How can we combine two following facts
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(1) $E|X| < \infty \iff \sum_{n=1}^{\infty}\mathbb{P}(|X| \geq n) \text{ converges}$
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(2) $\forall_{\epsilon > 0} \sum_n \mathbb{P}(|X_n| > \epsilon) \text{ converges} \implies X_n \to 0 \text{ almost surely}$
to prove:
If $E|X| < \infty$ and $X_n$ is sequence of random variables with same distribution as X then $\frac{X_n}{n} \to 0$ almost surely?
I have been trying to wrap my mind around this, but can't find the correct proof. This is probably a one-liner solution though…
Best Answer
Let $\varepsilon>0$ be given and note that
$$ \sum_{n=1}^{\infty}\mathbb{P}(|X_n/n|\geq \varepsilon)=\sum_{n=1}^{\infty} \mathbb{P}(|X_n/\varepsilon|\geq n)=\sum _{n=1}^{\infty} \mathbb{P}(|X/\varepsilon|\geq n)<\infty, $$ since $\mathbb{E}|X/\varepsilon|=\frac{\mathbb{E}|X|}{\varepsilon}<\infty$.