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.
 A: Let $f(r) =\sum_{n=1}^\infty r^n |X_n|$. This is a random power series.
Clearly, $E|f (r)|<\infty$ when $|r|<1$.  Therefore  $f(r)$ converges for all   rational $r$ a.s.   Next, let $\epsilon$ be such that $P(|X_1|>\epsilon)>0$. Then $P(|X_n|>\epsilon\mbox{ i.o.})=1$, by BC-II. Therefore $f(r)$ diverges for all rational $|r|\ge 1$ a.s.
As $f$ is a power series, this implies that with probability $1$, $f$ converges for all $|r|<1$ and diverges for all $|r|>1$. Therefore, a.s.  $R$, the radius of convergence $f$, is equal to $1$. Now the formula for radius of convergence of $f$ is  $\frac1R = \limsup |X_n|^{1/n}$.
A: Yes, indeed. But instead of showing $\mathbb P( \limsup |X_n|^{\frac{1}{n}} > 1) =0$ (resp. $<1$), it would be easier to show that given $\varepsilon > 0$ you have $\mathbb P( \limsup |X_n|^{\frac{1}{n}} > 1+\varepsilon) = 0$ (resp. $< 1-\varepsilon$). Take any $\varepsilon > 0$. By Bernoulli inequality (i.e $(1+\varepsilon)^n \ge 1+n\varepsilon$), you get $$ \mathbb P(|X_n|^{\frac{1}{n}} > 1+\varepsilon) = \mathbb P(|X_1| > (1+\varepsilon)^n ) \le \mathbb P(|X_1| > 1+n\varepsilon) = \mathbb P( \frac{|X_1| - 1}{\varepsilon} > n), $$ hence the series $\sum_n \mathbb P(|X_n|^{\frac{1}{n}} > 1+\varepsilon)$ is convergent, so by Borel-Cantelli $\limsup |X_n|^{\frac{1}{n}} \le 1 + \varepsilon$ almost surely, and since $\varepsilon > 0$ was arbitrary we get $\limsup |X_n|^{\frac{1}{n}} \le 1$ almost surely (for example by taking $\varepsilon = \varepsilon(m) = \frac{1}{m} \to 0$).
Now, for the other part, we want to show that with probability $1$, $\limsup |X_n|^{\frac{1}{n}} \ge 1-\varepsilon$. So, we would like to show that the series $\sum_n \mathbb P(|X_n|^{\frac{1}{n}} > 1-\varepsilon)$ is divergent (we use independence here). We get $$ \mathbb P(|X_n|^{\frac{1}{n}} > 1-\varepsilon) = \mathbb P(|X_1| > (1-\varepsilon)^n).$$ Since $(1-\varepsilon)^n \to 0$, hence for any $a>0$ and $n \ge N(a)$, we have $\mathbb P(|X_n|^{\frac{1}{n}} > 1-\varepsilon) \ge \mathbb P(|X_1| > a)$. Now, take $a>0$ such that $\mathbb P(|X_1| > a) > 0$ (it is doable due to the assumption $X_1 \not \equiv 0$). Then $$ \sum_n \mathbb P(|X_n|^{\frac{1}{n}} > 1-\varepsilon) \ge \sum_{n=N(a)}^{\infty} \mathbb P(|X_1| > (1-\varepsilon)^n ) \ge \sum_{n=N(a)}^\infty \mathbb P(|X_1| > a) = \infty, $$ and we conclude similarly as before (but now, by second Borel-Cantelli) that $\limsup |X_n|^{\frac{1}{n}} \ge 1-\varepsilon$ almost surely, and since $\varepsilon > 0$ was arbitrary we conclude $\limsup |X_n|^{\frac{1}{n}} \ge 1$ a.s.
A: 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.
