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 $x \to \infty$?
 A: I'll answer my own question:
The Abel Summation formula gives
$$A(x)=\frac{M(x)}{x}+ \int_1^x \frac{M(u)}{u^2} du = \frac{M(x)}{x}+\int_1^\infty \frac{M(u)}{u^2} du-\int_x^\infty \frac{M(u)}{u^2} du.$$
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 $n >0$.
A: The inequality (and the ensuing equalities) of your second line is unfortunately meaningless. However, using the bound you give on $M(x)$ in the integral, one gets easily
$$\left\vert \int_x^\infty \frac{M(u)}{u^2} du \right\vert =O\left(\sqrt{\log x}e^{-c \sqrt{\log x}}\right),$$
which ultimately does not change your conclusion.
Cheers
