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> > Let $(X\_n)$ be a sequence of identically distributed integrable random variables.
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My aim is to show for an arbitrary $\varepsilon>0$, that

$$
\sum\_{n=1}^{\infty}\mathbb{P}\left(\frac{1}{n}\lvert X\_n\rvert>\varepsilon\right)<\infty.
$$

I would like to make it as straightforward as possible.

I started with the following:

I call the probability space $(\Omega,\mathfrak{A},\mathbb{P})$.

Because the $X\_n$ are identically distributed it is

$$
\sum\_{n=1}^{\infty}\mathbb{P}\left(\frac{1}{n}\lvert X\_n\rvert>\varepsilon\right)=\sum\_{n=1}^{\infty}\mathbb{P}\left(\frac{1}{n}\lvert X\_1\rvert>\varepsilon\right).
$$
Then my next step was to write
$$
\sum\_{n=1}^{\infty}\mathbb{P}\left(\frac{1}{n}\lvert X\_1\rvert>\varepsilon\right)=\mathbb{E}\left(\sum\_{n=1}^{\infty}1\_{\frac{1}{n}\lvert X\_1\rvert>\varepsilon}\right).
$$

But now, I do not know how to continue it a way that makes sense...

Do you have an idea?

With greetings