In a [paper](https://www.researchgate.net/publication/226613874_A_Simple_Proof_of_Two_Generalized_Borel-Cantelli_Lemmas) in *Lecture Notes in Mathematics vol. 1874*, Yan states the Kochen-Stone theorem in the following form, where $A\_n$ is a sequence of events such that $\sum\_{n=1}^\infty P(A\_n) = \infty$:


$$ 
P(A\_n~\mbox{i.o.})
 \geq \limsup\_{n \to \infty} \frac{(\sum\_{k=1}^n P(A\_k))^2}{\sum\_{i,k=1}^n P(A\_i A\_k)}
= \limsup\_{n \to \infty} \frac{\sum\_{1 \le i < k \le n} P(A\_i)P(A\_k)}{\sum\_{1 \le i < k \le n}P(A\_iA\_k)} \tag\*{$(2)$}
$$


((2) is his numbering).


I understand his proof of the inequality in (2), but I can't make sense of his proof of the equation (which states that the diagonal terms in the sums in the first fraction in (2) are negligible).
He says *"Since $\sum\_{k=1}^\infty P(A\_k) = \infty$ and*
$$
\left(\sum\_{k=1}^n P(A\_k)\right)^2 \le 2\sum\_{1 \le i < k \le n}P(A\_i)P(A\_k) + \sum\_{k=1}^n P(A\_n) \tag\*{$(3)$}
$$


*we have*
$$
\lim\_{n \to \infty}\frac{\sum\_{k=1}^n P(A\_k)}{\sum\_{1 \le i < k \le n}P(A\_i)P(A\_k)} = 0 \tag\*{$(4)$}
$$
*Thus equality (2) holds."* ((3) and (4) are my numbering.)


I see how to get from (3) to (4) (by dividing by $\sum\_{k=1}^n P(A\_k)$ and noting that the left-hand side of the resulting inequality diverges), but I can't see how to get the equality in (2) from (4). Any suggestions on how to see this or any alternative proofs will be gratefully received as I am doing some work that relies heavily on the equation and am reluctant to use a result that I don't understand. Any relevant references would also be much appreciated.



Diagonal terms in the Kochen Stone inequality
Let
$$S\_n=\sum\_{k=1}^n P(A\_k),\quad T\_n:=\sum\_{1\le i<k\le n}P(A\_i)P(A\_k),$$
$$R\_n=\sum\_{i,k=1}^n P(A\_iA\_k),\quad U\_n:=\sum\_{1\le i<k\le n}P(A\_iA\_k).$$
Then $2T\_n\le S\_n^2\le2T\_n+S\_n$, and $S\_n<<S\_n^2$ (because $S\_n\to\infty$); we write $a<<b$ or, equivalently, $b>>a$ to mean $a=o(b)$; all the limits are taken as $n\to\infty$. Thus,
$$S\_n^2\sim2T\_n.\tag{\*}$$
Also, $2U\_n\le R\_n\le2U\_n+S\_n$, $S\_n<<S\_n^2$, and $S\_n^2\le (1+o(1))R\_n$ (by the inequality in (2)). Thus,
$$R\_n\sim2U\_n.\tag{\*\*}$$
From (\*) and (\*\*), we get
$$\frac{S\_n^2}{R\_n}\sim\frac{T\_n}{U\_n},$$
which immediately yields the equality in (2).


