How to prove that $E[|X|]$ is finite iff $\sum\limits_{n\ge1}\frac{1}{n^2}E[|X|^2I_{|X|\le n}]$ converges Let a random variables $X$. Show that
$$
E[|X|]<+\infty\Longleftrightarrow \sum_{n>1}\dfrac{1}{n^2}E[|X|^2I_{|X|\le n}]<+\infty
$$
where the $I_{|x|\le n}$ is the indicator function. 
My try: since
$E[|X|]<+\infty$
then we have
$$\int_{R}|X|d P_{X}(x)<+\infty$$
and then I want use this central limit Theorem ,and I can't.
Thank you 
 A: You can use "summation by parts:"
$$
\sum_{n>0}\dfrac{1}{n^2}E[|X|^2I_{|X|\le n}] = \sum_{n=1}^\infty \sum_{m=1}^n \dfrac{1}{n^2}E[|X|^2I_{m-1 < |X|\le m}] = \sum_{m=1}^\infty \sum_{n=m}^\infty \dfrac{1}{n^2} E[|X|^2I_{m-1 < |X|\le m}]
$$
Now approximate:
$$ \sum_{n=m}^\infty \dfrac{1}{n^2} \approx \frac1m ,$$
and
$$ E[|X|^2I_{m-1 < |X|\le m}] \approx m^2 \Pr(|X| \in (m-1,m]) .$$
Now use the kind of argument shown here: Show that $\sum_{n=1}^{\infty}\mathbb{P}\left(\frac{1}{n}\lvert X_n\rvert>\varepsilon\right)<\infty$.
(Notation: when I say $A \approx B$, I mean that $A/B$ and $B/A$ are bounded by universal constants that don't depend upon the parameters used to describe $A$ and $B$.)
A: Define $A_j:=\{\omega, j-1\lt j\leqslant j\}$.


*

*Show that "$X$ integrable" is equivalent to $\sum_j j\mu(A_j)$ is convergent. 

*Show that there exists a constant $C$ such that 
$$\frac 1C\sum_{j=1}^\infty \sum_{n=1}^j\mathbb E[X^2\chi_{A_j}]\leqslant \sum_{j=1}^\infty j\mu(A_j)\leqslant C\sum_{j=1}^\infty \sum_{n=1}^j\mathbb E[X^2\chi_{A_j}].$$

*Conclude.
A: Summing up inequalities $\frac{1}{n}-\frac{1}{n+1} \leq \frac{1}{n^2} \leq \frac{1}{n-1}-\frac{1}{n}$ for $n \geq 2$, we see that on the event $\{|X| \geq 2\}$, one has
$$
\tag{$\ast$}
|X| = \frac{|X|^2}{|X|} \leq \sum_{n > 1} \frac{1}{n^2} |X|^2I_{|X| \leq n} = |X|^2\sum_{n \geq |X|} \frac{1}{n^2} \leq \frac{|X|^2}{|X|-1} \leq |X|+2
$$
and of course $E(|X|) < \infty \iff E(|X|I_{|X| \geq 2}) < \infty$.
The result follows by taking expectations in ($\ast$) and using the Fubini-Tonelli theorem.
