4
$\begingroup$

Let $M(x) = \sum_{n\leq x} \mu(n)$ and $m(x) = \sum_{n\leq x} \frac{\mu(n)}{n}$, where $\mu(n)$ is the Möbius function. We know that (it is not the best known bounds): $$\limsup_{x \to \infty} M(x)x^{-1/2} > 1$$ $$\liminf_{x \to \infty} M(x)x^{-1/2} < -1$$

Are there any references showing that (if true) we also have for some $c>0$: $$\limsup_{x \to \infty} m(x)x^{1/2} > c$$ $$\liminf_{x \to \infty} m(x)x^{1/2} < -c$$

Improve this question
$\endgroup$
4
  • 2
    $\begingroup$ See mathoverflow.net/questions/153695 for upper bounds on $|m(x)|$, but that doesn't address lower bounds. (I mistakenly thought this was a duplicate of that one, but it isn't.) $\endgroup$
    – David E Speyer
    Commented Jul 30 at 14:48
  • $\begingroup$ For the Mertens function $M(x)$, Kotnik & te Riele established: $\limsup M(x)x^{-1/2} > 1.218$, $\liminf M(x)x^{-1/2} < -1.229$. However, analogous tight bounds for $m(x)x^{1/2}$ are unknown to me. $\endgroup$
    –  Babar
    Commented Jul 30 at 15:25
  • $\begingroup$ @Babar The inequalities by Kotnik & te Riele are not tight. It is expected that $\limsup M(x)x^{-1/2}=\infty$ and $\liminf M(x)x^{1/2}=-\infty$. By following my answer you should be able to find a numeric positive lower bound for $\limsup m(x)x^{1/2}$ and a numeric negative upper bound for $\liminf m(x)x^{1/2}$. $\endgroup$
    – GH from MO
    Commented Jul 30 at 15:29
  • 1
    $\begingroup$ @GH from MO Thanks, I expressed myself poorly. I meant to say "the best known result" instead of "tight". $\endgroup$
    –  Babar
    Commented Jul 30 at 15:33

1 Answer 1

Reset to default
5
$\begingroup$

This is true by Exercise 4(b) for Section 15.1.1 in Montgomery-Vaughan: Multiplicative number theory I (CUP, 2006). The inequality contained therein can surely be proved in much the same way as (15.13) in the book, with the help of Lemma 15.1.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .