Estimation of a sum involving Moebius function

Let $\mu$ be the Mobius function. In his paper "Explicit estimates on several summatory functions involving the Moebius function", Olivier Ramaré proves the following effective bound: $$\left|\sum_{n\leq x} \frac{\mu(n)}{n}\right|\log{x}\leq 1/69,$$ When $x\geq 96955.$ Unfortunately, I couldn't access to his paper since it is published in American Mathematical Society so I just take the result from the free abstract on the Web site of this journal. My question is the following: Is there an estimation of the sum $\sum_{n\leq x} \frac{\mu(n)}{n}$ in termes of $x$ which means with Big $O$ error term $(\sum_{n\leq x} \frac{\mu(n)}{n}=A(x)+O(B(x)))$. I guess it exists in the Ramaré's paper and from this estimation he derived his upper bound mentioned above.
Many thanks.

• Theorem 1.2 of the paper you mention by Ramaré shows that your sum is bounded above by $(0.0144\log x - 0.1)/(\log x)^2$ for $x\geq 463,421$. Is that the type of bound you are looking for? Also, the paper seems to be available on the author's website: math.univ-lille1.fr/~ramare/Maths/mqdex-3-6.pdf – Ben Linowitz Jul 31 '16 at 21:05
• @Bin Thank you for the paper! I am searching an estimate of this form – Khadija Mbarki Jul 31 '16 at 21:40
• Would you not just be happy with $\sum_{n \leq x} \frac{\mu(n)}{n} = O\left(\exp\left(-c\sqrt{\log x}\right)\right)$? Because this is a consequence of the prime number theorem. – Peter Humphries Jul 31 '16 at 23:23
• @PeterHumphries thank you for your answer and what is the main term in this formula? I mean the expression of $A(x)$ such that this sum is equal to $A(x)+O(\exp(-c\sqrt{\log{x}}))$ – Khadija Mbarki Aug 1 '16 at 11:26
• There is no main term: $\sum_{n \leq x} \frac{\mu(n)}{n} = o(1)$, i.e. it tends to zero as $x$ tends to infinity. – Peter Humphries Aug 1 '16 at 11:27

$$\bigg|\sum_{n\leq x} \frac{\mu(n)}{n}\bigg|\leq \bigg(\frac{3}{2} +o(1)\bigg) \exp \bigg(-\max_{x^{7/8}\leq t \leq x} \log \frac{2+|M(t)|}{t}\bigg)+O(x^{-1/4})$$
The sharpest one which works for all $x \geq 1$ seems to be (as of 2015)
$$\bigg|\sum_{n\leq x} \frac{\mu(n)}{n}\bigg|\leq \frac{726}{(\log x)^2}$$
Peter Humphries has alredy mentioned a (much cheaper!) estimate with $A(x)=0$ in the comments.