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## 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?

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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 $n\to \infty$.

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 Thank you ${}{}{}$ – joro Jul 27 at 8:52

Don't blame Maple (blame the user).

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 Thank you! I confirm this on version 13. Can you try without "MultiSeries" and writing Li in terms of Ei (that is what worked for me). – joro Jul 27 at 8:51 $$\tt{> G := -n+Ei(1/4*((log(log(n))-0.975)/log(n)+2)^2*n*log(n))-Ei(log(2)):}$$ $$\tt{> limit(G,n=infinity);}$$ $$\tt{Float(infinity)}$$ – Gerald Edgar Jul 27 at 14:13 Thank you. The last result is certainly different in version 13. – joro Jul 27 at 14:27