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Feb 25, 2017 at 22:45 comment added reuns If $M(n) = O(n^{a})$ then there is $N,C$ such that for every $n > N$ : $|M(n)| < C n^{a}$. Of course $M(n) > n^{1-\epsilon}$ for infinitely many $n$ contradicts this statement
Jun 29, 2016 at 21:14 comment added Jérôme JEAN-CHARLES OK I agree that my question are a bit simplistic, I thought a mere specific epsilon was enough. Considering how far we are ( N/logN) I feel justified in my mistake. Sorry for the disturbance though.
Jun 29, 2016 at 4:43 history closed Franz Lemmermeyer
Peter Humphries
Jan-Christoph Schlage-Puchta
Wolfgang
Ryan Budney
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S Jun 29, 2016 at 0:13 history edited Pedro Lauridsen Ribeiro CC BY-SA 3.0
spurious ) brackets plus characters to meet 6 requirement, bold font fixed, grammar improved, removed spurious closing parentheses
S Jun 29, 2016 at 0:13 history suggested Henry CC BY-SA 3.0
spurious ) brackets plus characters to meet 6 requirement
Jun 29, 2016 at 0:07 review Suggested edits
S Jun 29, 2016 at 0:13
Jun 28, 2016 at 19:50 answer added aosjckajsd timeline score: 4
S Jun 28, 2016 at 18:30 history suggested Martin Sleziak CC BY-SA 3.0
Riemmann -> Riemann
Jun 28, 2016 at 18:18 review Suggested edits
S Jun 28, 2016 at 18:30
Jun 28, 2016 at 18:17 answer added GH from MO timeline score: 5
Jun 28, 2016 at 17:53 comment added Peter Humphries Yes to Q1, no to Q2, of course, because then the statement that $M(x) \ll_{\varepsilon} x^{1/2 + \varepsilon}$ would be false, and this is equivalent to RH. Basically, if $\limsup_{x \to \infty} |M(x)|x^{-\Theta} > 0$ for some $\Theta > 1/2$, then RH is false.
Jun 28, 2016 at 17:38 review Close votes
Jun 29, 2016 at 4:43
Jun 28, 2016 at 16:47 history asked Jérôme JEAN-CHARLES CC BY-SA 3.0