Let $c>0$, $n \in \mathbb N$ and $q>1$. How to get the following approximating inequality when $n$ is large, please? To be more specific, I cannot see how to get rid of the square root.
$$ \frac{ce^{-\log(n\log q)}}{\sqrt{2\log(n\log q)}} > \frac{\tilde c}{n\log n}, $$
where $\tilde c$ is another constant. In addition, why the summation of the right side over $n$ does not converge, please? Thank you!
Because it suffices to prove that $$c\frac{(n\log{q})^{-1}}{\sqrt{2\log{(n\log{q})}}}\cdot (n\log{n})>\tilde{c}.$$ Since $\log{n}/\sqrt{\log{n}+\alpha}\sim\sqrt{\log{n}}\to\infty$, so for any $\tilde{c},\alpha$ and sufficiently large $n$, we have $$\frac{\log{n}}{\sqrt{\log{n}+\log\log q}}>\frac{\sqrt{2}\tilde{c} \log{q}}{c}.$$ which is exactly the same as before. For the divergence of $\frac{1}{n\log{n}}$, we may use the inequality $\frac{1}{n\log{n}}>\frac{1}{2^{k+1}\log{2^{k+1}}}$ when $2^k\le n <2^{k+1}$. Thus $$\sum_{n=2}^\infty\frac{1}{n\log{n}}=\sum_{k=1}^\infty\sum_{2^k\le n<2^{k+1}}\frac{1}{n\log{n}}>\sum_{k=1}^\infty\frac{2^k}{2^{k+1}\log{2^{k+1}}}=\sum_{k=1}^\infty \frac{1}{2k\log{2}}\to\infty$$