# Dubious step in 'order of mobius' proof which may lead to an interesting fact

Let $\mu(n)$ be the Moebius function, let $M(x)=\sum_{n\leq x} \mu(x)$ be the Mertens function and let $A(x)=\sum_{n\leq x}\tfrac{\mu(n)}{n}$ be the truncation of the Dirichlet series expansion of $1/\zeta(s)$ at $s=1$.

Question: is it true that $A(x)\geq \tfrac{M(x)}{x}$ for all integer $x$?

I provide some facts and context.

By partial summation [3] the difference is $A(x)-\tfrac{M(x)}{x} = \tfrac 1x\int_0^x A(t) dt$. In an elementary [3] way $|A(x)|\leq 1$ for all $x$. It is also elementary [1] [2] to prove the equivalences $$A(x)=o(1)\iff M(x)=o(x)\iff \text{Prime Number Theorem}.$$

In fact, I was reading on ProofWiki [2] a proof of the fact that $M(x)=o(x)$. At a certain point of the proof it is claimed that clearly $A(x)\geq \tfrac{M(x)}{x}$.

My considerations are: (i) it is not clear to me; (ii) but maybe I'm missing something obvious? (iii) you could think it were obvious if $\mu(n)$ didn't change sign; (iv) the statement is true for $x<1000$.

[1] Diamond, H. G. (1982). Elementary methods in the study of the distribution of prime numbers. Bulletin of the American Mathematical Society, 7(3), 553-589.

[3] Apostol, T. M. (2013). Introduction to analytic number theory. Springer Science & Business Media.

• That proofwiki article looks really sloppy to me and I would be a little dubious of trusting it for anything. – Steven Stadnicki Jul 17 '18 at 1:25
• In any case it raises an interesting question in the spirit of "things that (don't?) change sign infinitely often". – Luca Ghidelli Jul 17 '18 at 1:46

It is false. Using PARI/GP I found:

$M(18798)=0$

$A(18798)*18798 =-0.3589042...$

• Bonus question: does the difference change sign infinitely often? – Luca Ghidelli Jul 17 '18 at 15:00
• I would expect so, but I would not be surprised if that is still unknown. – Daniel Fischer Jul 17 '18 at 15:17
• @LucaGhidelli, it can be shown to change sign infinitely often, using the tricks in Chapter 15 of Montgomery and Vaughan. – Peter Humphries Jul 20 '18 at 9:35
• I will be happy to learn it, thank you for the reference! @Peter Humphries – Luca Ghidelli Jul 20 '18 at 9:46
• I learnt something, thank you! Lemma 15.1 applied to the function 1/{s(s-1)zeta(s)} = \int_0^\infty [A(x)-M(x)/x]x^{-s} (no need to do better) does the job perfectly, because it is analytic for real s>0 but not on all the semiplane Re s>0 ! Thank you, would you like to write it as an answer? – Luca Ghidelli Jul 20 '18 at 19:43