Let $X_n$ be i.i.d. sequence of random variables and $p\in(0,2)$. \begin{equation} \lim_{n\to\infty} \frac{1}{n^{1/p}}\sum_{k=1}^n X_k \label{limit}\tag{1} \end{equation} exists and finite with probability one, if and only if $E|X|^p<\infty$ and either $p\leq 1$ or $EX_k=0$.
Small Step If (\ref{limit}) exists finite, and $p>1$, \begin{equation} \lim_{n\to\infty} \frac{n}{n^{1/p}}\frac{1}{n}\sum_{k=1}^n X_k \end{equation} since $n^{1-1/p}\to\infty$, $\frac{1}{n}\sum X_k\to 0=EX_k$.
Question
I am not sure what is the relation between (\ref{limit}) and $E|X|^p$.
I tried to use the fact that $n^{-1}\sum |X_k|^p \to E|X|^p$ however couldn't bound it.
Statement suggests that if $p\leq 1$, $E|X|^p<\infty$ iff (\ref{limit}) exists immediately hence I expect it to be rather straightforward. I appreciate if one can advise a path for it.
If (1) holds with a finite limit, then $\left(X_n/n^{1/p}\right)_{n\geqslant 1}$ converges to $0$ almost surely. Using the version of the Borel-Cantelli lemma for independent events, we derive the convergence of $\sum_{n\geqslant 1}\Pr\left(\left\lvert X_n\right\rvert\gt n^{1/p}\right)$ from which finiteness of $\mathbb E|X|^p$ follows.