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.

5$\begingroup$ 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.univlille1.fr/~ramare/Maths/mqdex36.pdf $\endgroup$– user1073Jul 31, 2016 at 21:05

$\begingroup$ @Bin Thank you for the paper! I am searching an estimate of this form $\endgroup$– Khadija MbarkiJul 31, 2016 at 21:40

1$\begingroup$ 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. $\endgroup$– Peter HumphriesJul 31, 2016 at 23:23

$\begingroup$ @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 $A(x)+O(\exp(c\sqrt{\log{x}}))$ $\endgroup$– Khadija MbarkiAug 1, 2016 at 11:26

1$\begingroup$ 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. $\endgroup$– Peter HumphriesAug 1, 2016 at 11:27
1 Answer
Actually Ramaré's paper is freely avaible at his Lille university page (link here).
The estimate that he uses to deduce his result is
$$\bigg\sum_{n\leq x} \frac{\mu(n)}{n}\bigg\leq \bigg(\frac{3}{2} +o(1)\bigg) \exp \bigg(\max_{x^{7/8}\leq t \leq x} \log \frac{2+M(t)}{t}\bigg)+O(x^{1/4})$$
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.