Let $A_n$ be a sequence of independent events with $$lim_{n \to \infty}P(A_n)=0.$$ Suppose that $$\sum_{n=1}^{\infty}P(A_n^c \cap A_{n+1}) < \infty.$$ Will the events $A_n$ cease to occur eventually with probability one?
I think that the question is asking to show that $$P(A_n\ i.o.)=0$$ And I tried to prove that $$\sum_{n=1}^{\infty}P(A_n)<\infty$$ So that I can use the Borel-Cantelli Lemma.
But I did not get nowhere. I am kind of confused about all these operations includes limits, sets, and interplays between them, as well as the $limsup$ and $liminf$ stuff. Hope someone could help me.
First, to take advantage of the $\sum_n P(A_{k+1}\cap A_k^c)<\infty$ assumption, write$A_{n+2}\cap A_n^c=(A_{n+2}\cap A_{n+1}^c\cap A_n^c)\cup (A_{n+2}\cap A_{n+1}\cap A_n^c)\subset (A_{n+2}\cap A_{n+1}^c)\cup (A_{n+1}\cap A_n^c)$, and proceeding inductively, $$ A_{n+k}\cap A_n^c\subset \cup_{i=n}^{k-1}(A_{i+1}\cap A_i^c). $$
Next, $$ \cup_{k=n}^\infty A_k=\cup_{k=n}^\infty (A_k\cap A_n)\cup (A_k\cap A_n^c)\subset A_n\cup \{\cup_{k=n}^\infty (A_{k+1}\cap A_k^c)\}. $$
The last relation follows from the preceding display. We are given that $\sum_n P(A_{k+1}\cap A_k^c)<\infty$, which by borel-cantelli implies $P(\cup_{k=n}^\infty (A_{k+1}\cap A_k^c))\to 0$ as $n\to\infty$. We are also given $P(A_n)\to 0$. Therefore $$ P(\limsup A_n)=P(\cap_{n=0}^\infty\cup_{k=n}^\infty A_n)=\lim_{n\to\infty} P(\cup_{k=n}^\infty A_n)\le \lim_{n\to\infty} P(A_n)+P(\cup_{k=n}^\infty (A_{k+1}\cap A_k^c))=0. $$