When is the Kochen-Stone inequality an equality? The Kochen-Stone theorem says that if $A_n$ is sequence of events with $\sum_{i=1}^{\infty} P(A_i) = \infty$, then:
$$
P(A_n \mbox{ i.o.}) \ge \limsup_{n \to \infty} \frac{(\sum_{i=1}^nP(A_i))^2}{\sum_{i, j= 1}^nP(A_i \cap A_j)}
$$
I am interested in cases where the $A_n$ are not mutually independent, but where the inequality is actually an equality. Any hints or references to results or examples of this kind would be much appreciated.
 A: In case it's of interest to others, I can now show that if $A_n$ is any increasing sequence of events then the Kochen-Stone inequality is an equality. This was enough for my application (I am looking at issues of computability and wanted to arrange for equality when the probabilities $P(A_n)$ form a Specker sequence).
To see why the Kochen-Stone inequality is an equality for increasing $A_n$, first note that in A Simple Proof of Two Generalized Borel-Cantelli Lemmas (Springer Lecture Notes in Mathematics vol. 1874), Yan showed that the diagonal terms in the sums in the Kochen-Stone inequality are negligible:
\begin{equation} \label{yan-eq}
\limsup_{n \to \infty} \frac{(\sum_{k=1}^n P[A_k])^2}{\sum_{i,k=1}^n P[A_i A_k]} = \limsup_{n \to \infty} \frac{\sum_{1 \le i < k \le n} P[A_i]P[A_k]}{\sum_{1 \le i < k \le n}P[A_iA_k]} \tag{1}
\end{equation}
Now assume $A_1 \subseteq A_2 \subseteq A_3 \subseteq \ldots$ and assume, for simplicity, that $P(A_1) > 0$.  Put $q_n = P(A_n)$. Then $P(A_n \mbox{ i.o.}) = q$ where $q = \lim_{n \to \infty}q_n$. Noting that $P(A_iA_k) = P(A_i)$ when $i < k$, we can write  the fraction on the right-hand side of equation (1) as $w_n = \frac{u_n}{v_n}$ where:
\begin{align}
    u_n &= q_1\sum_{k= 2}^n q_k + q_2\sum_{k=3}^n q_k + \ldots + q_{n-1}q_n\\
    v_n &= (n-1)q_1 + (n-2)q_2 + \ldots + q_{n-1}\\
\end{align}
and then the Kochen-Stone inequality is $q \ge \limsup_{n \to \infty} w_n$. I claim that, in fact, $w_n \to q$ as $n \to \infty$, so that $q = \lim_{n \to \infty} w_n = \limsup_{n \to \infty} w_n$. To see this, note that:
\begin{align}
    \sum_{k=i}^nq_k &= q\left(n - i + 1 - \frac{1}{q}\sum_{k=i}^n (q - q_k) \right)
\end{align}
Let us write $\sigma_i^j$ for $\sum_{k=i}^j(q - q_k)$. From the above, multiplying by $q_{i-1}$ and summing for $i$ from $2$ to $n$, we have:
\begin{equation}
    u_n = qv_n - \sum_{i=2}^{n}  q_{i-1}\sigma_i^n
\end{equation}
Define the sequence $r_n$ by:
\begin{align}
    r_n &= \frac{\sum_{i=2}^{n}  q_{i-1}\sigma_i^n}{v_n} \tag{2}
    % %
% &= \frac{q_1\left(\sum_{k=2}^{n}(q - q_k)\right) + q_2\left(\sum_{k=3}^{n}(q - q_k)\right) + \ldots + q_{n-1}(q - q_{n-1})}
    % {q_1 (n-1) + q_2 (n-2) + \ldots + q_n}    
\end{align}
I claim that $r_n \to 0$ as $n \to \infty$ so that $w_n = u_n/v_n = q - r_n \to q$, which is what we want to prove. To see this, given $\epsilon > 0$, put $\epsilon_0 = \frac{q_1}{4}\epsilon$. Then there is an $N$ such that for all $n > N$, $q - q_n < \epsilon_0$. Then for $n > N$, we have:
\begin{equation}
r_n = \frac{C}{v_n} + \frac{s_n}{v_n}
\end{equation}
where $C$ and $s_n$ are obtained by grouping the monomials $q_iq_j$ in the numerator of the fraction on  the right-hand side of equation (2) so that $C$ comprises the $q_iq_j$ for which $i$ and $j$ are most $N$ and $s_n$ contains all the rest. Observing that $s_n$ contains at most $(n-1)(n - N)$ monomials all of which evaluate to at most $\epsilon_0$ and that $v_n > q_1((n-1) + (n - 2) + \ldots + 1)= q_1\frac{1}{2}(n-1)n$, we have:
\begin{equation}
r _n \le \frac{C}{v_n} + \frac{1}{q_1}\cdot\frac{(n-1)(n-N)\epsilon_0}{\frac{1}{2}(n-1)n} \to  \frac{2}{q_1}\epsilon_0 = \frac{\epsilon}{2}\mbox{ as $n \to \infty$}
\end{equation}
So there is an $M > N$ such that for all $n > M$, $|r_n - \frac{\epsilon}{2}| < \frac{\epsilon}{2}$, but then $r_n < \epsilon$ and we are done.
