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.
Best Answer
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 $\frac1R = \limsup |X_n|^{1/n}$.