Upper Bounds on a Ratio Concerning the Sum of Divisors

Question

Given that $n$ is an highly abundant number, what is a function $f(x)$ that is a tight upper bound on the ratio: $$\frac{\sigma_1(n)}{n}$$ where $\sigma_1(n)$ is the sum of divisors of $n$ (including $n$ itself). I also want to know an upper bound if $n$ is a superabundant number

$f(x)$ doesn't have to reach equality on certain numbers, but it would be optimal if it did.

Answer 1

umm. For $n \geq 16,$ we get $\log \log n > 1, $ using log base $e = 2.71828...$ Then for $n \geq 16,$ $$ \frac{\sigma(n)}{n \log \log n} $$ does have an upper bound.

From Robin we have an unconditional result (1984) that says that your ratio of interest is, for $n \geq 13,$ smaller than $$ \frac{\sigma(n)}{n \log \log n} < e^\gamma + \frac{0.64821364942...}{(\log \log n)^2},$$ with the constant in the numerator giving equality for $n=12.$ Note $\gamma \approx 0.5772156649,$ while $e^\gamma \approx 1.7810724$

In particular, for $n \geq 268,$ we get $ \frac{\sigma(n)}{n \log \log n} < 2$ Below are the ratios bigger than $1.9$ with $n \geq 3.$

  14.17718374918197           3
   5.357674360663709          4
   2.521617642462445          5
   3.429366566700587          6
   2.561127739941722          8
   2.158189420874225         10
   2.563440313761713         12
   2.041357661635523         18
   1.913982525796606         20
   2.162126621182643         24
   1.960580021444964         30
   1.980481186664786         36
   1.908540572163114         48
   1.98636974867664          60
   1.915700702020492        120

Answer 2

There is no upper bound. The sum of divisor function is multiplicative. If $p$ is prime, we have $\frac {\sigma_1(p^n)}{p^n}=\frac {p^{n+1}-1}{p^n(p-1)}\gt 1+\frac 1p$. If we multiply a lot of these together for different primes we get $$\left(1+\frac 1p\right)\left(1+\frac 1q\right)\left(1+\frac 1r\right)\ldots \gt 1+\frac 1p + \frac 1q + \frac 1r +\ldots$$
and we know the sum of the inverses of the primes diverges.