This is a typical problem in probability theory, including Ph.D Qualifying Exam.
Suppose $\{A_n,n\in \mathbb{N}\}$ is an infinite family if independent events with $\mathbf{P}(A_n)<1$. Show that $\mathbf{P}(\cup_{n\geq 1}A_n)=1$ iff $\mathbf{P}(A_n\text{ i.o.})=1$.
My attempt: ($\Leftarrow$) part is easy, since $(A_n\text{ i.o.})\subset \cup_{n\geq 1}A_n.$
For the converse, I found a proof using the fact that $\cup_{n=1}^N A_n$ and $\cup_{n=N+1}^\infty A_n$ are independent for every $N$, and so $\mathbf{P}(\cup_{n=N+1}^\infty A_n)=1$ for every $N$.
However, I want to know how to prove the same statement using Borel-Cantelli lemma.
From the assumption, $\mathbf{P}(\cap_{n\geq 1}A_n^c)=0$, which implies $\prod_{n\geq 1}(1-\mathbf{P}(A_n))=0.$ But how can I derive $\sum_{n\geq 1}\mathbf{P}(A_n)=\infty$ from this fact? Does anyone have ideas?
Thanks in advance!