15
$\begingroup$

The Mertens function is the partial sums of the Moebius function: $M(x)=\sum_{n\leq x}\mu(n)$ Since the zeta-function has a zero on the critical line it follows that $M(x)\ne O(x^\theta)$ for any $\theta<\frac 12$.

Does anyone know if there is an elementary proof of this statement? (By elementary I mean a proof which does not depend on complex analysis, in particular the existance of a zero of $\zeta$). even an elementary proof of $M(x)$ being unbounded would be interesting to me.

$\endgroup$
1

1 Answer 1

4
$\begingroup$

As far as I know, even an elementary proof that $M(x)$ is unbounded is not known.

$\endgroup$
1
  • 12
    $\begingroup$ Note that the function field analogue of M(x) is bounded (no zeroes for the analogous zeta function), so this already rules out a fairly large class of elementary proofs; one needs to somehow use a property of the rational integers that is not shared by the polynomials over a finite field. $\endgroup$
    – Terry Tao
    Nov 26, 2010 at 21:32

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.