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All:

I saw one form of Riemann Hypothesis, it says: $$ \lim ∑(μ(n))/n^σ $$ Converges for all σ > ½

Is this statement same as the order of Mertens function is less than square root of n ?

nilo de roock
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david
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1 Answers1

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Yes, since $\frac{1}{\zeta(\sigma)} = \sum{\frac{\mu(n)}{n^\sigma}}$, this is equivalent to the more canonical statement of RH that $\zeta$ has no zeroes to the right of the critical line.

You also mention $M(x) = O(x^{\frac{1}{2}+\epsilon})$, you can use the Mellin transform to show this is also equivalent to RH.

Dan Brumleve
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    RH is actually equivalent to $M(x)=O(x^{1/2+\epsilon})$, the form you cited is stronger. – Markus Shepherd Jan 08 '15 at 16:25
  • Is it true that $M(x)=o(x^{\frac{1}{2}+\epsilon})$ is equivalent to RH? – mkultra May 23 '19 at 12:33
  • @mkultra the $O$-estimates and $o$-estimates are equivalent because they are quantified over all $\varepsilon > 0$: show as $x\to\infty$ that $f(x)=O(x^\varepsilon)$ for all $\varepsilon > 0$ if and only if $f(x)=o(x^\varepsilon)$ for all $\varepsilon > 0$ and apply that to the function $f(x) = M(x)/\sqrt{x}$. – KCd Jul 23 '21 at 19:06