Proving a statement about probability theory Let X be arandom variable.
Consider any constant $c\gt 0$
how to prove the following satement
$$\sum P(|X|\ge cn) \lt \infty \iff E(|X|)\lt \infty $$
My answer trail:
$E[|X|]=\sum_X|X|P_x(X)\lt \infty$
Borel Cntelly Lemma:
Suppose $A_1,...$ is a squence of events if $\sum P(A_n)\lt \infty$
Then, $P(\lim sup A_n)=0$
note: I dont know whether can I use this lemma or not. :(
Also by writting in this way, Can I further my answer?
$P(|X_n|\ge n \ \ \ \ i.o)=0 iff E(|X_n|)\lt \infty$
please help me. thank you for helping :)
 A: If $Y$ is a nonnegative random variable then $\mathbb{E}Y=\int_{0}^{\infty}P\left\{ Y\geq y\right\} dy$ 
(Are you familiar with that?)
From $P\left\{ Y\geq\lceil y\rceil\right\} \leq P\left\{ Y\geq y\right\} \leq P\left\{ Y\geq\lfloor y\rfloor\right\} $
it follows that 
$$\sum_{n=1}^{\infty}P\left\{ Y\geq n\right\} =\int_{0}^{\infty}P\left\{ Y\geq\lceil y\rceil\right\} dy\leq\int_{0}^{\infty}P\left\{ Y\geq y\right\} dy\leq\int_{0}^{\infty}P\left\{ Y\geq\lfloor y\rfloor\right\} dy=\sum_{n=0}^{\infty}P\left\{ Y\geq n\right\}$$
Apply this on $Y=|X|/c$.

Addendum:
If $Y$ is a nonnegative random variable then applying Fubini we find:
$$\mathbb{E}Y=\int_{0}^{\infty}xdF_{Y}\left(x\right)=\int_{0}^{\infty}\int_{0}^{\infty}1_{\left[0,x\right]}\left(y\right)dydF_{Y}\left(x\right)=$$$$\int_{0}^{\infty}\int_{0}^{\infty}1_{\left[0,x\right]}\left(y\right)dF_{Y}\left(x\right)dy=\int_{0}^{\infty}P\left\{ Y\geq y\right\} dy$$

Second addendum:
Alternative route. For convenience I restrict to $c=1$ and let $X$ be nonnegative.
Define $X_n:=1_{[n,\infty)}(X)$ for $n=0,1,2,\dots$. Then $X_0=1$ and $\sum_{n=1}^{\infty}X_n=\lfloor X\rfloor$ so that 
$$\sum_{n=1}^{\infty}X_{n}=\lfloor X\rfloor\leq X<\lfloor X\rfloor+1=\sum_{n=0}^{\infty}X_{n}$$
Taking expectations we find:$$\sum_{n=1}^{\infty}P\left(X\geq n\right)=\sum_{n=1}^{\infty}\mathbb{E}X_{n}\leq\mathbb{E}X\leq\sum_{n=0}^{\infty}\mathbb{E}X_{n}=\sum_{n=0}^{\infty}P\left(X\geq n\right)$$
