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 $X_n \sim Bernoulli(\frac{1}{n})$ independently, then I know that because $\sum_{i=1}^{\infty} P(X_n = 1) = \infty$. Then by Borel-Cantelli, I know that $P(\limsup_{n \to \infty}\{X_n=1\}) = 1$. This means that the event $\{X_n = 1\}$ happens infinitely often with probability 1. However, I do not know how to rigorously use this to prove that $X_n$ doesn't converge to $0$ almost surely. I know that the definition of almost sure convergence is:

$$P(\{\omega \in \Omega: \lim_{n \to \infty}X_n(\omega) = X\}) = 1$$

How can I use the above to show that this doesn't hold?

• You have a sequence of 0 and 1 with infinitely many 1 and you are asking why this sequence does not converge to 0, right? Is the solution clearer using this reformulation?
– Did
Oct 5 '16 at 7:21
• You are done: You showed that sequence $X_n$ doesn't converge to 0 almost everywhere, since the set on which it doesn't converge to $0$ has probability $1$. Oct 5 '16 at 8:46
• I am wondering how this can be proved explicitly using the definition of almost sure convergence. Oct 5 '16 at 23:39

Note that $$\mathbb P\left(\left\{\omega\in\Omega:\lim_{n\to\infty} X_n(\omega)=X(\omega)\right\} \right)=1$$ is equivalent to $$\mathbb P\left(\left\{\omega\in\Omega:\liminf_{n\to\infty} |X_n(\omega)-X(\omega)|<\varepsilon \right\}\right)=1, \text{ for all }\varepsilon > 0.$$ This follows directly from the definition of convergence of a sequence of real numbers: $$\omega\in\left\{\lim_{n\to\infty} X_n= X\right\}\implies \omega\in\bigcup_{n=1}^\infty\bigcap_{k=n}^\infty\{|X_n-X|<\varepsilon \}, \text{ for all }\varepsilon > 0.$$ Now, since \begin{align} \left(\limsup_{n\to\infty}\ \{X_n=1\}\right)^c &= \left(\bigcap_{n=1}^\infty\bigcup_{k=n}^\infty \{X_k=1\} \right)^c\\ &= \bigcup_{n=1}^\infty\bigcap_{k=n}^\infty \{X_k=0\}\\ &= \liminf_{n\to\infty}\ \{X_n=0\}, \end{align} we see that $$\mathbb P\left(\liminf_{n\to\infty}\ \{X_n=0\}\right)=0.$$ From this the result follows immediately.

• @Did When you edit my posts, would you mind noting the changes in the edit summary? Oct 7 '16 at 11:44
• Why not.  
– Did
Oct 7 '16 at 11:46
• Can you elaborate why the third line equation holds? I understand that if the limit exists, then an element in the limit is in the limit infimum, but I thought that this wasn't true the opposite way around, that if an element is in the limit infimum it doesn't have to be in the limit? Oct 9 '16 at 4:52
• @Math1000 Doesn't the third line not hold for the opposite direction? Meaning it is not an if and only if statement but rather only for one-direction? Oct 9 '16 at 23:55

Let us call $A$ the event $\{\limsup_{n \to \infty}X_n=1\}$ (shorthand notation for $\{\omega : \limsup_{n \to \infty}X_n(\omega)=1\}$). You have already proved that $P(A)=1$.

Let us call $B$ the event $\{\lim_{n \to \infty}X_n=0\}$ (shorthand notation for $\{\omega : \lim_{n \to \infty}X_n(\omega)=0\}$). Almost sure convergence is equivalent (according to your definition) to $P(B)=1$

But the events $A$ and $B$ are disjoint (this is not probability anymore, just the mere deterministic fact that a sequence that has infinitely many ones can't converge to zero).

So $B \subset A^C$ so $P(B) \leq P(A^C) = 1-P(A) = 1-1 = 0$. Hence $P(B)=0$ so the definition of almost sure convergence is not met.

That was the developed argument, once you get more accustomed to probability you'd say : Almost surely the sequence has infinetely many ones, so almost surely it does not converge to 0, so there can't be almost sure convergence.

Last remark, you haven't just proven no almost sure convergence, but the stronger statement of almost-sure non convergence. Try to wrap your head around these two statements, and to write their formal definition yourself as a good exercise.