Convergence in probability vs almost surely using Borel-Cantelli Lemma

Question

Looking at a sequence of independent rvs $$ Z_n = \Bigg\{ \begin{array}{lr} 1 & w.p. \frac{1}{n}\\ 0 & w.p. 1-\frac{1}{n} \end{array} $$ It is easy to see that $$ P(|Z_n - 0|>\varepsilon) = P(Z_n = 1) = \frac{1}{n} \to_n 0 $$ so $Z_n \to_n 0$ in probability. At the same time $$ P(Z_n = 1 \ i.o.) = P(\limsup Z_n=1) = \sum_{n=1}^{\infty} \frac{1}{n} = H_n \to \infty \\ P(Z_n = 0 \ i.o.) = P(\limsup Z_n=0) = \sum_{n=1}^{\infty} \bigg(1- \frac{1}{n} \bigg) = \lim_{n \to \infty} n \bigg(1- \frac{H_n}{n} \bigg) \to \infty $$ so both $P(Z_n=1 \ i.o.) = P(Z_n=0 \ i.o.) = 1$. Does this mean the $Z_n$ doesn't converge almost surely at all? Is there something faulty with my reasoning here?

Answer 1

The Borel-Cantelli lemma doesn't tell you that $P(\limsup_n A_n) = \sum_{n=1}^\infty P(A_n)$ but rather that $P(\limsup_n A_n) = 0$ or $1$ according as $\sum_n P(A_n)$ converges or diverges (assuming the $A_n$ are independent). In your case, this means $Z_n = 1$ happens infinitely often so $Z_n$ cannot converge a.s.

Note that since you already showed that $Z_n \to 0$ in probability, the only possible a.s.-limit is $0$ so really you only need to consider $P( Z_n = 1 \ \text{i.o.})$.