Let $(A_n)$ be a sequence of independent events with $\mathbb P(A_n)<1$ and $\mathbb P(\cup_{n=1}^\infty A_n)=1$. Show that $\mathbb P(\limsup A_n)=1$.
It looks like the problem is practically asking to apply the Borel-Cantelli. Yet the suggested solution went differently: via $\prod_{n=1}^\infty \mathbb P( A_n^c)=0$.
How can we apply the Borel-Cantelli lemma here? I.e. how to show that $\sum_{n=1}^\infty \mathbb P( A_n)= \infty$?
What you are asked to show:
If $\mathbb P(A_n)\lt1$ for every $n$ and $\prod\limits_{n=1}^\infty \mathbb P( A_n^c)=0$ then $\sum\limits_n\mathbb P( A_n)$ diverges.
Thus, Borel-Cantelli lemma is not involved in the proof that the series $\sum\limits_n\mathbb P( A_n)$ diverges, which is purely a problem of real analysis. In full generality:
Consider some nonnegative sequence $(x_n)$ such that $x_n\lt1$ for every $n$ and $\prod\limits_{n=1}^\infty (1-x_n)=0$ then the series $\sum\limits_nx_n$ diverges.
Can you think of a simple approach to show this?
If $x_n\geqslant\frac12$ infinitely often then $\sum\limits_nx_n$ diverges. Otherwise, $x_n\leqslant\frac12$ for every $n$ large enough, say, for every $n\geqslant N$, and $\prod\limits_{n=N}^\infty (1-x_n)=0$ (this is where we use that $x_n\ne1$ for every $n$).
For every $x$ in $[0,\frac12]$, $1-x\geqslant\mathrm e^{-cx}$ for some suitable $c$ hence $\prod\limits_{n=N}^\infty (1-x_n)\geqslant\exp\left(-c\sum\limits_{n=N}^\infty x_n\right)$, which shows that $\sum\limits_{n=N}^\infty x_n$ diverges, QED. (Exercise: Find $c$.)
You want to show $P(\cap_{n=1}^{+\infty}\cup_{k=n}^{+\infty}A_k) = 1$, which is equivalent to $P(\cup_{n=1}^{+\infty}\cap_{k=n}^{+\infty}A_k^c) = 0$.
At the same time, we have $$P(\cup_{n=1}^{+\infty}\cap_{k=n}^{+\infty}A_k^c) = \lim_{n\to +\infty } P(\cap_{k=n}^{+\infty}A_k^c) = \lim_{n\to +\infty } \prod_{k=n}^{+\infty}P(A_k^c)$$
Remark that $\prod_{n=1}^{+\infty}P(A_n^c) = 0$ implies $\prod_{k=n}^{+\infty}P(A_k^c) = 0$ for any $n$ (since all $P(A_n^c) > 0$), so the above limit is $0$.
We've proven $P(\cap_{n=1}^{+\infty}\cup_{k=n}^{+\infty}A_k) = 1$, in consequence we have $\sum_{n=1}^{+\infty}P(A_n) = +\infty$
For a direct proof, see the related question
