# $\frac{\sigma(n)}{n} < e \ln \ln (n)$ is True?

In G. Robin, Grandes valeurs de la fonction somme des diviseurs et hypothèse de Riemann, J. Math. Pures Appl. 63 (1984), 187–213

A results is:

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: $$\frac{\sigma(n)}{n} < e \ln \ln (n)$$ ?

• 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 – Gerry Myerson Oct 8 at 6:22
• @The Company , I wish that i didn't change the meaning of your question . – zeraoulia rafik Oct 8 at 19:05
• This makes me want to define the "Robin constant" $R$ as the infimum of the $C>0$ such that $\frac{\sigma(n)}{n}<e^{C}\log\log n, n\geq 5041$. – Sylvain JULIEN Oct 8 at 21:34
• @zeraouliarafik you have introduced the condition $n\ge5041$, which was not in the original question. – Gerry Myerson Oct 8 at 21:56

Wikipedia says Robin proved unconditionally that the inequality $${\sigma(n)\over n} holds for all $$n\ge3$$. I believe this is in the same paper as the one cited in the body of the question.
• Tks! $e^{\gamma} \ln \ln (n) + \frac{0.6483}{ \ln \ln (n)} < e \ln \ln (n)$ for all $n>9$ – The Company Oct 8 at 11:10