= Solution
The <arithmetic-geometric mean inequality> gives
$$
\sqrt{a_na_{n+1}}\leq\frac{a_n+a_{n+1}}2.
$$
Thus convergence of $\sum_na_n$ implies convergence of $\sum_n\sqrt{a_na_{n+1}}$ by the <comparison test for series>.
The converse is false. Define
$$
a_{2n}=\frac1n,
\qquad
a_{2n-1}=\frac1{n^3}.
$$
The even terms make $\sum_na_n$ diverge, while
$$
\sqrt{a_{2n-1}a_{2n}}=\frac1{n^2},
\qquad
\sqrt{a_{2n}a_{2n+1}}=\frac1{\sqrt{n(n+1)^3}}\leq\frac1{n^2},
$$
so the neighboring geometric-mean series converges.
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