Prove that a $\limsup$ related with the sum of divisor function equals $e^{\gamma}$ I want to show that $$\limsup_{n \rightarrow \infty} \frac{\sigma(n)}{n \log \log n} = e^{\gamma}.$$
where $\sigma(n):=$ sum of positive divisor function and $\gamma$ is the Euler constant (the one appear in Merten's estimates $(3)$ https://en.wikipedia.org/wiki/Mertens%27_theorems)
I try to use Merten estimate since it appear $\gamma$ in the limit valaue. 
Since $$\sigma(p_1^{a_1}...p_k^{a_k})= \frac{\prod_{i=1}^k (p_i^{a_i} -1/p_i)}{\prod_{i=1}^k (1-1/p_i)}.$$ The reciprocal might be modified to use Merten reciprocal form $$\prod_{p \leq x; p\ \ prime} (1-1/p)^{-1} = e^{\gamma} \log x (1 + O(1/\log x)).$$
I guess that $\log \log$ thing, and $\log$ and error term in Merten should be somehow cancel, but really not sure why.
 A: In order to exploit Mertens' theorem, you just have to show that $\frac{\sigma(n)}{n}$ is maximized when $n$ is a primorial.  By the PNT, $\sum_{p\leq x}\log p = x+o(x)$, hence if we take $n$ as $\prod_{p\leq x}p$, we have
$$ \frac{\sigma(n)}{n\log\log n}\approx \frac{1}{\log x}\prod_{p\leq x}\left(1+\frac{1}{p}\right)\approx\frac{1}{\log x}\prod_{p\leq x}\left(1-\frac{1}{p}\right)^{-1}.$$
A: Let $N_x = \prod_{p < x} p$ the primorial. By the Mertens theorems we have
$$\frac{\sigma(N_x)}{N_x} = \prod_{p < x} (1+\frac{1}{p} )=  e^{\gamma+\ln \ln x+o(1)}$$
Now $\ln N_x = \sum_{p < x} \ln p $ so that by Chebyshev's work (*) $x/2 < \ln N_x <2x$ and $\ln x =\ln  \ln N_x+ \mathcal{O}(1)$, $\ln \ln x =\ln \ln  \ln N_x+ o(1)$
and we get
$$\frac{\sigma(N_x)}{N_x} =  e^{\gamma+\ln \ln \ln N_x+o(1)} = e^{\gamma+o(1)}\ln \ln N_x$$
and finish with
$$\lim \sup_{n \to \infty} \frac{\sigma(n)}{n \ln \ln n} = \lim \sup_{x \to \infty} \frac{\sigma(N_x)}{N_x \ln \ln N_x} = e^{\gamma}$$

The Mertens theorem we need is proved  there, but I can try to make it shorter :
Let $\Lambda(p^k) = \ln p$ if $p$ is prime, $\Lambda(n) = 0$ otherwise.
By a simple sieving/combinatoric we have
$x!= \prod_{p^k \le x} p^{ \lfloor x /p^k \rfloor}$ so that
$$\sum_{n \le x} \lfloor x /n\rfloor \Lambda(n)  = \sum_{p^k \le x} \lfloor x /p^k \rfloor \ln p = \ln x!= x \ln x+ \mathcal{O}(x)$$
by Stirling's approximation.
Hence
$$\sum_{n \le x} \frac{\Lambda(n)}{n} = \frac{1}{x} \sum_{n \le x} \Lambda(n)(\lfloor x /n\rfloor+\mathcal{O}(1)) = \ln x+ \mathcal{O}(1)$$
where we used that $\sum_{n \le x}\Lambda(n) = \mathcal{O}(x)$
Finally with $\ln x = \mathcal{O}(1) +\sum_{n \le x} \frac{1}{n}$ we get
$$\sum_{n < x} \frac{\Lambda(n)-1}{n} = \mathcal{O}(1)$$
Summing by parts (mistakes here)
$$\mathcal{O}(1) = \sum_{n =2}^x \frac{\Lambda(n)-1}{n \ln n}\ln n = \ln x\sum_{n \le x} \frac{\Lambda(n)-1}{n \ln n}+\sum_{2 \le k < x}(\ln k-\ln (k+1))\sum_{2\le n < k} \frac{\Lambda(n)-1}{n \ln n} = \ln x\sum_{n \le x} \frac{\Lambda(n)-1}{n \ln n} + \sum_{2 \le k < x} \mathcal{O}(1/k)$$
so that $\sum_{n \le x} \frac{\Lambda(n)-1}{n \ln n} = \mathcal{O}(1/\ln n)$ so we can refine $\sum_{2 \le k < x} \mathcal{O}(1/k)$ to $\sum_{2 \le k < x} \mathcal{O}(\frac{1}{k \ln k}) = \mathcal{O}(\ln \ln x)$ and get
$$\sum_{n \le x} \frac{\Lambda(n)-1}{n \ln n} = o(1)$$
Finally
$\sum_{n \le x} \frac{\Lambda(n)}{n \ln n} = o(1)-\sum_{p \le x} \ln(1-\frac{1}{p})$ and
$$\sum_{p \le x}\ln(1+\frac{1}{p}) = o(1)+\sum_{p \le x} (\ln(1-\frac{1}{p^2})-\ln(1-\frac{1}{p})) = - \ln \zeta(2)+o(1)+\sum_{n \le x} \frac{\Lambda(n)}{n \ln n} =- \ln \zeta(2)+o(1)+\sum_{2 \le n \le x} \frac{1}{n \ln n} =\gamma+o(1)+\ln \ln x$$
where I used $\sum_{2 \le n \le x} \frac{1}{n \ln n} = \ln \ln(x)+\gamma-\zeta(2)+o(1)$, no reference for the $\gamma-\zeta(2)$ constant
