I am referring to a question that has already been asked, but not completely answered (Why almost sure convergence holds if $X_n = n$ w.p. $1/n$?).
For $X_1,\dots, X_n$ independent distributed with $P(X_n=0) = 1-\frac1n; P(X_n=n) = \frac1n$, does $X_n$ converge to $0$ almost surely?
My guess would be that is does not since we can apply the second Borel-Cantelli Lemma like in this very similar and well known example: If $(X_n)$ is independent with $X_n$ Bernoulli of parameter $\frac1n$, how to show that $X_n$ doesn't converge almost surely to $0$?
If it does in fact not converge almost surely to $0$ is it also the case that is does not converge to zero almost surely?
Best Answer
Yes, you are right: the falseness of the claim follows from the second Borel-Cantelli Lemma. Since the $X_n$ are independent and $$\sum_n \mathbb{P}(X_n=n)=\sum_n \frac{1}{n}=+\infty$$ then with probability $1$ we have $X_n=n$ infinitely often, which means that in fact almost surely $X_n$ does not converge to $0$.