= Solution
The two-variable <arithmetic-geometric mean inequality> follows from a square:
$$
0\leq(\sqrt a-\sqrt b)^2=a+b-2\sqrt{ab}.
$$
Hence \b[$2\sqrt{ab}\leq a+b$], with equality exactly when $a=b$.
For the <series>, apply that same inequality with $a=a_n$ and $b=n^{-2}$ to obtain
$$
0\leq\frac{\sqrt{a_n}}n\leq\frac12\left(a_n+\frac1{n^2}\right).
$$
The comparison <series> converges: its first summand has finite sum by hypothesis, and the inverse-square <series> converges, for example by the telescoping estimate $n^{-2}\leq1/[n(n-1)]=1/(n-1)-1/n$ for $n\geq2$. Thus the <comparison test for series> proves
$$
\boxed{\sum_{n=1}^\infty\frac{\sqrt{a_n}}n<\infty.}
$$
Equivalently, its positive partial sums are increasing and bounded above by $\tfrac12\sum a_n+\tfrac12\sum n^{-2}$, so the <bounded monotone sequence theorem> proves convergence directly. This is the case $p=1$ of <weighted square roots of a summable sequence>.
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