For any sequence of complex numbers $(a_n)$, an application of the CauchySchwarz inequality gives $$\left\sum_{m=1}^{n}a_m\right\leq \sqrt {n\sum_{m=1}^{n}a_m^2}.$$ Putting $a_n=\mu(n)/\sqrt n$, one (trivially) finds that $$\sum_{m=1}^{n}\frac{\mu(m)}{\sqrt m}= O(\sqrt{n\log n}).$$ Is a better unconditional estimate known to hold for this sequence? In other words, is Cauchy Schwarz the most we know about this sequence?
Take the 2minute tour
×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $\epsilon >0$. 

