In Guy Robin, Grandes valeurs de la fonction somme des diviseurs et hypothèse de Riemann, J. Math. Pures Appl. 63 (1984), 187–213 (pdf) we find the following result:

If the Riemann hypothesis is true and $n ≥ 5041$, $\frac{\sigma(n)}{n} < e^\gamma \ln \ln (n)$

We also know that $e^\gamma < e$. Now my question here is:

Question: Without using the Riemann hypothesis, is it possible to show that $\frac{\sigma(n)}{n} < e \ln \ln (n)$ ?

  • 5
    $\begingroup$ Aleksander Grytczuk, Upper bound for sum of divisors function and the Riemann Hypothesis, Tsukuba J. Math. vol. 31 No. 1 (2007), 67–75, proved if $m$ is odd, $m>(1/2)3^9$, then $\sigma(2m)/(2m)<(39/40)e^{\gamma}\log\log2m$. See projecteuclid.org/download/pdf_1/euclid.tkbjm/1496165115 $\endgroup$ Oct 8, 2020 at 6:22
  • $\begingroup$ @The Company , I wish that i didn't change the meaning of your question . $\endgroup$ Oct 8, 2020 at 19:05
  • $\begingroup$ This makes me want to define the "Robin constant" $R$ as the infimum of the $C>0$ such that $\frac{\sigma(n)}{n}<e^{C}\log\log n, n\geq 5041$. $\endgroup$ Oct 8, 2020 at 21:34
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    $\begingroup$ @zeraouliarafik you have introduced the condition $n\ge5041$, which was not in the original question. $\endgroup$ Oct 8, 2020 at 21:56
  • 1
    $\begingroup$ Please do not vandalize your own questions. $\endgroup$
    – Wojowu
    Nov 14, 2020 at 17:47

1 Answer 1


Wikipedia says Robin proved unconditionally that the inequality $${\sigma(n)\over n}<e^{\gamma}\log\log n+{0.6483\over\log\log n}$$ holds for all $n\ge3$. I believe this is in the same paper as the one cited in the body of the question.

  • $\begingroup$ Tks! $e^{\gamma} \ln \ln (n) + \frac{0.6483}{ \ln \ln (n)} < e \ln \ln (n)$ for all $n>9$ $\endgroup$ Oct 8, 2020 at 11:10
  • $\begingroup$ When said unconditionally, does it imply that it does not depend on HR? $\endgroup$ Oct 13, 2020 at 5:11
  • $\begingroup$ Yes. "Unconditionally" means not depending on any unproven statements (and that would include the Riemann Hypothesis, which I assume is what you mean by "HR"). But as you're interested in this, why not get the paper, and see for yourself? $\endgroup$ Oct 13, 2020 at 11:10
  • $\begingroup$ thanks, it has been difficult to find the bibliographical references of G. Robin! $\endgroup$ Oct 13, 2020 at 11:12
  • $\begingroup$ Huh? The full bibliographical reference is in the body of the question. $\endgroup$ Oct 13, 2020 at 11:16

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