The answer is given in Theorem 5 below. This post is essentially a reiteration of [1]'s first section ("Generalization of the Borel-Cantelli lemma").
Let $(\Omega, \mathcal{F}, P)$ be a probability space. Let $A_1, A_2, \dots \in \mathcal{F}$ be an infinite sequence of positive events (i.e. $P(A_n) > 0$ for every $n \in \{1, 2, \dots\}$). Note that this implies, in particular, that for every $k \in \{1, 2, \dots\}$, $E\big(\mathbb{1}_{A_k}\big) > 0$. For every $n \in \{1, 2, \dots\}$ define $X_n:\Omega\rightarrow\mathbb{R}$ to be the random variable $X_n := \sum_{k = 1}^n \mathbb{1}_{A_k}$.
Lemma 1
For every $n \in \{1, 2, \dots\}$, $ \frac{\sum_{i=1}^n\sum_{j=1}^nP(A_i\cap A_j)}{\big(\sum_{k=1}^nP(A_k)\big)^2} = \frac{E\big(X_n^2\big)}{E^2(X_n)}. $
Proof
Let $n \in \{1, 2, \dots\}$. Then $ \begin{align} \sum_{k = 1}^nP(A_i) &= \sum_{k = 1}^nE\big(\mathbb{1}_{A_i}\big) \\ &= E\bigg(\sum_{i = 1}^n\mathbb{1}_{A_i}\bigg) \\ &= E(X_n) \\ \sum_{i = 1}^n\sum_{j = 1}^nP(A_i\cap A_j) &= \sum_{i = 1}^n\sum_{j = 1}^nE\big(\mathbb{1}_{A_i\cap A_j}\big) \\ &= \sum_{i = 1}^n\sum_{j = 1}^nE\big(\mathbb{1}_{A_i}\mathbb{1}_{A_j}\big) \\ &= E\Bigg(\sum_{i = 1}^n\sum_{j = 1}^n\mathbb{1}_{A_i}\mathbb{1}_{A_j}\Bigg) \\ &= E\Bigg(\Bigg(\sum_{k = 1}^n\mathbb{1}_{A_k}\Bigg)^2\Bigg) \\ &= E\big(X_n^2\big) \end{align} $
Q.E.D.
Lemma 2
For every $n \in \{1, 2, \dots\}$, $ \frac{E\big(X_n^2\big)}{E^2(X_n)} = 1 + \frac{V(X_n)}{E^2(X_n)}. $
Proof
Let $n \in \{1, 2, \dots\}$. The conclusion is a consequence of the fundamental variance identity $V(X_n) = E\big(X_n^2\big) - E^2(X_n)$.
Q.E.D.
Lemma 3
If $\sum_{n = 1}^\infty P(A_n) = \infty$ and $ \underset{n\rightarrow\infty}{\underline{\lim}}\frac{V(X_n)}{E^2(X_n)} = 0, $ then $P\bigg(\underset{n\rightarrow\infty}{\overline{\lim}}A_n\bigg) = 1$.
Proof
Choose an increasing sequence $n_1 < n_2 < \dots$ in $\{1, 2, \dots\}$ such that $ \sum_{k=1}^\infty \frac{V\big(X_{n_k}\big)}{E^2\big(X_{n_k}\big)} < \infty. $ Then for every $\varepsilon \in (0,1)$, $ \begin{align} \sum_{k = 1}^\infty P\Big((1-\varepsilon)E\big(X_{n_k}\big) \geq X_{n_k}\Big) &= \sum_{k = 1}^\infty P\big(E\big(X_{n_k}\big) - X_{n_k} \geq \varepsilon E\big(X_{n_k}\big)\big) \\ &\leq \sum_{k = 1}^\infty P\Big(\big|X_{n_k} - E\big(X_{n_k}\big)\big| \geq \varepsilon E\big(X_{n_k}\big)\Big) \\ &\overset{\text{Chebyshev}}{\leq} \sum_{k = 1}^\infty \frac{V\big(X_{n_k}\big)}{\varepsilon^2\ E^2\big(X_{n_k}\big)} \\ &< \infty. \end{align} $ Hence by the Borel-Cantelli lemma $ P\bigg(X_{n_k} > (1-\varepsilon)E\big(X_{n_k}\big)\ \text{eventually}\bigg) = 1. \tag{*}\label{star} $ Since $ \begin{align} \sup_{n \in \{1, 2, \dots\}}E\big(X_n\big) &= \sup_{n \in \{1, 2, \dots\}}E\bigg(\sum_{k = 1}^n\mathbb{1}_{A_k}\bigg)\\ &= \sup_{n \in \{1, 2, \dots\}} \sum_{k = 1}^nE\big(\mathbb{1}_{A_k}\big) \\ &= \sup_{n \in \{1, 2, \dots\}} \sum_{k = 1}^nP(A_k) \\ &= \sum_{k = 1}^\infty P(A_k) \\ &= \infty, \end{align} $ and since $X_1 \leq X_2 \leq \cdots$, we have $\sup_{k \in \{1, 2, \dots\}}E\big(X_{n_k}\big) = \infty$. Therefore \eqref{star} implies that $ P\bigg(\sup_{k\in\{1, 2, \dots\}} X_{n_k} = \infty\bigg) = 1. $ It remains to observe that $ \bigg\{\sup_{k\in\{1, 2, \dots\}} X_{n_k} = \infty\bigg\} = \underset{k\rightarrow\infty}{\overline{\lim}}A_{n_k} \subseteq \underset{n\rightarrow\infty}{\overline{\lim}}A_n. $
Q.E.D.
Corollary 4
For every $n \in \{1, 2, \dots\}$, $ \frac{\sum_{i=1}^n\sum_{j=1}^nP(A_i\cap A_j)}{\big(\sum_{k=1}^nP(A_k)\big)^2} \geq 1. $
If $\sum_{n = 1}^\infty P(A_n) = \infty$ and $ \underset{n\rightarrow\infty}{\underline{\lim}} \frac{\sum_{i=1}^n\sum_{j=1}^nP(A_i\cap A_j)}{\big(\sum_{k=1}^nP(A_k)\big)^2} = 1, $ then $P\Big(\underset{n\rightarrow\infty}{\overline{\lim}}A_n\Big) = 1$.
Proof
By Lemma 1 and Lemma 2.
By Lemma 1, Lemma 2, and Lemma 3.
Q.E.D.
Theorem 5
If $\sum_{n = 1}^\infty P(A_n) = \infty$ and if the $A_n$'s are negatively correlated in the sense that for every $i, j \in \{1, 2, \dots\}$ such that $i\neq j$, $P(A_i\cap A_j) \leq P(A_i)P(A_j)$, then $P\Big(\underset{n\rightarrow\infty}{\overline{\lim}}A_n\Big) = 1$.
Proof
By Corollary 4 it suffices to show that $ \underset{n\rightarrow\infty}{\underline{\lim}} \frac{\sum_{i=1}^n\sum_{j=1}^nP(A_i\cap A_j)}{\big(\sum_{k=1}^nP(A_k)\big)^2} \leq 1. $ This is a consequence of the following observations. $ \begin{align} \frac{\sum_{i=1}^n\sum_{j=1}^nP(A_i\cap A_j)}{\big(\sum_{k=1}^nP(A_k)\big)^2} &\leq \frac{\sum_{\substack{i,j\in\{1,\dots,n\}\\i\neq j}}P(A_i)P(A_j) + \sum_{k=1}^nP(A_k)}{\big(\sum_{k=1}^nP(A_k)\big)^2} \\ &= \frac{\big(\sum_{k = 1}^nP(A_k)\big)^2}{\big(\sum_{k=1}^nP(A_k)\big)^2} + \frac{\sum_{k = 1}^nP(A_k)\big(1 - P(A_k)\big)}{\big(\sum_{k=1}^nP(A_k)\big)^2} \\ &\leq 1 + \frac{\sum_{k = 1}^nP(A_k)}{\big(\sum_{k=1}^nP(A_k)\big)^2} \\ &= 1 + \frac{1}{\sum_{k=1}^nP(A_k)}. \end{align} $
Q.E.D.
Works cited
[1] P. Erdős and A. Rényi. On Cantor's Series with Convergent $\sum\frac{1}{q_n}$. Mathematical Institute, Eötvös Loránd University, Budapest. Received 6 April, 1959.