The asymptotic growth rate of the sigma function can be expressed by :
$$\lim \sup\frac{\sigma(n)}{n\log(\log(n))}=e^{\gamma}$$ $$n \to\infty$$
according to the above limit , Is this : $$\lim \sup\frac{\sigma(n)}{n}$$ $$n \to\infty$$ has a finite limit ?
Note : I just would like to know the behavior of sigma function for large $n$ and to know the evaluation of second limit .
Thank you for any help .
the lim sup you are asking about is infinite
zeraoulia, the article you should read is Alaoglu and Erdos
http://www.renyi.hu/~p_erdos/1944-03.pdf
The numbers that are most efficient at giving large ( and increasing without bound) $\sigma(n)/n$ are called Colossally Abundant Numbers there. See https://mathoverflow.net/questions/79927/which-n-maximize-gn-frac-sigmann-log-log-n
I should also point out that programming the CA numbers is not so easy. However, the approximate factorization of one of these is very similar to a fairly easy sequence, $$ \operatorname{lcm} \; \{1,2,3,4,5, \ldots, N \} $$ Note that this number increases only when $N$ is a prime or prime power, so make the sequence under consideration from those.
Furthermore, you ask about the ratio $\sigma(n)/n$ by itself. This is, as pointed out, unbounded. As you can see at How would I find a number where $\sum_{d\mid n}d > 100n$? we can make arbitrarily large values from either the factorials or the primorials. I decided to do it with CA numbers, I did get the ratio up to $30$ and gave up. I think someone dedicated could do the task there with the LCM numbers I describe above.
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Decided to do it myself this way with the LCM, and see how far I can get. Below is an abbreviated printout. I told it to just print when the ratio became at least as large as the next integer...
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