What are the fallacies that this RH inequality may fail at most finitely often? 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?

 A: I suspect the mistake is in relying on Sage or Maple in the last step. Instead use the asymptotic expansion of li(x) ( http://en.wikipedia.org/wiki/Logarithmic_integral_function )
$$\operatorname{li}(x)=\frac x {\log x}+\frac{x}{\log^2 x}+O \left(\frac x {\log^3 x} \right)$$
to obtain an asymptotic expression of $G(n)$. Letting 
$$x=\frac 1 4 \left(\frac {\log \log n-0.975}{\log n}+2 \right)^2 n \log n$$ we see that
$$x = n \log n \left(1+\frac{\log \log n}{\log n}- \frac{0.975}{\log n} +O \left( \frac{\log \log n}{\log^2 n} \right) \right) $$
and that
$$\log x=\log n+\log \log n+\frac{\log \log n}{\log n}-\frac{0.975}{\log n}+O \left(\frac {\log \log n}{\log^2 n} \right).$$
With some more calculation we get
$$\frac x {\log x}=n-\frac{0.975 n}{\log n}+O \left(\frac {n \log \log n}{\log^2 n} \right) $$
By the first two terms in the asymtotic expansion of li(x) we get that
$$ \operatorname{li} (x)=n+\frac{0.025n}{\log n}+O \left(\frac {n \log \log n}{\log^2 n} \right) $$
and thus $$G(n)=\frac{0.025 n}{\log n}+O \left(\frac {n \log \log n}{\log^2 n} \right) $$ and $G(n) \to \infty$ as $n\to \infty$.
A: Don't blame Maple (blame the user).


