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My previous question here on stack exchange has caused a follow-up question.

The Borel-Cantelli lemma tells us that, if for any $\varepsilon > 0$ we have

$\sum_{n=1}^\infty \mathbb{P}(\vert X_n - X \vert > \varepsilon) < \infty$, then $X_n \to X$ a.s.

Especially, if I can show that

$\sum_{n=1}^\infty \mathbb{P}\left(\frac{\vert X_n\vert}{n} > \varepsilon\right) < \infty$ for all $\varepsilon > 0$, then $X_n/n \to 0$ a.s.

But, if I can only show that

$\sum_{n=1}^\infty \mathbb{P}\left(\frac{\vert X_n\vert}{n} > \varepsilon\right) < \infty$ holds for a fixed $\varepsilon > 0$,

then I can only conclude that $\lim_{n \to \infty} \vert X_n \vert/n \leq \varepsilon$ a.s. Hence I still know, that the sequence $\vert X_n \vert/n$ is converging almost surely, but I lose information about the exact value of the limit, is that true?

Davide Giraudo
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student7481
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1 Answers1

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Indeed, $$ \sum_{n=1}^\infty \mathbb{P}\left(\frac{\vert X_n\vert}{n} > \varepsilon\right) < \infty$$ only for some fixed $\varepsilon$ only implies that $\limsup_{n\to\infty}\lvert X_n\rvert/n\leqslant\varepsilon$. The example by Steven Mai, namely, $X_n=(-1)^n n$ and $\varepsilon=2$ shows that we cannot deduce that $X_n/n\to 0$ almost surely.

Davide Giraudo
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