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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,

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

Please help me showing these two parts. Thank you.

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    $\begingroup$ No, the Borel-Cantelli lemma doesn't say (all of) that, at least, not without further assumptions. $\endgroup$ – cardinal Mar 29 '15 at 20:27
  • $\begingroup$ @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 $\endgroup$ – user315 Mar 29 '15 at 21:33
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    $\begingroup$ Added one "further assumption". $\endgroup$ – Zen Mar 30 '15 at 2:13
  • $\begingroup$ 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 $\endgroup$ – jld Nov 15 '16 at 19:56
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None of the assertions are true.

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 $\lim_n P(A_n,A_{n+1}^c)\rightarrow 0$.

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