About $\lim\dfrac{\sigma(n)}{n}$, where $\sigma(n)$ is the sum of divisor of $n$

Question

Denote $p\equiv 1 \mod 3$ is a prime number and $n$ is any number that divisible by $p$. Denote $\sigma(n)$ is the sum of divisor of $n$. We know that $\dfrac{\sigma (n)}{n}>1$ and $$\dfrac{\sigma (n)}{n}=\prod\limits_{p_i\mid n}\dfrac{p_i^{\alpha_i+1}-1}{p_i^{\alpha+1}-p_i^\alpha}<\prod_\limits{p_i\mid n\\p_i\neq p}\dfrac{p_i}{p_i-1}.\dfrac{p}{p-1}$$ When $p$ goes to infinitely then $n\rightarrow \infty$, we have $\dfrac{p}{p-1}\rightarrow 1$. But does $\dfrac{\sigma (n)}{n}\rightarrow 1$ when $p\rightarrow \infty$, since we can't control other $p_i$ and the number of divisor of $n$ ($n$ is arbitrary and $p\mid n$).

Answer 1

The limit of $\dfrac {\sigma(n)}{n}$ does not exist, since for any multiple of $6p$, i.e. $n = 6pk$ we have

$$\frac {\sigma(n)}n > \frac {pk + 2pk + 3pk + 6pk}{6pk} = 2$$

while for $n = p^k$, we have $$\frac {\sigma (n)}n = \frac {p^{k+1} - 1}{p^k(p-1) } < \frac {p^{k+1} + p^k(p-2)}{p^k(p-1)} = 2$$

Moreover, the ratio for prime powers tends to $1$, so for $n$ an arbitrary multiple of $p$, the limit of the abundancy index does not exist, no matter if $n \to\infty$ or $p \to \infty$.