Question

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 $$\frac x {\log x}=n-\frac{0.975 n}{\log n}+O \left(\frac {\log n \log \log n}{\log^2 n} \right) $$ By the first two terms in the asymtotic expansion of li(x) we get that $$ \operatorname{li} (x)=n+\frac{0.025}{\log x)=n+\frac{0.025n}{\log n}+O \left(\frac {\log n \log \log n}{\log^2 n} \right) $$ and thus $$G(n)=\frac{0.025}{\log $G(n)=\frac{0.025 n}{\log n}+O \left(\frac {\log n \log \log n}{\log^2 n} \right) $$ and $G(n) \to \infty$ as $n\to \infty$.