= Solution
The <comparison test for series> says that an eventual upper bound by a convergent positive <series> proves convergence, while an eventual lower bound by a divergent positive <series> proves divergence. The <p-series> $\sum n^{-s}$ converges exactly when $s>1$.
If $p>1$, then $(\log n)^q\geq1$ for all sufficiently large $n$, so
$$
0<\frac1{n^p(\log n)^q}\leq\frac1{n^p},
$$
and the <series> converges. If $0<p<1$, choose $\eta=(1-p)/2>0$. The allowed logarithmic bound, applied with exponent $\eta/q$, gives $(\log n)^q<n^\eta$ eventually. Thus
$$
\frac1{n^p(\log n)^q}>\frac1{n^{p+\eta}},\qquad p+\eta=(p+1)/2<1,
$$
and the <series> diverges.
For $p=1$, use the <integral test for convergence>: for an eventually positive decreasing function $g$, $\sum g(n)$ and $\int g(x)\,dx$ either both converge or both diverge. Here $g(x)=1/[x(\log x)^q]$ is decreasing for $x>1$, and the substitution $u=\log x$ gives
$$
\int_2^\infty\frac{dx}{x(\log x)^q}
=\int_{\log2}^\infty u^{-q}\,du.
$$
This integral converges exactly when $q>1$. Hence the <polynomial-logarithmic series convergence> classification is
$$
\boxed{\text{Convergence iff }p>1\text{ or }(p=1\text{ and }q>1).}
$$
All other allowed positive $p,q$ give divergence.
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