I am working on this question: $P(X_n=n^{\alpha}）=\frac{1}{n}$, $P(X_n=0)=1-\frac{1}{n}$, for what values of $\alpha$ such that $X_n$ converges almost surely to $0$?
Here is what I think: $X_n$ converges almost surely to $0$ is equivalent to prove $P(|X_n|>\epsilon\, \text{ i.o.})=0$ for any $\epsilon>0$; and by B-C lemma, this is equivalent to $\sum P(|X_n|>\epsilon)<\infty$, for any $\epsilon>0$, and I am confusing about figuring out the $P(|X_n|>\epsilon)$, how can I express the probability here?
Fix $\varepsilon>0$. Note that for $\alpha\geq 0$, $\sum P(|X_n|>\varepsilon)=\sum_{n=m}^\infty 1/n=\infty$ where $m$ is chosen so that, $n^\alpha>\epsilon$ for $n\geq m$. For $\alpha<0$, $\sum P(|X_n|>\varepsilon)<\infty$ since $n^{\alpha}<\epsilon$ for sufficiently large $n$.
The Borel Cantelli lemma implies that $X_{n}\to 0$ a.s. iff $\alpha<0$.