Define $N\_k \geq 6$ to be the $k-th$ primorial number and let $\sigma(n)$ be the divisor function.


It *seems* that $u\_k = \dfrac{\sigma(N\_k)}{N\_k \log\log N\_k}$ is a decreasing function ?


By computation, we find that the first few values of $u\_k$ (to 2dp) are 
 $u\_1 = 3.42, u\_2 = 1.96, u\_3 = 1.63, u\_{4} = 1.46, u\_5 = 1.38 \cdots$


Is there a known result from which this follows ?


As far as i'm aware, the only existing result *along these lines* is Gronwall's theorem, which says that $\lim$ $\sup u\_k = e^{\gamma}$, where $\gamma$ is the Euler-Mascheroni constant.


EDIT: By direct arithmetic that $u\_k > u\_{k+1}$ and simplifying, the problem reduces to showing that $\dfrac{\log\log N\_{k+1}}{\log\log N\_k} > \dfrac{1+p\_{k+1}}{p\_{k+1}}$, where $p\_n$ denotes the $n-th$ prime.



A decreasing sequence involving the divisor function?
The answer is probably no. Assume that $u\_{k+1}<u\_k$ for all $k$. Then
$$ \frac{1}{p\_{k+1}}<\frac{\log\log N\_{k+1}}{\log\log N\_k}-1=\frac{\log\frac{\log N\_{k+1}}{\log N\_k}}{\log\log N\_k}<\frac{\frac{\log N\_{k+1}}{\log N\_k}-1}{\log\log N\_k}=\frac{\log p\_{k+1}}{\log N\_k\log\log N\_k}.$$
In particular,
$$ \theta(p\_k)\log\theta(p\_k)<p\_{k+1}\log p\_{k+1},$$
whence
$$ \theta(p\_k)<p\_{k+1}. $$
By Littlewood's theorem, the left hand side exceeds $p\_k+p\_k^{1/2}$ infinitely often, while it is widely believed that the right hand side is smaller than $p\_k+p\_k^{1/3}$ for every large $k$. So, if we believe the last upper bound, there are infinitely many counterexamples (but these might occur very rarely).


**Added.** By quoting Littlewood's theorem a bit more precisely, we get a contradiction already by Legendre's conjecture that there is a prime number between any two large consecutive squares.


