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