Intuition of LLN's condition Let $X_1,X_2,...$ be independent r.v. with finite expectation
If there is a constant $p\in[1,2]$ s.t.
$$\lim_{n\rightarrow\infty}\frac{1}{n^p}\sum_{i=1}^nE|X_i|^p=0$$
then $$\frac{1}{n}\sum_{i=1}^n(X_i-EX_i)\rightarrow_p0$$
If there is a constant $p\in[1,2]$ s.t.
$$\sum_{i=1}^\infty\frac{E|X_i|^p}{i^p}<\infty$$
then $$\frac{1}{n}\sum_{i=1}^n(X_i-EX_i)\rightarrow_{a.s.}0$$
My question is that:
1.Does it means that $\sum_{i=1}^\infty\frac{E|X_i|^p}{i^p}<\infty$ implies $\lim_{n\rightarrow\infty}\frac{1}{n^p}\sum_{i=1}^nE|X_i|^p=0$? If so, how to prove it?
I think if $p=1$, then we can directly use Kronecker's lemma to prove it, but what if $p\in(1,2]$?


*How to understand these conditions? For example,

If $X_1,X_2,...$ be iid r.v.
then $$\lim_{n\rightarrow\infty}nP(|X_1|>n)=0$$ iff
$$\bar{X}-E(X_1I_{|X_1|\le n})\rightarrow_p 0$$,
so what's the relationship between $\lim_{n\rightarrow\infty}\frac{1}{n^p}\sum_{i=1}^nE|X_i|^p=0$ and $\lim_{n\rightarrow\infty}nP(|X_1|>n)=0$? Say if we consider "iid" in the independent WLLN, and let $p=1$
$$\lim_{n\rightarrow\infty}\frac{1}{n^p}\sum_{i=1}^nE|X_i|^p=0$$ becomes
$$\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{i=1}^nE|X_i|=0$$
It doesn't make sense, does it means that "iid WLLN" is not the special case of "independent WLLN" ?
3 Similarly, If $X_1,X_2,...$ be iid r.v.
then $$E(|X_1|)<\infty$$ iff
$$\bar{X}-E(X_1)\rightarrow_{a.s.} 0$$
what's the relationship between $E(|X_1|)<\infty$ and $\sum_{i=1}^\infty\frac{E|X_i|^p}{i^p}<\infty$? For example, if we consider "iid" in the independent SLLN and let $p=1$
$$\sum_{i=1}^\infty\frac{E|X_i|^p}{i^p}<\infty$$ implies $$\sum_{i=1}^\infty\frac{E|X_1|}{i}<\infty$$
compare with $$E|X_1|<\infty$$, let $E|X_1|=c<\infty$
$$\sum_{i=1}^\infty\frac{E|X_1|}{i}=\sum_{i=1}^\infty\frac{c}{i}=c\sum_{i=1}^\infty\frac{1}{i}$$ divergence. So $E|X_1|<\infty$ with iid condition does not imply $$\sum_{i=1}^\infty\frac{E|X_i|^p}{i^p}<\infty$$
on the other hand, if we have $$\sum_{i=1}^\infty\frac{E|X_i|^p}{i^p}<\infty$$
can we prove $E|X_1|<\infty$ with iid condition?
 A: *

*It cannot be deduced directly that $\sum_{i=1}^\infty\frac{E|X_i|^p}{i^p}<\infty$ implies $\lim_{n\rightarrow\infty}\frac{1}{n^p}\sum_{i=1}^nE|X_i|^p=0$ because we do not know a priori whether the results are optimal. For example, the condition $\lim_{n\rightarrow\infty}\frac{1}{n^p}\sum_{i=1}^nE|X_i|^p=0$ could be replaced by something much more restrictive. In order words, we can have $P_1\Rightarrow Q_1$, $P_2\Rightarrow Q_2$ and $Q_1\Rightarrow Q_2$ but it does not mean that $P_1\Rightarrow Q_2$.
However, that $\sum_{i=1}^\infty\frac{E|X_i|^p}{i^p}<\infty$ implies $\lim_{n\rightarrow\infty}\frac{1}{n^p}\sum_{i=1}^nE|X_i|^p=0$ follows but some calculus: let $c_i=E|X_i|^p$ and $s_k=\sum_{i=1}^kc_i$. Using Abel's transformation, $\sum_{i=1}^\infty\frac{E|X_i|^p}{i^p}<\infty$ is equivalent to $\sum_k s_k\left(\frac 1{k^p}-\frac 1{(k+1)^p}\right)<\infty$ and we get $\sum_k s_k\frac 1{k^{p+1}}<\infty$. Summing on intervals of the form $[2^n+1,2^{n+1})$ gives $\sum_n s_{2^n}2^{-np}<\infty$ which gives $s_{2^n}2^{-np}\to 0.


*Indeed, if $p=1$ and the $X_i$ have the same distribution, then both conditions fails. They are actually more appropriated in the case $p>1$, where, as you noticed, the conditions take place provided that $\mathbb E\lvert X_1\rvert^p<\infty$. These conditions are also useful in the non-identically distributed case, when the moment of order $p$ of $X_i$ is even allowed to increase with $i$, but of course not too fast.
