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If I define $X_n=\text{Binomial}(2,1/n)$ then $P(X_n>\epsilon)=P(\text{Binomial}(2,1/n)>0)=1-P(\text{Binomial}(2,1/n)=0)=1-(1-1/n)^2$ Clearly $X_n$ converges to $0$ in probability But I can't show(using Borel cantelli Lemma) that $X_n$ converges to $X$ almost surely since $\sum_{n=1}^\infty P(X_n>0)=\sum_{n=1}^\infty 1-(1-1/n)^2=\sum_{n=1}^\infty (1/n)(2-2/n) \ge \sum_{n=1}^\infty (1/n)=\infty$ and hence I cant conclude that the lim sum of $(X_n>\epsilon)$ has measure zero and therefore almost sure convergence.

Are my calculations correct? Is it even true that there is almost sure convergence? If not can you give a counter example?

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Using the second Borel-Cantelli Lemma, you can in fact prove it does not converge almost surely to $0$, assuming the $X_n$ are independent. Actually, almost surely, it does not converge to zero. Defining $E_n=\{X_n\neq 0\}$, you have shown that $\sum_{n=1}^{\infty}\mathbb{P}(E_n)=\infty$. The second B-C Lemma then says that $\mathbb{P}(\limsup_{n\to \infty} E_n)=1$, so almost surely $X_n\neq 0$ infinitely often, so almost surely $X_n$ does not converge to zero.

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