Bounds on a sum involving the Möbius function

In Apostol's Analytic Number Theory, Apostol defines $$A(x):= \sum_{n \leq x} \frac{\mu(n)}{n}$$ and proves that $A(x)=o(1)$ implies the Prime Number Theorem, by showing that $$\frac{M(x)}{x}=A(x)-\frac{1}{x}\int_1^x A(t)dt,$$ in which $M(x):=\sum_{n \leq x} \mu(n)$ is the summatory function for the Möbius function (Theorem 4.16). What are some known error bounds for the function $A(x)$? In particular, do we have $A(x)= o(1/\log x)$ as $x \to \infty$?

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Terence Tao has a rather interesting paper about this : 'A remark on partial sums involving the Mobius function'. See too his blog. –  Raymond Manzoni Dec 7 '12 at 1:20
Schoenfeld's paper, Marraki's one and Cohen's paper seem interesting too (summatory function is much more studied than the $\zeta(1)$ formula). –  Raymond Manzoni Dec 7 '12 at 1:25
$$A(x)=\frac{M(x)}{x}+ \int_1^x \frac{M(u)}{u^2} du = \frac{M(x)}{x}+\int_1^\infty \frac{M(u)}{u^2} du-\int_x^\infty \frac{M(u)}{u^2} du.$$ As $A(x)=o(1)$, the right-hand side of the above must tend to $0$. We have $M(x)/x \to 0$, and the estimate $$\left\vert \int_x^\infty \frac{M(u)}{u^2} du \right\vert \leq \int_x^\infty \frac{\vert M(u)\vert}{x^2} du =\frac{1}{x^2}O(xM(x))=O(M(x)/x)$$ implies that the rightmost integral of our first line tends to $0$ as well. Thus $$\int_1^\infty \frac{M(u)}{u^2} du=0,$$ and $A(x)=O(M(x)/x)$. In particular, we can answer our question by simply bounding the growth of Mertens' function $M(x)$. We have $$M(x)=O\left(xe^{-c\sqrt{\log x}}\right)$$ for some positive constant $c$. (I believe this follows from the classical bounds in the PNT but am unable to find a proper reference. Edit: I found a mention of the process here.) Then $A(x) =O(e^{-c \sqrt{\log x}})$, and since $$\lim_{x \to \infty} \frac{(\log x)^n}{e^{c \sqrt{\log x}}}=0$$ for all $n$, we find $A(x)=o((\log x)^{-n})$ for all $n >0$.