Let $H_{n}$ denote the outcome of the $n$-th trial of flipping a coin. Suppose $P(H_{n}=0)=\sin\left(\frac{1}{n}\right)$ for $n = 1,2, \cdots$. I need to determine the probability of observing infinitely many heads.
Now, $\sum_{n=1}^{\infty} P(H_{n}=0) = \sum_{n=1}^{\infty}\sin \left( \frac{1}{n}\right)$.
I was able to show by limit comparison with the series $\sum_{n=1}^{\infty}\frac{1}{n}$ that $\sum_{n=1}^{\infty}\sin \left( \frac{1}{n}\right) = \infty$.
And since each coin toss is independent, I can apply the Second Borel-Cantelli Lemma, which states as follows:
For a sequence of mutually independent events $A_{1},A_{2},\cdots, $ $$P\left(\limsup_{n \to \infty}A_{n} \right) = 1, \, \text{whenever} \, \sum_{i=1}^{\infty}P(A_{i})=\infty.$$
So, let's denote $A_{n}:=\,\{\text{event where}\,H_{n}=0\}$.
Then, what I think the Lemma is saying is that the probability of $H_{n}=0$ is $1$. And since $H_{n}=0$ is the probability of zero heads (I think), then the probability of infinitely many heads would be $1-P(A_{n}) = 0$. Is this correct?
I'm a little confused about this...
The lemma applied to $H_n=0$ states that for almost all coins, a tail occurs infinitely often.
Note that $P[H_n = 1] = 1-\sin {1 \over n}$ and $\sum_n P[H_n = 1] = \infty$, so that for almost all coins, a head will occur infinitely often as well.