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The Mertens function is defined as follows:

(1) $\quad\mathcal{M}(N)=\sum_{n=1}^N\mu(n)$

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 $\mathcal{M}(N)$ 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 $\mathcal{M}(N)=0$?

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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.

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  • $\begingroup$ Interesting proof. I have a further question: assuming RH and given the informations about the distributions of zeroes of $\zeta$ along the critical line provided by the Riemann-Von Mangoldt theorem, is it possible to estimate how often the Mertens function has to exhibit a sign change? $\endgroup$ – Jack D'Aurizio Jul 3 '17 at 23:17
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    $\begingroup$ @JackD'Aurizio I would try using the explicit formulas $$ \psi(x)-x-\log 2\pi = \sum_{|Im(\rho)| < T} \frac{x^\rho}{\rho}+ \mathcal{O}(x^{1/2} \log^2x/T),\qquad M(x)+2 = \sum_{|Im(\rho)| < T} \frac{x^\rho}{\rho \zeta'(\rho)}+ \mathcal{O}(x^{1/2} \log^2x/T)$$ (if the zeros are simple) $\endgroup$ – reuns Jul 3 '17 at 23:34
  • $\begingroup$ @JackD'Aurizio Using the density of zeros, we can try finding a simple function $h(x)$ such that $(\psi(x)-x)h(x)$ is almost always non-negative, so that $\int_1^\infty (\psi(x)-x) h(x) x^{-s-1}dx$ has a singularity at $s =\sigma$ the abscissa of convergence of $\int_1^\infty (\psi(x)-x) x^{-s-1}dx$ $\endgroup$ – reuns Jul 3 '17 at 23:40

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