= Solution
Another application of the <Cauchy-Schwarz inequality> gives
$$
\sum_{n=1}^N\sqrt{a_n}\,n^{-p}
\leq
\left(\sum_{n=1}^Na_n\right)^{1/2}
\left(\sum_{n=1}^Nn^{-2p}\right)^{1/2}.
$$
When $p>1/2$, the second factor is bounded because $2p>1$, so the <series> converges.
At the endpoint, take $a_1=1$ and, for $n\geq2$,
$$
a_n=\frac1{n(\log n)^2}.
$$
Then $\sum a_n$ converges, while
$$
\sqrt{a_n}\,n^{-1/2}=\frac1{n\log n},
$$
whose <series> diverges. This proves both parts of <weighted square roots of a summable sequence>.
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