4
$\begingroup$

Let $\mu$ be the Mobius function. In his paper "Explicit estimates on several summatory functions involving the Moebius function", Olivier Ramaré proves the following effective bound: $$\left|\sum_{n\leq x} \frac{\mu(n)}{n}\right|\log{x}\leq 1/69,$$ When $x\geq 96955.$ Unfortunately, I couldn't access to his paper since it is published in American Mathematical Society so I just take the result from the free abstract on the Web site of this journal. My question is the following: Is there an estimation of the sum $\sum_{n\leq x} \frac{\mu(n)}{n}$ in termes of $x$ which means with Big $O$ error term $(\sum_{n\leq x} \frac{\mu(n)}{n}=A(x)+O(B(x)))$. I guess it exists in the Ramaré's paper and from this estimation he derived his upper bound mentioned above.
Many thanks.

$\endgroup$
  • 5
    $\begingroup$ Theorem 1.2 of the paper you mention by Ramaré shows that your sum is bounded above by $(0.0144\log x - 0.1)/(\log x)^2$ for $x\geq 463,421$. Is that the type of bound you are looking for? Also, the paper seems to be available on the author's website: math.univ-lille1.fr/~ramare/Maths/mqdex-3-6.pdf $\endgroup$ – Ben Linowitz Jul 31 '16 at 21:05
  • $\begingroup$ @Bin Thank you for the paper! I am searching an estimate of this form $\endgroup$ – Khadija Mbarki Jul 31 '16 at 21:40
  • 1
    $\begingroup$ Would you not just be happy with $\sum_{n \leq x} \frac{\mu(n)}{n} = O\left(\exp\left(-c\sqrt{\log x}\right)\right)$? Because this is a consequence of the prime number theorem. $\endgroup$ – Peter Humphries Jul 31 '16 at 23:23
  • $\begingroup$ @PeterHumphries thank you for your answer and what is the main term in this formula? I mean the expression of $A(x)$ such that this sum is equal to $A(x)+O(\exp(-c\sqrt{\log{x}}))$ $\endgroup$ – Khadija Mbarki Aug 1 '16 at 11:26
  • 1
    $\begingroup$ There is no main term: $\sum_{n \leq x} \frac{\mu(n)}{n} = o(1)$, i.e. it tends to zero as $x$ tends to infinity. $\endgroup$ – Peter Humphries Aug 1 '16 at 11:27
1
$\begingroup$

Actually Ramaré's paper is freely avaible at his Lille university page (link here).

The estimate that he uses to deduce his result is

$$\bigg|\sum_{n\leq x} \frac{\mu(n)}{n}\bigg|\leq \bigg(\frac{3}{2} +o(1)\bigg) \exp \bigg(-\max_{x^{7/8}\leq t \leq x} \log \frac{2+|M(t)|}{t}\bigg)+O(x^{-1/4})$$

The sharpest one which works for all $x \geq 1$ seems to be (as of 2015)

$$\bigg|\sum_{n\leq x} \frac{\mu(n)}{n}\bigg|\leq \frac{726}{(\log x)^2}$$

Peter Humphries has alredy mentioned a (much cheaper!) estimate with $A(x)=0$ in the comments.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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