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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.

flag	asked Nov 26 2010 at 15:24 Alastair Irving 301●4

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You might enjoy reading the answers to mathoverflow.net/questions/11074/… – David Speyer Nov 26 2010 at 18:17

engelbrekt · Answer 1 · 2010-11-26 18:47:53Z UTC

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

answered Nov 26 2010 at 18:47

engelbrekt
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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. – Terry Tao Nov 26 2010 at 21:32

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

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