As Alexey has pointed out the problem can be reduced, via [summation by parts][1], to understanding the asymptotic of Mertens' sum 
$$M(x) := \sum_{n\leq x} \mu(n).$$
Conditional on the Riemann hypothesis, the best bound to date on Mertens' sum, [due to Soundararajan][2] in 2009, is
$$M(x) \ll x^{1/2} e^{(\log(x))^{1/2} (\log\log(x))^{14}}.$$
In particular, this implies
$$\sum_{n\leq x} \frac{\mu(n)}{n} \ll \frac{e^{(\log(x))^{1/2} (\log\log(x))^{14}}}{x^{1/2}}.$$


  [1]: http://en.wikipedia.org/wiki/Summation_by_parts
  [2]: http://www.ams.org/mathscinet-getitem?mr=2542220