This is inspired by the Question on coefficient of $\exp(H_n).\log(H_n)$ in Lagarias equivalence of RH , an answer and some comments there.
For $n\geqslant2$ denote $$ L(n):=\frac{\sigma(n)-H_n}{\exp(H_n)\log(H_n)}, $$ where $\sigma$ is the divisor sum and $H_n$ is the $n$th harmonic number.
As Emil Jeřábek explains in one of the comments there, $\limsup_{n\to\infty}L(n)\geqslant1$. The result of Lagarias says that the Riemann Hypothesis is equivalent to $L(n)\leqslant1$ for all $n$, so that there exists a sequence $a_k$ with $L(a_k)$ growing monotonously and $\lim_{k\to\infty}a_k=1$. Clearly there is a smallest such sequence: take $a_1=2$ and $a_{k+1}$ the smallest number $n$ with $L(n)\geqslant L(a_k)$.
Here are the few starting values (up to ten decimal places): $$ \begin{array}{cl} n & L(n)\\ \hline \color{red} 2 &\color{red}{0.82546003} \\ 3 & 0.571499183 \\ \color{red} 4 &\color{red}{0.83408973} \\ 5 & 0.458908111 \\ \color{red} 6 &\color{red}{0.91966830} \\ 7 & 0.424541931 \\ 8 & 0.81094546 \\ 9 & 0.57778078 \\ 10 & 0.74961989 \\ 11 & 0.39656917 \\ \color{red}{12} &\color{red}{0.98723811} \\ 13 & 0.38886853 \\ 14 & 0.68121794 \\ 15 & 0.62448763 \\ 16 & 0.77143689 \\ 17 & 0.37810207 \\ 18 & 0.86099035 \\ 19 & 0.37404301 \\ 20 & 0.82141873 \\ \end{array} $$ After that, I checked for $n$ up to more than one million, $L(n)$ never exceeds $L(12)$.
In a comment, Alexander Kalmynin indicated that $L(12)$ is exceeded by the 100th Colossally Abundant Number (CAN in what follows). Searching through their list at OEIS A004490 I found that the first CAN with $L$ exceeding $L(12)$ is the 58th one, $$ 2^9\cdot3^5\cdot5^3\cdot7^3\cdot11^2\cdot13^2\cdot17^2\cdot19\cdot23\cdot\ldots\cdot167 $$ (all primes in between, $\approx5.94632\times10^{76}$). In fact it seems that, denoting the $k$th CAN by $C_k$, $L(C_k)$ grows monotonously starting from $C_{29}$. The line through $(C_k,L(C_k))$, $k\leqslant150$, looks like
Thus one such sequence is $2=C_1$, $4$, $6=C_2$, $12=C_3$, $C_{58}$, $C_{59}$, ..., $C_{150}$, ...
But is the sequence of $L(C_k)$ really monotonously growing? Is it the smallest such sequence? Is in fact $C_{58}$ the smallest $n$ with $L(12)<L(n)$? Can there be some $m\ne n$ with $L(m)=L(n)$? And, is there an efficient method to check all this computationally?
