= Polynomial-logarithmic series convergence
{title2=$\sum_{n\geq2}n^{-p}(\log n)^{-q}$}
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, $\sum_{n\geq3}[n(\log\log n)^r]^{-1}$ diverges for every finite $r>0$, since $(\log\log n)^r<\log n$ eventually and $\sum[n\log n]^{-1}$ diverges.
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