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