Is there an elementary proof that the Mertens function is not $O(x^\theta)$ if $\theta <1/2$?

2010-11-26T15:24:15Z

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.

Answer by engelbrekt for Is there an elementary proof that the Mertens function is not $O(x^\theta)$ if $\theta <1/2$?

2010-11-26T18:47:53Z

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

Is there an elementary proof that the Mertens function is not $O(x^\theta)$ if $\theta <1/2$? - MathOverflow

Is there an elementary proof that the Mertens function is not $O(x^\theta)$ if $\theta <1/2$?

Answer by engelbrekt for Is there an elementary proof that the Mertens function is not $O(x^\theta)$ if $\theta <1/2$?