Durret 2.5.8: Question about probability of limsup I am interested in this problem by Durret in Probability: Theory and Examples.

Let $X_1,X_2, \dots$ be i.i.d. and not $\equiv0$. If $E|X_1| < \infty$, show that $$\limsup_{n \to \infty} |X_n|^{1/n} = 1$$ almost surely.

One idea I had, was trying to show that $P(\limsup |X_n|^{1/n} > 1) = 0$ and $P(\limsup |X_n|^{1/n} < 1) = 0$. This would give us the result. Now for the first one, we have $$P(\limsup |X_n|^{1/n} > 1) \leq P(|X_n|^{1/n} > 1 \text{ i.o.}).$$ I want to show that the last probability above is $0$ maybe by Borel-Cantelli and using $E|X_1| < \infty$ if and only if $\sum_nP(|X_1| \geq n) < \infty$. I am not sure if this is the right directon. Please help.

From the last two lines of the OP, it follows that
$$\sum_nP[|X_n|>n]<\infty$$
(for $X_n\stackrel{law}{=}X_1$). Borel-Cantelli direct lemma  implies that $P[\{X_n>n,\, \text{i.o}\}]=0$ and so for $p$-almost all $\omega$, there is $N_\omega$ such that $|X_n(\omega)|\leq n$ for all $n\geq N_\varepsilon$. Hence $\limsup_n|X_n|^{1/n}\leq \lim_n\sqrt[n]{n}=1$ almost surely.
As $P[|X_1|>0]>0$, there is $\delta>0$ such that $P[|X_1|>\delta]>0$; hence
$\sum_nP[|X_n|>\delta]=\infty$ and so, by the converse Borel-Cantelli (independence is used here) $P[\{X_n>\delta,\,\text{i.o}\}]=1$; hence for almost all $\omega$, there is an increasing  sequence $n_k(\omega)$ of integers such that $|X_{n_k(\omega)}(\omega)|>\delta$ whence it follows that $\limsup_n|X_n|^{1/n}\geq\lim_n\sqrt[n]{\delta}=1$ almost surely.
