= Solution
For $p\leq0$, the terms $k^{-p}$ do not tend to zero, so the series diverges. For $p>0$, group the positive terms into the dyadic blocks $2^j\leq k<2^{j+1}$. Each block contains $2^j$ terms, and monotonicity gives
$$
2^{-p}\,2^{j(1-p)}\leq
\sum_{k=2^j}^{2^{j+1}-1}k^{-p}
\leq2^{j(1-p)}.
$$
When $p>1$, the upper bounds form a convergent <geometric series> with ratio $2^{1-p}<1$, proving convergence. When $0<p\leq1$, the lower bounds do not tend to zero, so infinitely many blocks each make a nonvanishing contribution and the sum diverges. Thus the <p-series> criterion is
$$
\boxed{\sum_{k=1}^{\infty}k^{-p}\text{ converges exactly when }p>1.}
$$
This dyadic comparison is the mechanism of the <Cauchy condensation test>. In particular $p=1$ is the divergent <harmonic series>.
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