# is this method of showing that the almost sure convergence fails correct, and can we use the Borel-Cantelli?

Consider a sequence of independant random variables such that $$X_{n} \in\{0,1\}$$ almost surely, $$\mathbb{P}\left(X_{n}=1\right)=\frac{1}{n}, \quad \text { and } \quad \mathbb{P}\left(X_{n}=0\right)=1-\frac{1}{n} \quad \forall n \geq 1$$ the sequence $$X_{n}$$ converges in probability towards $$0,$$ but not almost surely.

My attempt :

$$\mathbb{P}(\lim_{k} X_k = 0) = \mathbb{P}(X_n = 0) < 1$$

my reasoning is that only $$0$$ can converge point-wise to zero, $$1$$ can't.

is this method correct ?

also I was told that one can use the Borel-Cantelli lemma, how so ? I'm only used to proving a.s comvergence using that, not disproving it.

thanks !

A sequence of $$0$$'s and $$1$$'s only converges to $$0$$ if it is eventually ALWAYS $$0$$.
So, almost sure convergence to $$0$$ is equivalent to saying that almost surely, the sequence contains only finitely many $$1$$'s.
So, to show that the sequence does not almost surely converge to $$0$$, it is sufficient to show that there is a non-zero probability that the sequence contains infinitely many $$1$$'s. You can prove this using the second Borel-Cantelli lemma.
• just making sure I get it, $\sum \mathbb{P}(X_n = 1) = +\infty$, therefore by the second Borel-Cantelli lemma, $\mathbb{P}(\limsup X_n = 1) = 1$, and so there is infinitely many one's. Feb 21 '20 at 18:33
• That's right, yes. As for your attempt: I can't quite make sense of it. Saying that $P(\lim_k X_k=0)=P(X_n=0)$ is not only false... it doesn't actually make sense. (For which $n$?) Feb 21 '20 at 19:13
• I get it now, I was confusing between the indices and the $\omega$'s Feb 21 '20 at 19:17