It is wellknown 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?
It is wellknown 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 $$ s^{1}f(s)=\int_1^{\infty} S(x)x^{s1}dx. $$ 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) nontrivial 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: 

