According to "EQUIVALENCES TO THE RIEMANN HYPOTHESIS p.4
Let $g(n)$ be the maximal order of a permutation of n objects
RH Equivalence 3.3. The Riemann Hypothesis is equivalent to $\log{g(n)} < Li^{-1} (n)$ for n large enough.
$Li$ is strictly increasing for $n>1$ and $\log{g(n)} \sim \sqrt{n \log{n}}$.
$\log{g(n)} \ge \sqrt{n \log{n}}$ for $n \ge 906$ and $\log{g(n)} \gg 1$.
Some equivalences using the fact that $Li$ and squaring preserve inequalities (because they are strictly increasing in the relevant intervals, easy to show) are:
$$ \log{g(n)} < \sqrt{Li^{-1}(n)} \iff (\log{g(n))^2 < Li^{-1}(n) \iff Li( (\log{g(n)})^2}) < n \qquad (1) $$
for $n$ large enough.
In Effective Bounds for the Maximal Order of an Element in the Symmetric Group Theorem 2, p. 2 the following bound for $g(n)$ is given unconditionally for $n \ge 3$:
$$\log{g(n)} \le F(n) = \sqrt{n \log{n}} \left( 1 + \frac{\log{\log{n}} - 0.975}{2 \log{n}}\right) $$
$$\log{g(n)} \le F(n) \iff (\log{g(n)})^2 \le F(n)^2 \iff Li((\log{g(n)})^2) \le Li(F(n)^2) \qquad (1a)$$
From (1) and (1a) showing $ Li( (\log{g(n)) ^2}) \le Li(F(n)^2) < n $ (if true) will prove (1).
Let $$ G(n) = Li(F(n)^2) - n = $$
$$ -n + {Li}\left(\frac{1}{4} \, {\left(\frac{\log\left(\log\left(n\right)\right) - 0.975}{\log\left(n\right)} + 2\right)}^{2} n \log\left(n\right)\right) =$$
$$ -n + {\rm Ei}\left(\log\left(\frac{1}{4} \, {\left(\frac{\log\left(\log\left(n\right)\right) - 0.975}{\log\left(n\right)} + 2\right)}^{2} n \log\left(n\right)\right)\right) - {\rm Ei}\left(\log\left(2\right)\right) $$
We must show $$G(n)<0 \qquad (2)$$ (if true) for $n$ large enough.
According to both sage and maple (using the $\rm Ei$ expression) $\lim_{n \to \infty} G(n) = -\infty$ so (2) may fail at most finitely often ($n \in \mathbb{N}$).
The derivative of $G(n)$ is of elementary functions only.
What are the mistakes and logical errors in this?
