$\mathbb E[ |X| ] < \infty \iff \forall \epsilon : \mathbb E[ |X / \epsilon | ] < \infty$ $X$ is a random variable.
$$ \mathbb E ( | X | ) < \infty \implies \forall \epsilon > 0, \sum_n \mathbb P ( | X | \geq n \epsilon ) < \infty. $$
Any help to prove this?
(This amounts to prove that $\mathbb E[ |X| ] < \infty \iff  \forall \epsilon :  \mathbb E[ |X / \epsilon | ] < \infty$).
---------------------------------------------------- original post
Since I know that  $$X \in L^1 \iff \sum \mathbb P ( | X | \geq n) < \infty$$ I was given the hint to try to prove the statement hereinafter:
Let us say that $X$ is a random variable.
$$\forall \epsilon > 0, \exists K>0, \forall n \in \mathbb N: \mathbb P( X \geq n \epsilon ) \leq K \mathbb P ( X \geq n), $$
is a useful fact that I need in some proof of my probability lecture, however I am quite unsure about how to prove such statement. Any idea?
---------------------------------------------------- comment
I guess that the original post is completly wrong because one cannot bound each term (cf. Ian comment), however bounding the whole sum is doable somehow.

Using the monotone convergence theorem, for any $\epsilon>0$,
$$
\mathsf{E}|X|=\lim_{M\to\infty}\mathsf{E}[|X|\wedge M]=\epsilon\lim_{M\to\infty}\mathsf{E}[|X/\epsilon|\wedge M/\epsilon]=\epsilon\,\mathsf{E}|X/\epsilon|.
$$
Thus,
$$
\mathsf{E}|X|<\infty \Leftrightarrow \mathsf{E}|X/\epsilon|<\infty \Leftrightarrow \sum_{n}\mathsf{P}(|X/\epsilon|\ge  n)<\infty.
$$
