If $X_n \to 0$ a.s. then this series converges for all $x > 0$. Let $(X_n)$ be a sequence of independent random variables such that $X_n \to 0$ almost surely. Prove that for every $x > 0$ we have $$\sum_{n=1}^\infty P(|X_n| \geq x) < \infty.$$
If $$\sum_{n=1}^\infty P(|X_n| \geq x) = \infty,$$ then BC gives that $P(|X_n| \geq x \text{ 
for infinitely many n}) = 1$. This implies that $X_n \not \to 0$ a.s., but this is not exactly the negative of the statement "$X_n \to 0$ a.s." Is there a way to fix this?
Edit: Is the converse result also valid btw? I think yes, but can I show this using a similar proof?
 A: Direct part: If $|X_n| >x$ i.o  then $X_n$ cannot tend to $0$.
Converse is also true and it follows from Bore-Cantelli Lemma: $\sum P(|X_n| >x)<\infty$ for all $x>0$ implies that for any $k$, $|X_n| \leq \frac 1 k$ for all $n$ sufficiently large, with probability $1$. This implies that $X_n \to 0$ almost surely.
A: You are right it is not the negation, but it is stronger, so it is alright. If you have almost surely that $X_n\not \to 0$, then you cannot have $X_n\to0$ almost surely. This contradicts your original assumption so you can conclude that the sum is finite.
The converse result is true. Indeed $\sum_{n=1}^\infty P(|X_n| \geq x)=\mathbb E[\sum_{n=1}^{+\infty}1_{\{\vert X_n\vert\ge x\}}]$. If the latter is finite, then $\sum_{n=1}^{+\infty}1_{\{\vert X_n\vert\ge x\}}$ is integrable and therefore almost surely finite. This implies that almost surely, $1_{\{\vert X_n\vert\ge x\}}\to0$ as $n\to+\infty$. Equivalently, $\vert X_n\vert<x$ for $n$ large enough (you could also have found that from the Borel-Cantelli lemma). You deduce that almost surely, for all $k\in\mathbb N^*$, for $n$ large enough, $\vert X_n\vert<\frac1k$. This implies that almost surely, $X_n\to0$ as $n\to+\infty$. Note that the independence of $(X_n)_{n\in\mathbb N}$ was not needed here.
