= Solution
The first <Borel-Cantelli lemma> says that $\sum_n\Pr(A_n)<\infty$ implies $\Pr(A_n\text{ infinitely often})=0$, without an <independence> assumption. Indeed, for every $N$,
$$
\Pr\left(\bigcup_{n\ge N}A_n\right)\le\sum_{n\ge N}\Pr(A_n)\longrightarrow0.
$$
The decreasing intersection of these tail unions is the infinitely-often event; continuity of <probability> from above proves the assertion. The second lemma says that for independent $A_n$, divergence of the <probability> sum implies $\Pr(A_n\text{ infinitely often})=1$.
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