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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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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 $n >0$.

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  • $\begingroup$ The link in this post is dead. A proper reference for both M(x) and A(x) is given in G.J.O. Jameson's The Prime Number Theorem, Thm. 5.1.9. page 186 (2003). An earlier source for M(x) is Landau's Handbuch vol.II, sec. 164, page 613 (1909) (the proof is in preceding pages). $\endgroup$ – daniel Feb 2 '15 at 19:17
  • $\begingroup$ It's also the error estimate de la Vallee Poussin found in his original proof of the PNT. Harold Edwards, Riemann's Zeta Function, p.82. $\endgroup$ – daniel Feb 3 '15 at 4:05
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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

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