Suppose $\sum P(A_k)=\infty$. Show that, if
$$\limsup_{n\to\infty}\Big(\sum_{k=1}^n P(A_k)\Big)^2\Big/\Big(\sum_{1\leq j,k\leq n} P(A_j\cap A_k)\Big)=\alpha>0$$
then $P(A_n \space\text{i.o.})\geq \alpha$
I know, $P(A_n \space\text{i.o.})=P(\limsup A_n)\geq \limsup P(A_n)$
Then if I show that $P(A_n)\geq \Big(\sum_{k=1}^n P(A_k)\Big)^2\Big/\Big(\sum_{1\leq j,k\leq n} P(A_j\cap A_k)\Big)$ for all $n$, it is done.
I know that I need to use this theorem: $Y\geq 0$ with $E\space Y^2<\infty$. Then, $P(Y>0)\geq \dfrac{(E\space Y)^2}{E\space Y^2}$ . But I do not understand how can I use that.
I took $N_n=\sum_{k=1}^{n}1_{A_k}$. Then, $E(N_n)=\sum_{k=1}^n P(A_k)$ and $E(N^2_n)=E(N_n)+\sum_{1\leq j,k\leq n} P(A_j\cap A_k)$.
I am stuck after that. Thanks for any hint or help.
Also I saw this solution, but could not understand.
After searching a lot, I found a similar problem, which states:
Suppose $\sum P(A_k)=\infty$. Show that, if
$$\limsup_{n\to\infty}\Big(\sum_{k=m+1}^n P(A_k)\Big)^2\Big/\Big(\sum_{m+1\leq j,k\leq n} P(A_j\cap A_k)\Big)=\alpha>0$$
then $P(A_n \space\text{i.o.})\geq \alpha$ for large $m$.
Now, I solved this one, in this way:
$N_n=\sum_{k=m+1}^{n}1_{A_k}$. Then, $E(N_n)=\sum_{k=m+1}^n P(A_k)$ and $E(N^2_n)=E(N_n)+\sum_{m+1\leq j,k\leq n} P(A_j\cap A_k)$
Since, $P(N_n>0)=P(\bigcup_{k=m+1}^{n}A_k)$ and, $P(N_n>0)\geq \dfrac{(E\space N_n)^2}{E\space N_n^2}$, taking limit over $n$, we get,
$$\lim_{n\to\infty}P(\bigcup_{k=m+1}^{n}A_k)=\frac{1}{\dfrac{1}{\sum_{k=m+1}^{\infty}P(A_k)}+\lim_{n\to\infty}\Big(\sum_{m+1\leq j,k\leq n} P(A_j\cap A_k)\Big)\big/\Big(\sum_{k=m+1}^n P(A_k)\Big)^2}=\lim_{n\to\infty}\Big(\sum_{k=m+1}^n P(A_k)\Big)^2\Big/\Big(\sum_{m+1\leq j,k\leq n} P(A_j\cap A_k)\Big)\space\space\big(\text{since $\sum P(A_k)=\infty$}\big)$$
Taking limit over $m$, we get:
$$\lim_{m\to\infty}\lim_{n\to\infty}P(\bigcup_{k=m+1}^{n}A_k)=\lim_{m\to\infty}\lim_{n\to\infty}\Big(\sum_{k=m+1}^n P(A_k)\Big)^2\Big/\Big(\sum_{m+1\leq j,k\leq n} P(A_j\cap A_k)\Big)=\limsup \Big(\sum_{k=m+1}^n P(A_k)\Big)^2\Big/\Big(\sum_{m+1\leq j,k\leq n} P(A_j\cap A_k)\Big)=\alpha$$
Again, since $P(\limsup A_n)\geq \limsup P(A_n)=\lim_{m\to\infty}\lim_{n\to\infty}P(\bigcup_{k=m+1}^{n}A_k)$. We finally have, $P(A_n \space\text{i.o.})\geq \alpha$, $\color{green}{\text{for large}\space m}$.
Is this okay? How can I argue the first problem from this problem? Thanks.