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?

7$\begingroup$ For starters, the sum is bounded by $\sum_{m=1}^n 1/\sqrt{m} = O(\sqrt{n})$, which is of strictly lower order than $O(\sqrt{n\log n})$, though still trivial. One can probably do a bit better using zerofree regions for the Riemann zeta function. $\endgroup$ – Noam D. Elkies Oct 25 '11 at 19:46

$\begingroup$ Thanks, Noam. Ok, I can see that $O(\sqrt{n})$ can be proved by EulerMaclaurin, or by writing $\sum_{m=1}^{n}1/\sqrt{m}$ as an inverse Mellin transform. Can the same be proved using CauchySchwarz? $\endgroup$ – Kevin Smith Oct 25 '11 at 20:57
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 $\epsilon >0$.

$\begingroup$ Great! Thanks, Pietro. I had thought the Prime Number Theorem would come into it. I guess that $o(\sqrt{n})$ is the best that is currently known then? I asked this question because I wondered if, when viewed as the distance from the origin, one can distinguish the growth of the sequence from a typical point in $\mathbb{R}^n$  it appears then that one can hardly do so. $\endgroup$ – Kevin Smith Oct 25 '11 at 21:47

5$\begingroup$ The specific versions of $o(\sqrt n)$ that are known are more precise than the statement $o(\sqrt n)$ itself. For example, the error term in the classical prime number theorem is $O\big(n\exp(c\sqrt{\log n})\big)$ for some constant $c>0$; the corresponding estimate for your $M(n)$ would be $O\big(\sqrt n\cdot\exp(c\sqrt{\log n})\big)$ (possibly with a slightly smaller $c$). $\endgroup$ – Greg Martin Oct 25 '11 at 23:33