TITLE: Does the Mertens function have an infinite number of zeros?
QUESTION [6 upvotes]: The Mertens function $M\colon\mathbb N\to\mathbb N$ is defined as
$$M(N)=\sum_{n=1}^N\mu(n).$$
I have a very simple question for which I cannot seem to find a definitive answer on the web. I've also consulted several books on number theory and can't even seem to find a conjecture much less a proven result.
Question 1: Does the Mertens function $M(N)$ evaluate to zero for an infinite number of integers $N$?
I also have a second related question.
Question 2: What is the largest known integer $N$ such that $M(N)=0$?

REPLY [12 votes]: Assume $M(x)$ has constant sign for $x > A$. Then 
$$\frac{1}{s \zeta(s)} = \int_1^\infty M(x)x^{-s-1}dx =g(s)+ \int_A^\infty M(x)x^{-s-1}dx $$
where $g(s) = \int_1^A M(x) x^{-s-1}dx$ is entire.
Hence 
$$|\frac{1}{s \zeta(s)}-g(s)| \le\int_A^\infty |M(x)x^{-s-1}|dx =  \pm \int_A^\infty M(x)x^{-\sigma-1}dx =|\frac{1}{\sigma \zeta(\sigma)}-g(\sigma)|$$
Proving $\frac{1}{s \zeta(s)}-g(s)$ is analytic  on $\Re(s) > \sigma$ and has a singularity at $s= \sigma$, where $\sigma$ is the abscissa of convergence of the integral (this would make the RH very easy to prove or disprove !)
But we know $\frac{1}{s \zeta(s)}$ is analytic on $(0,\infty)$ and has a pole at $s \approx 1/2+i14.134725$
Qed.