant $C_0 \neq 0$ (essentially $1/\zeta^{\prime}(1/2+i\gamma)$). Letting $\sigma \to 0$ from above, we get a contradiction.

Note that the same heuristics underlying Gonek's conjecture should also suggest that $M_0(x)$ grows like $(\log \log \log x)^{5/4}$. I'm sure all this is classical, but I don't know a reference offhand.

$\endgroup$
2
  • 4
    $\begingroup$ This proof seems to work for any Dirichlet series whose analytic continuation has a pole, that is: Ordinary summation of a Dirichlet series must fail on the vertical line containing the pole. $\endgroup$ Oct 3, 2014 at 14:18
  • 4
    $\begingroup$ Lucia, let me ask a silly question : assuming the RH, on what depends the growth rate of $M_0(x)$ the more ? the asymptotic density of zeros ? their order ? the modulus of the residues ? $\endgroup$
    – reuns
    Aug 3, 2016 at 15:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.