IF $\sum_{n=1}^{\infty}P(|X_n|>c)=\infty\forall c>0$, then $\limsup_{n\rightarrow\infty}|S_n|=\infty$ a.s.

Question

If $S_n=\sum_{i=1}^n X_i$ where $X_i,i\ge 1$ are iid and $\sum_{n=1}^{\infty}P(|X_n|>c)=\infty\forall c>0$, then $\limsup_{n\rightarrow\infty}|S_n|=\infty$ a.s.

I tried this but got stuck: Let $A_n=\{|X_n|>c\}$.

$\sum_{n=1}^{\infty}P(A_n)=\infty$ so by Borel-Cantelli, $P(|X_n|>c, i.o.)=1$

So $|X_n|>c$ a.s.

Then $\sum_{k=1}^n |X_k|>nc$ a.s.

But I do I relate $|S_n|=|\sum_{k=1}^nX_k|$ to this with the inequality in the direction I want? The triangle inequality doesn't help here.

Answer 1

So... almost surely, $|X_n|\gt c$ for infinitely many $n$. Note that if $|X_n|\gt c$, then $|S_n|\gt\frac12c$ or $|S_{n-1}|\gt\frac12c$. Hence $|S_n|\gt\frac12c$ for infinitely many $n$, in particular, $\limsup\limits_n|S_n|\geqslant\frac12c$, almost surely. This holds for every $c$, say, every integer $c$, hence $\limsup\limits_n|S_n|=+\infty$, almost surely.