Let $(X, A, P)$ be a probability space and $(A_n)_{n\in \mathbb N}$ a sequence of events in $A$ such that $P(A_n) = \frac{1}{7^n}$ for $n \in \mathbb N$. Determine the probability of the event $\limsup A_n$.
I've come across this question on a past final for a measure theory course I'm taking. It has me stumped and I'm not even quite sure where to begin. I know there's the Borel-Cantelli lemma which shows that if $\displaystyle\sum_{n=1}^\infty P(A_n)$ converges then $P(\limsup A_n) =P([A_n \text{ i.o.}])=0$. Would this proof be useful when answering my question? Any help appreciated, thanks in advance!
Let me summarize the comments, and add more information. For each $\omega \in X$, let $N(\omega) = \#\{n: \omega \in A_n\}$ (that is, the number of $n$ so that $\omega \in A_n$). Note: $\omega \in \limsup A_n$ if and only if $N(\omega) = \infty$.
We may write the random variable $N$ as $$ N(\omega) = \sum_{n=1}^\infty \mathbf1_{A_n}(\omega) $$ Here $\mathbf1_{A_n}$ is the indicator of the set $A_n$. Compute $$ \mathbb E[N] = \sum_{n=1}^\infty \mathbb E[\mathbf1_{A_n}] =\sum_{n=1}^\infty \mathbb P[A_n] = \sum_{n=1}^\infty\frac{1}{7^n} = \frac{1}{6} . $$ In probability-speak: for a random point $\omega \in X$, the expected number of $n$ so that $\omega \in A_n$ is $1/6$.
A random variable with finite expectation must be finite a.s. So $\mathbb P [N = \infty] = 0$. Equivalently, $\mathbb P [\limsup A_n] = 0$.