Summing by parts and using the bound on the Mertens function $M(n)=o(n)$ (which is equivalent to the Prime Number Theorem) one gets for your sum $S(n)=o(\sqrt n)$. Better bounds on the order of magnitude of $M(n)$ of course give a better one for $S(n)$ (see e.g. the *Handbook of Number Theory*, by J. Sándor, Dragoslav S. Mitrinović, B. Crstici). And, since $O(n^{1/2+\epsilon})$ for all $\epsilon >0$ is equivalent to the Riemann hypothesis, I guess one can't expect for your sum better than $O(n^{\epsilon})$ for all $\epsilon >0$.