It is well-known that Mertens function $$M(N)=\sum_{i=1}^N\mu(n)$$ changes sign infinitely many times when $N\rightarrow +\infty$.

Question: Is there a proof of this statement without Riemann zeta function?


I am not an expert, but I do not think so. The proofs I have seen use Dirichlet series $f(x)=\sum_{n\le x} a_n n^{-s}$, with $S(x)=\sum_{n\le x}a_n$, to consider $$ s^{-1}f(s)=\int_1^{\infty} S(x)x^{-s-1}dx. $$ This leads to information on sign changes of $\sum_{i=1}^N a_n$. For the $\mu$-function, this gives of course $f(s)=\sum_{n=1}^{\infty} \mu (s)n^{-s}=(\zeta(s))^{-1}$, and the formula $$ (s\zeta(s))^{-1}- C(s-\lambda)^{-1} =\int_1 ^{\infty} \lbrace M(x)-Cx^{\lambda}\rbrace x^{-s-\lambda}dx. $$ For details on the notations and the precise arguments see "OSCILLATION THEOREMS OF ARITHMETICAL FUNCTIONS" by Emil Grosswald from $1967$.

I have the feeling that every (sufficently) non-trivial statement on $\mu(n)$ will always involve $\zeta(s)$, directly or indirectly.

Update: Of course one should mention here the reference of Odlyzko and te Riele, "Disproof of the Mertens Conjecture", which contains many more informations on the behavior of $M(x)$, which is in fact determined by the zeros of the zeta function. They refer to the following reference:
J. Pintz, Oscillatory properties of $M(x)$, Acta Arith. 43 (1984), 105-113.

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    $\begingroup$ I have no idea how to define a "sufficiently non-trivial statement". What about the elementary proof of PNT? It is likely to be a non-trivial statement(without zeta function in its proof), I think. $\endgroup$ – Y. Zhao Jun 7 '13 at 1:05

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