Showing $X_n\sim \operatorname{Bin}\left(1,\frac{1}{n}\right)$ almost surely does not converge to $0$ I want to show that $ X_n\sim \operatorname{Bin}\left(1,\frac{1}{n}\right)$  almost surely does not converge to $0$; $X_n$'s are independent.
Therefore I got the hint to use Borel-Cantelli and showed that
$\sum_{n=1}^{\infty}P(X_n=1)=\sum_{n=1}^{\infty}\frac{1}{n}=\infty$ and so
$P(\limsup\limits_{x\rightarrow\infty}X_n)=1$.
Can I now follow that it is not possible that $X_n$ converges almost sure to $0$ because the probability that infinity $X_n=1$ happens is $1$ ?
Maybe you can help me with that
 A: If $X_n$ are independent, the second Borel-Cantelli lemma says $\mathbb P\left(\limsup_{n \to \infty} X_n =1\right) = 1$, i.e. that with probability $1$, infinitely many $X_n = 1$. But if $X_n \to 0$, there would have to be some $N$ such that $|X_n| < 1/2$ for all $n > N$.  Since $X_n$ can only take values $0$ and $1$, that means $X_n = 0$ for all $n > N$.
But that would say only finitely many $X_n = 1$.
So with probability $1$, that doesn't happen.
A: I believe you have a Bernoulli, not Binomial. First, show convergence in probability (by Markoff inequality):
$$
P(|X_n-0|>\varepsilon) \leq \frac{\mathbf{E} X_n}{\varepsilon} = \frac{1}{n \varepsilon} \to 0
$$
so $X_n \to_p 0$. Now, for convergence a.s. you need $P(\limsup X_n \neq 0) = 0 \Leftrightarrow\sum_{k=1}^{\infty}P(|X_k -0|> \varepsilon) < \infty$:
$$
P(\limsup X_n \neq 0) \Leftrightarrow \sum_{k=1}^{\infty}P(|X_k-0|> \varepsilon) \geq \sum_{k=1}^{\infty}P(X_k> \varepsilon) =  \sum_{k=1}^{\infty}\frac{1}{k} \to \infty
$$
Since the measure on the $\limsup$ on this set is 1, $X_n \not\to X$ a.s.
