A question related to Borel-Cantelli Lemma

Note:

Borel-Cantelli Lemma says that

$$\sum_{n=1}^\infty P(A_n) \lt \infty \Rightarrow P(\lim\sup A_n)=0$$

$$\sum_{n=1}^\infty P(A_n) =\infty \textrm{ and } A_n\textrm{'s are independent} \Rightarrow P(\lim\sup A_n)=1$$

Then,

if $$\sum_{n=1}^\infty P(A_nA_{n+1}^c )\lt \infty$$

by using Borel-Cantelli Lemma

I want to show that

firstly,

$\lim_{n\to \infty}P(A_n)$ exists

and secondly,

$\lim_{n\to \infty}P(A_n) =P(\lim\sup A_n)$

• No, the Borel-Cantelli lemma doesn't say (all of) that, at least, not without further assumptions. Mar 29, 2015 at 20:27
• @cardinal well, how can i show these two statements? please can you explain it to me? i dont have any enough idea. i'll be glad if you'll show a solutin way:) thank you
– 1190
Mar 29, 2015 at 21:33
– Zen
Mar 30, 2015 at 2:13
• Minor note: as mentioned here, for instance, we can get by with only pairwise independence of the $A_n$ in the second part of the lemma
– jld
Nov 15, 2016 at 19:56

Let $A_n$ be the chance of heads in a coin flip, with probability $1/n^2$ when $n$ is odd and $1-\frac{1}{n^2}$ when $n$ is even. Then:
$$\sum_{n=1}^\infty P(A_n,A_{n+1}^c)=\sum_{odd \ n}^\infty \frac{1}{n^2}\left(1-\frac{1}{(n+1)^2}\right)+\sum_{even \ n}\frac{1}{n^2}\left(1-\frac{1}{(n+1)^2}\right)<\sum_{n=1}^\infty \frac{1}{n^2}<\infty.$$
However, $\lim_nP(A_n)$ clearly does not exist. The best you can conclude is $\lim_n P(A_n,A_{n+1}^c)\rightarrow 0$.