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.

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.

Note:

Borel-Cantelli Lemma says that

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

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

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.

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.