Timeline for Estimation of a sum involving Moebius function
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Aug 1, 2016 at 14:25 | vote | accept | Khadija Mbarki | ||
| Aug 1, 2016 at 13:36 | answer | added | Myshkin | timeline score: 1 | |
| Aug 1, 2016 at 11:27 | comment | added | Peter Humphries | 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. | |
| Aug 1, 2016 at 11:26 | comment | added | Khadija Mbarki | @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 $A(x)+O(\exp(-c\sqrt{\log{x}}))$ | |
| Jul 31, 2016 at 23:23 | comment | added | Peter Humphries | 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. | |
| Jul 31, 2016 at 21:55 | history | edited | GH from MO |
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| Jul 31, 2016 at 21:46 | history | edited | Khadija Mbarki | CC BY-SA 3.0 |
added 36 characters in body
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| Jul 31, 2016 at 21:40 | comment | added | Khadija Mbarki | @Bin Thank you for the paper! I am searching an estimate of this form | |
| Jul 31, 2016 at 21:05 | comment | added | user1073 | 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 | |
| Jul 31, 2016 at 20:47 | history | asked | Khadija Mbarki | CC BY-SA 3.0 |