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Could someone please provide information about the best possible known bounds of the sum $$A(x)=\sum_{n\leq x}\frac{\mu(n)}{n}?$$ Unconditionally, $A(x)=O(e^{-c\sqrt{\log x}})$ is known to me. Does there exist any better bound conditionally or unconditionally?

I am expecting a result like $A(x)=O(\frac1{\sqrt x})$ if Riemann Hypothesis is assumed. Is this true? What I could get is up on truth of RH $A(x)=O(x^{-1/2+\epsilon})$, which is not very hard to prove.

Any reference regarding conditional (RH) bound of $A(x)$ will be highly appreciated. Thanks.

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    $\begingroup$ You probably meant to write $A(x)=O(\exp(-c\sqrt{\log x}))$ unconditionally. The Vinogradov zero-free region should improve that to $O(\exp(-(\log x)^{\frac 23-\epsilon}))$ (indeed something a little more precise). Mark Lewko has supplied the best known result on RH $\endgroup$ – Lucia Jan 6 '14 at 3:49
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As Alexey has pointed out the problem can be reduced, via summation by parts, to understanding the asymptotic of Mertens' sum $$M(x) := \sum_{n\leq x} \mu(n).$$ Conditional on the Riemann hypothesis, the best bound to date on Mertens' sum, due to Soundararajan in 2009, is $$M(x) \ll x^{1/2} e^{(\log(x))^{1/2} (\log\log(x))^{14}}.$$ In particular, this implies $$\sum_{n\leq x} \frac{\mu(n)}{n} \ll \frac{e^{(\log(x))^{1/2} (\log\log(x))^{14}}}{x^{1/2}}.$$

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    $\begingroup$ Using Soundararajan's method, the power of $\log\log x$ in this estimate has been improved by Balazard & Roton: arxiv.org/abs/0810.3587 $\endgroup$ – Micah Milinovich Jan 6 '14 at 15:30
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See Mertens conjecture. Summation by parts gives your sum from Mertens conjecture.

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    $\begingroup$ This is technically true, but Mertens conjecture has been disproved... $\endgroup$ – Terry Tao Dec 22 '17 at 22:50

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