[proofplan] We bound $\mathbb{P}(\limsup A_n)$ using countable subadditivity by the tail sum $\sum_{m=n}^\infty \mathbb{P}(A_m)$, and then observe that the convergence of the series forces its tails to zero. No independence hypothesis is needed. [/proofplan] [step:Bound $\mathbb{P}(\limsup A_n)$ by the tail sum using subadditivity] For each $n \in \mathbb{N}$, $\limsup_{k} A_k \subseteq \bigcup_{m \geq n} A_m$, so by countable subadditivity of $\mathbb{P}$, \begin{align*} \mathbb{P}(\limsup_k A_k) \leq \mathbb{P}\left(\bigcup_{m \geq n} A_m\right) \leq \sum_{m=n}^\infty \mathbb{P}(A_m). \end{align*} [/step] [step:Send $n \to \infty$ and use convergence of the series to conclude] Since $\sum_{n=1}^\infty \mathbb{P}(A_n) < \infty$, the tail sum $\sum_{m=n}^\infty \mathbb{P}(A_m) \to 0$ as $n \to \infty$. The left-hand side $\mathbb{P}(\limsup_k A_k)$ is a fixed non-negative number bounded above by every tail sum, so $\mathbb{P}(\limsup_k A_k) = 0$. [/step]