Let $\sigma(n)$ be the sum of divisor function. We know, by Gronwall's Theorem, that
$${\lim \text{sup}}_{n \to \infty} \frac{\sigma(n)}{n \log \log{n}} =e^\gamma$$
And the Riemann Hypothesis (Robin's Inequality) states that:
$$\frac{\sigma(n)}{n \log \log{n}} <e^\gamma \: \forall n>5040$$
But, regarding the following equality:
$$\sigma(n)-e^\gamma n \log \log{n}= O(f_1(n))$$
• Do we know anything about what is supposed to be $f_1(n)$?
• Do we have any (either formal or heuristic) argument to defend any special function?
• When trying to go beyond that $f_1(n)$, what would be the next step? Finding a function $f_2(n)$ such that
$$\sigma(n)-e^\gamma n \log \log{n}-f_1(n)= O(f_2(n))$$
? If so, do we know/believe anything about it?
• Do we know/believe at which "step" $k$ (if any) we would find something of the form:
$$\sigma(n)-e^\gamma n \log \log{n}-f_1(n)- \cdots -f_k(n)= O(1)$$ ?
One can show $$ \sigma(n) \le n e^\gamma \log \log n + O(n). $$ On the other hand, when $n$ is prime, we have $$ \sigma(n) = n+1. $$ so, in fact, $$ n e^\gamma \log \log n -\sigma(n) = n e^\gamma \log \log n -n -1 $$ infinitely often. The sum of divisors function is too erratic to say anything more unless you want to talk about average orders.
Here is a plot of $\sigma(n)-e^\gamma n \log \log n$ for $n<10^5$.
Here is a plot of $\frac{\sigma(n)}{e^\gamma n \log \log n}$ for $n<10^5$.