# Does $\text{Binomial}(2,1/n)$ converge to zero almost surely?

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?

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.