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The sums $$S_N=\sum_{n=1}^N\frac{\mu(n)}{n},$$ where $\mu$ is the Moebius function, are known to tend to 0 as $N\to+\infty$. As far as I remember, there was an estimate on $S_N$ equivalent to the Riemann Hypothesis. If so, could anyone please remind me the exact formulation and give the reference?

Also, what is the best known estimate?

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  • $\begingroup$ The reference I have about a equivalence of the Riemann hypothesis (RH) and the sum of some type of Moebius terms is: Define $M(N)=\sum_{n\leq N}\mu(n)$ then $M(N)\ll N^{1/2+\varepsilon}$ if and only if (RH) holds. You can find a proof in Ivic's book; The Riemann Zeta-Function. For your sum, I think we can use Abel summation formula to get an estimate equivalent to the given in this comment (and then equivalent to (RH)). I'll think about it and post it if I get a positve anwser. $\endgroup$
    – MrSelberg
    Aug 28 '15 at 20:58
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Define $M(N)=\sum_{n\leq N}\mu(n)$. Then $M(N)\ll N^{1/2+\varepsilon}$ is equivalent to the Riemann Hypothesis (Aleksandar Ivic, The Riemann Zeta-Function, page 47). Define $S(N)=\sum_{n\leq N}\frac{\mu(n)}{n}$. We will prove


$M(N)\ll N^{1/2+\varepsilon}$ if and only if $S(N)\ll N^{-1/2+\varepsilon}$.

Proof: Suppose $M(N)\ll N^{1/2+\varepsilon}$. Then by partial summation $$ \begin{align*} S(N)&=\frac{1}{N}\sum_{n\leq N}\mu(n)+\int_1^N\frac{1}{t^2}\sum_{n\leq t}\mu(n)\,d t\\ &=\frac{M(N)}{N}+\int_1^N\frac{M(t)}{t^2}\,dt\\ &\ll N^{-1/2+\epsilon} \end{align*} $$ Suppose $S(N)\ll N^{-1/2+\varepsilon}$. Again, by partial summation $$ \begin{align*} M(N)=\sum_{n\leq N}\frac{\mu(n)}{n}n&=N\sum_{n\leq N}\frac{\mu(n)}{n}-\int_1^N\sum_{n\leq t}\frac{\mu(n)}{n}\,d t\\ &\ll NS(N)+\left|\int_1^N S(t)\,dt\right|\\ &\ll N^{1/2+\epsilon}. \end{align*} $$ This Completes the proof


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    $\begingroup$ Thanks a lot for the answer! $\endgroup$
    – 0-0
    Aug 29 '15 at 21:10

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