Equality about limsup. Suppose $\sum_{n=1}^\infty \mathbb P(A_n)=\infty$,then:
$$\limsup_{n\to\infty}\frac{(\sum_{k=1}^n \mathbb P(A_k))^2}{\sum_{i,k=1}^n\mathbb P(A_i\cap A_k)}=\limsup_{n\to\infty}\frac{\sum_{1\le i<k\le n}\mathbb P(A_i)\mathbb P(A_k)}{\sum_{1\le i<k\le n}\mathbb P(A_i\cap A_k)}$$
I have tried:
$$(\sum_{k=1}^n \mathbb P(A_k))^2=2\sum_{1\le i<k\le n}\mathbb P(A_i)\mathbb P(A_k)+\sum_{k=1}^n \mathbb P^2(A_k)\le 2\sum_{1\le i<k\le n}\mathbb P(A_i)\mathbb P(A_k)+\sum_{k=1}^n \mathbb P(A_k)$$
$$\sum_{k=1}^n \mathbb P(A_k)\le  \frac{2\sum_{1\le i<k\le n}\mathbb P(A_i)\mathbb P(A_k)}{\sum_{k=1}^n \mathbb P(A_k)}+1$$
Then:
$$\lim_{n\to\infty}\frac{\sum_{k=1}^n \mathbb P(A_k)}{\sum_{1\le i<k\le n}\mathbb P(A_i)\mathbb P(A_k)}=0$$
How to do next?

It is a part of Kochen-Stone lemma :
Suppose $\sum_{n=1}^\infty \mathbb P(A_n)=\infty$,then:
$$\mathbb P(A_n ,\mathrm{i.o.})\ge\limsup_{n\to\infty}\frac{(\sum_{k=1}^n \mathbb P(A_k))^2}{\sum_{i,k=1}^n\mathbb P(A_i\cap A_k)}=\limsup_{n\to\infty}\frac{\sum_{1\le i<k\le n}\mathbb P(A_i)\mathbb P(A_k)}{\sum_{1\le i<k\le n}\mathbb P(A_i\cap A_k)}$$
the first inequality is proved,now I want to prove the second equality.
 A: You can rewrite the right side as
\begin{equation*}
\frac{\sum_{i<j}\mathbb{P}(A_{i})\mathbb{P}(A_{j})}{\sum_{i<j}\mathbb{P}(A_{i}\cap A_{j})}= 
\frac{\sum_{i\neq j}\mathbb{P}(A_{i})\mathbb{P}(A_{j})}{\sum_{i\neq j}\mathbb{P}(A_{i}\cap A_{j})}=
\frac{(\sum_{i}\mathbb{P}(A_{i}))^{2}-\sum_{i}\mathbb{P}(A_{i})^{2}}{\sum_{i,j}\mathbb{P}(A_{i}\cap A_{j})-\sum_{i}\mathbb{P}(A_{i})}
\end{equation*}
One one hand $\sum_{i}\mathbb{P}(A_{i})^{2} \leqslant \sum_{i}\mathbb{P}(A_{i})= o((\sum_{i}\mathbb{P}(A_{i}))^{2})$ since $\sum_{i=1}^{n}\mathbb{P}(A_{i}) \rightarrow 
+\infty$. This is what you already showed.
I suppose the real difficulty is showing that $\sum_{i}\mathbb{P}(A_{i})=o(\sum_{i,j}\mathbb{P}(A_{i}\cap A_{j}))$. (If we can do so then we can take away the negliglible terms without changing the limsup and get the desired equality).
I don't know how to do so from nothing, but I can deduce it from the inequality $\displaystyle\mathbb{P}(\limsup A_{n}) \geqslant \limsup \frac{(\sum_{i}\mathbb{P}(A_{i}))^{2}}{\sum_{i,j}\mathbb{P}(A_{i}\cap A_{j})}$ you gave.
Indeed you have $1 \geqslant \mathbb{P}(\limsup A_{n})$ so that  $\sum_{i,j}\mathbb{P}(A_{i}\cap A_{j})\geqslant 2 (\sum_{i}\mathbb{P}(A_{i}))^{2}$ for $n$ large enough and $\sum_{i}\mathbb{P}(A_{i})=o((\sum_{i}\mathbb{P}(A_{i}))^{2})=o(\sum_{i,j}\mathbb{P}(A_{i}\cap A_{j}))$
