# Convergence almost sure of sequence random variables with Bernoulli distribution

i stacked in one example.

We have independent sequence of random variables $$X_{n}$$, where $$X_{n}$$ has Bernoulli distribution with parameter $$\frac{1}{n}$$, so $$X_{n} \sim Bernoulli(\frac{1}{n})$$ and then $$P(X_{n} = 1) = \frac{1}{n}$$ and $$P(X_{n} = 0) = 1-\frac{1}{n}$$. Show, that this convergence does not apply $$X_{n} \xrightarrow{a.s} 0$$.

I know, that converges a.s is defined like this:

$$P(\lim_{n\to\infty} |X_{n} - X| = 0) = 1$$.

So in my case i need to prove that:

$$P(\lim_{n\to\infty} |X_{n}| = 0) \neq 1$$.

So i know that $$\lim_{n\to\infty} X_{n} = 1$$, if $$X_{n} = 0$$ and $$\lim_{n\to\infty} X_{n} = 0$$, if $$X_{n} = 1$$. And then $$P(\lim_{n\to\infty} |X_{n}| = 0) = 1$$. But this is wrong. Can you please tell me where i made a mistake ? Thank you!

• $\lim_{n \to \infty} X_n \to 0$ does not make sense. It should be $\lim_{n \to \infty} X_n = 0$. Do not use an arrow when you have taken a limit. Commented Oct 21, 2018 at 12:29

Your mistake is taking limits of random variables. $$\lim_{n \to \infty} X_n = 1$$ does not make sense. $$X_n$$ are random. You cannot just assert the limit is 1 or 0. If you do take a limit you need to state that it is almost surely or with probability 1. But even then, what you write really doesn't make sense.

Let $$0<\epsilon <1,$$. Note that $$X_n$$ are independent, so if you can show that

$$\sum_n P(|X_n - 0|> \epsilon) = \infty$$

then it will follow by the second Borel-Cantelli theorem that

$$P(|X_n - 0|>\epsilon \text{ i.o }) = 1$$

And hence $$X_n$$ does not converge to $$0$$ almost surely. But clearly

$$\sum_n P(|X_n - 0|>\epsilon) = \sum_n P(X_n = 1) = \sum_n \frac{1}{n} = \infty$$

Hence $$X_n$$ does not converge to $$0$$ almost surely.