Asymptotic divisor function / primorials

Question

Let $p_n\#\equiv\prod_{k=1}^{n}p_k$ (primorial), and $\sigma(n)=\sum_{d|n}^{}d$ (divisor function).

Could someone please tell me what the general asymptotic of $\dfrac{\sigma(p_n\#)}{p_n\#}$ is? It looks like $O \log(n)$. Is there anything more precise?

Update

Thank you to Daniel Fischer for his answer. Out of interest, I include the plot of $\dfrac{\sigma(p_n\#)}{p_n\#}$ against $\dfrac{6 e^{\gamma}}{\pi^2} \log p_n$. The primorial curve is surprisingly smooth though - why is this?

Answer 1

The divisor sum function is multiplicative, and $\sigma(p) = p+1$ for a prime $p$. So

$$\frac{\sigma(p_n\#)}{p_n\#} = \prod_{k=1}^n \left( 1 + \frac{1}{p_k}\right) = \frac{\prod_{k=1}^n \left(1 - \frac{1}{p_k^2}\right)}{\prod_{k=1}^n\left(1 - \frac{1}{p_k}\right)}.$$

By Mertens' third theorem, we have

$$\prod_{k=1}^n \left(1 - \frac{1}{p_k}\right) \sim \frac{e^{-\gamma}}{\log p_n},$$

and by Euler's product formula

$$\lim_{n\to\infty} \prod_{k=1}^n \left(1 - \frac{1}{p_k^2}\right) = \frac{1}{\zeta(2)} = \frac{6}{\pi^2}.$$

Therefore, we have the asymptotics

$$\frac{\sigma(p_n\#)}{p_n\#} \sim \frac{6 e^{\gamma}}{\pi^2} \log p_n.$$

Since $p_n \sim n\log n$, we have $\log p_n \sim \log n$, but $\log n + \log \log n$ is a little more accurate.

Answer 2

Daniel's answer sems to cover it. Note gronwall's theorem, Thm 323 on page 266 of Hardy nad Wright, $$ \limsup \frac{\sigma(n)}{n \log \log n} = e^\gamma. $$ H+W use the sequence of numbers $$\operatorname{lcm} \{1,2,3, \ldots, \} $$ to show this, prrof pages 353-354.

Daniel can confirm whether the primorials give a good sequence for Gronwall's.

Meanwhile, the best numbers are not primorials, although they are the product of primorials, meaning that the exponents in the prime factorization decraese, or at least do not increase. The best numbers here are the colossally abundant numbers, by Erdos and Alaoglu, but actually cut out of the original Ramanujan article owing to paper shortages at the time. (Don't ask me, i wasn't there).

Also note that the criterion of Nicolas (1983) is much easier to deal with, as the primorials are the genuine test case there. Answer with proof: Is the Euler phi function bounded below? ... other answer Euler's Phi Function Worst Case