Sign of Mertens function 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?
 A: 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.
