Is $\limsup_n \frac{\sigma(n)}{n \log p(n)} <\infty$, where $p(n)$ is the greatest prime factor of $n$ and $\sigma(n)=\sum_{d | n} d$?

Let $\sigma(n)=\sum_{d | n} d$ and $p(n)$ be the greatest prime factor of $n$.

Can we prove that $$\limsup_n \frac{\sigma(n)}{n \log p(n)} <\infty ?$$

I know that $$\limsup_n \frac{\sigma(n)}{n \log \log n} <\infty.$$

I came up with this question when thinking of sequences $\{n_k\}$ such that $\frac{\sigma(n_k)}{n_k} \to \infty$, e.g., $n_k=(p_k)!$ or $p_1 p_2 \cdots p_k$, where $p_k$ is the $k$-th prime.

And, in fact, those examples where all satisfying $\lim \frac{\sigma(n_k)}{n_k \log p(n_k)} <\infty$.

I appreciate any references or ideas on that.

Let $2=p_{1}<p_{2}<p_{3} ...$ be the list of all the consecutive primes

Suppose $m_{k}$ is a natural number that has prime divisors atmost $p_{k}$ with $p_{k}|m_{k}$. Note $m_{k}$ is of the form $\prod_{j=1}^{k}p_{j}^{\alpha_{j}}$. We calculate

$$\frac{\sigma(m_{k})}{m_{k}\log(p_{k})} = \frac{\prod_{j=1}^{k}\frac{(\sum_{u=0}^{\alpha_{j}}p_{j}^{u})}{p_{j}^{\alpha_{j}}}}{\log(p_{k})} \leq \frac{\prod_{j=1}^{k}(1+\sum_{u=1}^{\infty}\frac{1}{p_{j}^{u}})}{\log(p_{k})} \leq \frac{\prod_{j=1}^{k}(1+\frac{1}{p_{j}-1})}{\log(p_{k})}$$

Set $S(k) = \log(\prod_{j=1}^{k}(1+\frac{1}{p_{j}-1})) = \sum_{j=1}^{k}\log(1+\frac{1}{p_{j}-1})$. For $x \geq 0$ we have the inequality $\log(1+x) \leq x$. Thus

$$S(k) \leq \sum_{j=1}^{k}\frac{1}{p_{j}-1} \leq \sum_{j=1}^{k}\frac{1}{p_{j}}+C$$

For some constant $C > 0$. Also by What is the sum of the reciprocal of primes? (Yes, it diverges..) there exists some constant $D > 0$ so that

$\sum_{j=1}^{k}\frac{1}{p_{j}} \leq \log(\log(p_{k}))+D$.

Thus $S(k) \leq \log(\log(p_{k}))+E$ where $E = C+D$.

Hence

$\log(\frac{\sigma(m_{k})}{m_{k}\log(p_{k})}) \leq S(k) - \log(\log(p_{k})) \leq E$.

Hence

$\frac{\sigma(m_{k})}{m_{k}\log(p_{k})} \leq e^{E}$ where $E$ is independent of $k$. This proves the result.

Edit: If you want the tightest bound then set

$$H = \max_{k \in \mathbb{N}}\left((\sum_{j=1}^{k}\frac{1}{p_{j}-1})-\log(\log(p_{k}))\right)$$ (which has to be finite because of the above argument) so that

$$\frac{\sigma(n)}{n\log(p(n))} \leq e^{H}$$

where $e^{H}$ is tight (left to the reader to confirm).

$$$$

Call

$$f(n)=\frac{\sigma(n)}{n\log p(n)}$$

and $2=p_1<p_2<\dots$, the list of consecutive primes. If we have a number $n=q_1^{a_1}q_2^{a_2}\dots$ where $a_1\geq a_2\geq\dots$ where $q_i$ are distinct primes we can produce another one $n'=p_1^{a_1}p_2^{a_2}\dots$ so that $f(n')\geq f(n)$. So to find the limit superior we only have to consider numbers of the form $p_1^{a_1}p_2^{a_2}\dots$ where $a_1\geq a_2\geq\dots$. So let n be of this form then

$$f(n)=\frac 1{\log p_k}\prod_{h=1}^k \frac{1-p_h^{-a_h-1}}{1-p_h^{-1}}$$

$$=\left(\prod_{h=1}^k 1-p_h^{-a_h-1}\right)\left(\log p_k\prod_{h=1}^k1-\frac 1{p_h}\right)^{-1}.$$

The first parenthesis goes to 1 as the exponents go to infinity and the second one we have by Mertens' third theorem goes to $e^{-\gamma}$ so $f(n)\to e^\gamma\approx 1.78$ if we let k go to infinity. But we don't have to do that, we can take $n=2^a$ in which case $f(n)$ approaches $\frac 2{\log(2)}\approx 2.88$. So far we have that the limit superior is attained in a sub-sequence of n with bounded prime divisors, and with this result, we have that

$$e^\gamma\left(1-\frac 1{2\log(p_k)^2}\right)^{-1}>\left(\log p_k\prod_{h=1}^k1-\frac 1{p_h}\right)^{-1}>e^\gamma\left(1+\frac 1{2\log(p_k)^2}\right)^{-1}$$

which for $p_k\geq 5$ gives $f(n)<2.20$, so you only need to check the $p_k=3$ case to conclude

$$\limsup_{n\to\infty} f(n)=\frac 2{\log(2)}.$$

I plotted $f(n)$ for n up to ten thousand, and you can see the powers of 2 clearly.