Prove that $P(\bigcup_{i=1}^{\infty} A_i) = 1$ $A_i$ $(i=1,2,...)$ are independent events
$\sum_{i=1}^{\infty}P(A_i) = \infty.$ 
Prove that: 
$P(\bigcup_{i=1}^{\infty} A_i) = 1 $
Can someone please help me out with this question?
 A: I believe this is related to the second Borel-Cantelli lemma. The proof goes as follows:
\begin{align}P\left( \bigcup_{i=1}^{\infty} A_i \right) &\geq P\left( \bigcup_{i=n}^{\infty} A_i \right)\\
&=\lim_{k \to \infty} P\left( \bigcup_{i=n}^{k} A_i \right)\\
&= 1-\lim_{k \to \infty} P\left( \bigcap_{i=n}^{k} A_i^C \right) \\
&= 1-\lim_{k \to \infty}\prod_{i=n}^k \left(1-P(A_i)\right)\\
&\geq 1-\lim_{k \to \infty} \exp\left(-\sum_{i=n}^k P(A_i)\right)\\
&= 1\\
\end{align}
Where the second-to-last inequality follows from the fact that $$-\ln(1-P(A_i))=\sum_{k=1}^{\infty}\frac{[P(A_i)]^k}{k}\geq P(A_i)$$
A: By virtue of the non-trivial part of Borel-Cantelli lemma, one is allowed to conclude that the event $\displaystyle \mathrm{A} = \{A_n, \mathrm{i.o.}\} = \bigcap_{k = 1}^\infty \bigcup_{n = k}^\infty A_n$ has total mass equal to one. Notice that $\mathrm{A}$ lies within the union of the $A_n$ and the exercise follows.
A: TL;DR BCL2 tells us $A_n$ occurs infinitely often. So obviously, at least one $A_n$ occurs.

Pf:
$$\sum_{i=1}^{\infty}P(A_i) = \infty$$
and
$A_n$ indp
By BCL2, we have $P(\limsup A_n) = 1$
Note that $$\limsup A_n \subseteq \bigcup_n A_n \tag{*}$$
QED

Intuition:
$(*)$ should be not so difficult to prove and should be easily understood intuitively:
In the sequence
$$A_1, A_2, A_3, A_4, ...$$
we deduce by BCL2 that for any day $m$, (let's say $A_m$ corresponds to day $m$), there is some future day $n \ge m$ that will occur eg on day 1, there is some day $n \ge 1$ s.t. $A_n$ will occur. On day $m=100$, there is some $n \ge 100$ s.t. $A_n$ will occur. We say that $A_n$ occurs infinitely often.
So if $A_n$ occurs infinitely often, then obviously at least one of those $A_n$'s will occur i.e. $$P(\bigcup_{n=1}^{\infty} A_n) = 1$$
A: If $P\left( \bigcup_{n=1}^\infty A_n \right) < 1,$ then $P\left( \bigcap_{n=1}^\infty (\text{not } A_n) \right)>0.$ By independence, we have
$$
\prod_{n=1}^\infty P(\text{not } A_n) > 0, \text{ so } \log\prod_{n=1}^\infty (1-P(A_n))>-\infty.
$$
$$
\sum_{n=1}^\infty -\log (1-P(A_n)) <\infty.
$$
If you can show that when $0\le p<1$ then $p\le -\log(1-p),$ then you get
$$
\sum_{n=1}^\infty P(A_n) < \infty.
$$
The inequality $p\le -\log(1-p)$ depends on the base of the logarithm being $\le e$ but $>1;$ however, this is not essential, because for any base $>1$ one can find a suitable positive constant and say $p\le -(\text{constant}\cdot\log(1-p)).$
Observe that if $\sum_{n=1}^\infty p_n = \infty$ then $\sum_{n=\mathbf N}^\infty p_n=\infty$ no matter how be $\mathbf N$ gets. Thus after any such index $\mathbf N$, one can find some $n$ for which $A_n$ occurs. Therefore, infinitely many of them occur. Consequently this whole argument amounts to a proof of one of the Borel–Cantelli lemmas.
