For independent random variables $X_n$, how to prove $X_n / n^p \rightarrow 0$ a.s. iff $E\left(\left|X_{n}\right|^{1 / p}\right)<\infty$ for $p>0$ I am working on the following problem.

Let $X_{n}, n=1,2, \ldots,$ be independent random variables with identical distribution. Show that $X_{n} / n^{p} \rightarrow 0$ a.s. if and only if $E\left(\left|X_{n}\right|^{1 / p}\right)<\infty$ for $p>0$.

The hint suggests that using Borel-Cantelli lemma, but I don't have any idea how to apply it. Any more hint is appreciated.


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One suggested the counterexample. How about the case if the random variables are identically distributed? 
 A: First, it suffices to solve the case $p=1$. Indeed, assume that we showed that $X_{n} / n  \rightarrow 0$ a.s. is equivalent to $E\left(\left|X_{1}\right| \right)<\infty$ for each i.i.d. sequence $(X_n)_n$. Then for a general $p$, apply this to $X'_n=\lvert X_n\rvert^{1/p}$ instead of $X_n$ and conclude with the equivalences
$$
X_n/n^p\to 0a.s.\Leftrightarrow {X'_n}^p/n^p\to 0 a.s.\Leftrightarrow {X'_n} /n\to 0 a.s.\Leftrightarrow E\left(\left|X'_{1}\right| \right)<\infty\Leftrightarrow E\left(\lvert X_1\rvert^{1/p} \right)<\infty.
$$
As a second step, note that $X_{n} / n  \rightarrow 0$ a.s. if and only if for each positive $\varepsilon$, $\sum_{n\geqslant 1}\mathbb P\left(\lvert X_n\rvert/n\gt \varepsilon\right)$ converges. This is a consequence of the Borel-Cantelli lemma.
To conclude, notice that since the random variable $X_n$ has the same law as $X_1$, $\mathbb P\left(\lvert X_n\rvert/n\gt \varepsilon\right)=\mathbb P\left(\lvert X_1\rvert/n\gt \varepsilon\right)$ and a random variable $X$ is integrable if and only if $\sum_{n\geqslant 1}\mathbb P\left(\lvert X\rvert \gt n\right)$ converges.
