Polynomial-logarithmic series convergence (${\sum }_{n\ge 2}{n}^{-p}(\log n{)}^{-q}$)

For $p,q>0$, the positive series converges precisely when $p>1$, or $p=1$ and $q>1$. For $p>1$ use comparison with a convergent p-series; for $0<p<1$, the bound $(\log n{)}^q<n^{(1-p)/2}$ gives a lower comparison with a divergent p-series. At $p=1$, the integral test for convergence and $u=\log x$ reduce the question to $\int u^{-q}du$. In contrast, ${\sum }_{n\ge 3}[n(\log \log n{)}^r{]}^{-1}$ diverges for every finite $r>0$, since $(\log \log n{)}^r<\log n$ eventually and $\sum [n\log n{]}^{-1}$ diverges.