[proofplan] We show $\mathbb{P}((\limsup_n A_n)^c) = 0$ by proving that for each fixed $n$, the probability of avoiding all $A_m$ with $m \geq n$ is zero. The key tools are: independence (to factor the probability of a finite intersection of complements), the elementary inequality $1 - x \leq e^{-x}$ (to convert the product into an exponential), and divergence of the series (to drive the exponential to zero). Continuity of $\mathbb{P}$ from below then gives the result. [/proofplan] [step:Express the complement of $\limsup A_n$ as an increasing union and reduce to showing each term vanishes] The complement of $\limsup_n A_n$ satisfies \begin{align*} (\limsup_n A_n)^c = \bigcup_{n=1}^\infty \bigcap_{m \geq n} A_m^c. \end{align*} The sets $B_n = \bigcap_{m \geq n} A_m^c$ are increasing in $n$ (intersecting over fewer indices gives a larger set), so by continuity of $\mathbb{P}$ from below, \begin{align*} \mathbb{P}((\limsup_n A_n)^c) = \lim_{n \to \infty} \mathbb{P}(B_n). \end{align*} It suffices to show $\mathbb{P}(B_n) = 0$ for every $n$. [/step] [step:Show $\mathbb{P}\!\left(\bigcap_{m=n}^\infty A_m^c\right) = 0$ using independence and the exponential bound] [claim:Finite Intersection Probability Vanishes] For each fixed $n \in \mathbb{N}$, $\mathbb{P}\!\left(\bigcap_{m=n}^\infty A_m^c\right) = 0$. [/claim] [proof] Fix $n$. For any $N \geq n$, the events $A_n, A_{n+1}, \dots, A_N$ are independent (as a sub-collection of the independent sequence $(A_k)$), so their complements $A_n^c, \dots, A_N^c$ are also independent. Therefore \begin{align*} \mathbb{P}\!\left(\bigcap_{m=n}^N A_m^c\right) = \prod_{m=n}^N \mathbb{P}(A_m^c) = \prod_{m=n}^N (1 - \mathbb{P}(A_m)). \end{align*} [claim:Exponential Bound] For any $x \in [0,1]$, $1 - x \leq e^{-x}$. [/claim] [proof] The function $\varphi: [0,1] \to \mathbb{R}$ defined by $\varphi(x) = e^{-x} - (1 - x)$ satisfies $\varphi(0) = 0$ and $\varphi'(x) = -e^{-x} + 1 \geq 0$ for $x \in [0,1]$ (since $e^{-x} \leq 1$). Hence $\varphi(x) \geq 0$ on $[0,1]$. [/proof] Applying this bound with $x = \mathbb{P}(A_m) \in [0,1]$ for each factor: \begin{align*} \prod_{m=n}^N (1 - \mathbb{P}(A_m)) \leq \prod_{m=n}^N e^{-\mathbb{P}(A_m)} = \exp\!\left(-\sum_{m=n}^N \mathbb{P}(A_m)\right). \end{align*} Since $\sum_{m=1}^\infty \mathbb{P}(A_m) = \infty$, the tail $\sum_{m=n}^N \mathbb{P}(A_m) \to \infty$ as $N \to \infty$. Therefore \begin{align*} \prod_{m=n}^N (1 - \mathbb{P}(A_m)) \leq \exp\!\left(-\sum_{m=n}^N \mathbb{P}(A_m)\right) \to 0 \quad \text{as } N \to \infty. \end{align*} The sets $\bigcap_{m=n}^N A_m^c$ are decreasing in $N$, so by continuity of $\mathbb{P}$ from above, \begin{align*} \mathbb{P}\!\left(\bigcap_{m=n}^\infty A_m^c\right) = \lim_{N \to \infty} \mathbb{P}\!\left(\bigcap_{m=n}^N A_m^c\right) = \lim_{N \to \infty} \prod_{m=n}^N (1 - \mathbb{P}(A_m)) = 0. \end{align*} [/proof] [/step] [step:Conclude $\mathbb{P}(\limsup_n A_n) = 1$] Since $\mathbb{P}(B_n) = 0$ for every $n$, the first step gives \begin{align*} \mathbb{P}((\limsup_n A_n)^c) = \lim_n \mathbb{P}(B_n) = 0, \end{align*} so $\mathbb{P}(\limsup_n A_n) = 1$. [/step]