Asymptotic divisor function / primorials 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?


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.
