Suppose $X_i$ is independently and identically distributed over $i$, and $E(|X_i|^d)$ with $d>0$ is not finite or undefined:
- I am wondering whether $\sum_{i=1}^N |X_i|^d$ is of any asymptotic order. Take Cauchy distribution as an example. If $d=1$, then $N^{-1}\sum_{i=1}^N X_i$ is still Cauchy and hence $$ \sum_{i=1}^N |X_i|\ \mbox{is}\ o_p(N^{1+\delta})\ \mbox{for any}\ \delta>0 $$ That is $N^{-(1+\delta)}\sum_{i=1}^N |X_i|$ converges in probability to $0$.
- If $d=2$, $E(|X_i|^2)$ is infinite. I am not sure whether there exists $\delta>0$ such that $N^{-(1+\delta)}\sum_{i=1}^N |X_i|^2$ converges in probability to $0$.
- Another interesting example is related to Central Limit Theorem (CLT): if $X_i$ follows student t distribution with $3$ degrees of freedom independently over $i$, then $E(X_i)=0$, $E(X_i^2)=3$ and any moments of order higher than 3 do not exist. Due to CLT, $N^{-\frac{1}{2}}\sum_{i=1}^N X_i$ converge in distribution to $N(0,3)$. Hence, $(N^{-\frac{1}{2}}\sum_{i=1}^N X_i)^3\leq 3!N^{-\frac{3}{2}}\sum_{i=1}\sum_{m=1}\sum_{j=0}X_iX_{i+m}X_{i+m+j}+N^{-\frac{3}{2}}\sum_{i=1}^NX_i^3=O_p(1)$. Since the third moment of normal distribution with zero mean is $0$, one could conjecture $E[(N^{-\frac{1}{2}}\sum_{i=1}^N X_i)^3]$ tends to $0$ (notation abuse due to the non-existence of $E(X_i^3)$). One should have $N^{-\frac{3}{2}}\sum_{i=1}^NX_i^3=o_p(1)$ or $\sum_{i=1}^NX_i^3=o_p(N^{\frac{3}{2}})$. Similarly, $\sum_{i=1}^NX_i^d=o_p(N^{\frac{d}{2}})$ with $d\geq3$ if the sequence $\lbrace X_i\rbrace$ satisfies the CLT conditions, though such asymptotic orders may not be on the boundary.