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.



Estimation of a sum involving Moebius function
Actually Ramaré's paper is freely avaible at his Lille university page ([link here](http://math.univ-lille1.fr/~ramare/Maths/mqdex-3-6.pdf)).


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](https://www.google.es/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&cad=rja&uact=8&ved=0ahUKEwintqaEpKDOAhXLvhQKHUgZBD4QFgghMAA&url=https%3A%2F%2Fcs.uwaterloo.ca%2Fjournals%2FJIS%2FVOL18%2FBordelles2%2Fbordelles21.pdf&usg=AFQjCNEG-JYtr919qx4YUcoYL1nH6Jh2Qw&sig2=JMzEUsKEwYDJ_xe9784L6w))


$$\bigg|\sum\_{n\leq x} \frac{\mu(n)}{n}\bigg|\leq \frac{726}{(\log x)^2}$$


Peter Humphries has alredy mentioned a (much cheaper!) estimate with $A(x)=0$ in the comments.


