I'm here for another problem. I need to show the following:
If $\{X_n\}$ is a sequence of independent random variables, and $S_n=\sum_{k=1}^{n}X_k$ then $$\limsup_{n}|S_n|=\infty \quad a.s. \text{ if } \quad \forall r>0: \sum_{n=1}^{\infty}\mathbb{P}\left[|X_n|>r\right]=\infty$$
And I had two ideas:
IDEA 1: Using Borel-Cantelli I can conclude that the event $$A_r=\{|X_n|>r \ \text{infinitely often}\}$$ has probability one. But this implies that $\bigcap_{r=1}^{\infty} A_r$ has probability one. So, I can choose a subsequence $\{X_{n_j}\}$ such that $|X_{n_j}|\rightarrow \infty$ as $j\rightarrow \infty$. How I can use this subsequence to show that $\limsup |S_n|=\infty$?
The trouble that I have with this is that I don't know the sign of $X_{n_j}$: If for all $j$ the $X_{n_j}$ have the same sign, then I'm done. The problem is when there exists a $j$ such that $X_{n_j}$ has different sign.
IDEA 2: The event $\{\limsup|S_n|=\infty\}$ is a tail event and since the $X_n$ are independent, it follows that it has probability zero or one. I have the idea to proceed by contradiction; but when I start, I don't see how to work with it.
If $\lim \sup |S_n| \leq r$ for some $r <\infty$ then $|X_n|=|S_n-S_{n-1}| \leq 2r$ for all $n$. But this has probability $0$ by Borel-Cantelli Lemma applied to the condition $\sum P(|X_n| >2r)=\infty$. Hence $(\lim \sup |S_n| <\infty)=\bigcup_{r=1}^{\infty} (\lim \sup |S_n| \leq r)$ has probability $0$.