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 $x \to \infty$?
Bounds on a sum involving the Möbius function
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2Terence Tao has a rather interesting paper about this : ['A remark on partial sums involving the Mobius function'](http://arxiv.org/abs/0908.4323). See too his [blog](http://terrytao.wordpress.com/2009/08/30/an-elementary-inequality-involving-the-mobius-function). – 2012-12-07
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0Schoenfeld's [paper](http://matwbn.icm.edu.pl/ksiazki/aa/aa15/aa15120.pdf), Marraki's [one](http://archive.numdam.org/ARCHIVE/JTNB/JTNB_1995__7_2/JTNB_1995__7_2_407_0/JTNB_1995__7_2_407_0.pdf) and Cohen's [paper](http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&id=pdf_1&handle=euclid.facm/1229618741) seem interesting too (summatory function is much more studied than the $\zeta(1)$ formula). – 2012-12-07
2 Answers
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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0The 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). – 2015-02-02
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0It'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. – 2015-02-03
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