Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $x \to \infty$?

share|improve this question
1  
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
add comment

1 Answer

up vote 5 down vote accepted

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 $n >0$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.