How to show that $p$th moment being finite is equivalent to a limit existing Let $p \in (0,2)$ and let $\xi_n, n \geq 1$, be iid random variables. Show that the following two conditions are equivalent:


*

*With probability one, the limit
$$ \lim_{n \rightarrow \infty} \frac{1}{n^{1/p}} \sum_{k=1}^n \xi_k$$
exists and is finite.

*$\mathbb{E}\lvert \xi_i \rvert^p < \infty$ AND either $\mathbb{E} \xi = 0$ or $p \leq 1$.
I have tried using Holder's inequality but didn't really get anywhere. I don't really have any idea how to approach the problem for either direction...
 A: Here's a proof that (1)$\implies$ (2). 
Letting $T_n=n^{-1/p}\sum_1^n \xi_n$, we have
$$
\frac{\xi_n}{n^{1/p}}=T_n - T_{n-1}\cdot \frac{(n-1)^{1/p}}{n^{1/p}}
$$
Letting $n\to\infty$ above, since $T_n\to T$ a.s, and $\frac{(n-1)^{1/p}}{n^{1/p}}\to 1$, we get 
$$\frac{\xi_n}{n^{1/p}}=T_n - T_{n-1}\cdot \frac{(n-1)^{1/p}}{n^{1/p}}\to T-T\cdot 1=0$$
 so that $\xi_n/n^{1/p}\to 0$ a.s. This means $P(|\xi_n|/n^{1/p}>1\text{ i.o.})=P(|\xi_n|^p>n\text{ i.o.})=0$, so (using Borel Cantelli on the last inequality),
$$
E|\xi|^p=\int_0^\infty P(|\xi|^p>t)\,dt\le \sum_{n\ge0} P(|\xi_n|^p>n)<\infty
$$
proving $E|\xi|^p<\infty$. Now, suppose by way of contradiction that $p>1$ and $E\xi\neq0$. Using Jensen's, $E|\xi|^p<\infty$ allows you to show $E|\xi|<\infty$, so by SLLN,
$$
\frac{\sum_{k=1}^n\xi_n}{n}\to E\xi\neq0
$$
almost surely as $n\to\infty$. We also have, since $p>1$, that
$$
\frac1{n^{1/p-1}}\to \infty
$$
Multiplying the two above limits implies that
$$
\frac{\sum_{k=1}^n\xi_n}{n^{1/p}}=T_n\to\infty\qquad\text{a.s.}
$$
contradicting that the limit was finite. Thus, we must have $p\le 1$ or $E\xi=0$.
