Suppose it holds: $P(|S_n|> \varepsilon )\leq \frac{1}{2^n \varepsilon^2} $, where $S_n $ is sequence of random variables.
Furthermore it holds: $\sum_{n \geq 1} P[|S_n|> \varepsilon ] <\infty $
An application of the Borel-Cantelli lemma implies that: $P(\lim\limits_{n \rightarrow \infty} |S_n|> \varepsilon )=0$ . Why does this imply $\lim\limits_{n \rightarrow \infty} S_n=0$ a.s., since I only have $P(\lim\limits_{n \rightarrow \infty} |S_n| \leq \varepsilon )=1$
Best Answer
$\sum P(|S_n| >\frac 1k )<\infty$ for each $k$ . By Borel-Cantelli Lemma $|S_n| \leq \frac 1 k$ for $n$ sufficiently large (or $\lim \sup_n |S_n| \leq \frac 1 k$ ) with probability $1$ (for each $k$) so $S_n \to 0$ almost surely.