Diagonal terms in the Kochen Stone inequality

Question

In a paper 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.

Answer 1

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).