Maybe you can look at the proof of BC's lemma with the previous examples in mind.
"For any n" => intersection
"there exits" => union
and you can use the equality
for independent events, and the inequality
Just an example. Translating into symbols, I suggested to first show that, for any
. In order to use independence, I must handle intersections, hence let's write
. And we have
(If you aren't used to applying independence to infinitely many events at once, you can say
. What about
? Switching to the complement:
(using what we got before). In conclusion,
. This is what you need.
You can remember this (this was the last part): a countable intersection of almost sure events is almost sure.