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I am learning about the Borel-Cantelli lemmas, which I think allow us to conclude that things converge "almost surely" to other things. I am having trouble with a homework question:

Show: If $\mathbb{P}(A_n) \rightarrow 0$ and $\sum_{n=1}^{\infty}\mathbb{P}(A_n^c\cap A_{n+1})<\infty$ then $\mathbb{P}(A_n$ $i.o.) = \mathbb{P}(\bigcap_{k=1}^\infty \bigcup_{n\geq k} A_n) = 0$

My solution thus far:

Edit: After reading Davide's hint

$\begin{align*} \bigcup_{n\geq k}A_n &= A_k \cup (\bigcup_{n\geq k}A_n^c\cap A_{n+1}) \\ \bigcap_{k=1}^{\infty}\bigcup_{n\geq k}A_n &= \bigcap_{k=1}^{\infty}(A_k \cup (\bigcup_{n\geq k}A_n^c\cap A_{n+1})) \\ \mathbb{P}(\bigcap_{k=1}^{\infty}\bigcup_{n\geq k}A_n) &= \mathbb{P}(\bigcap_{k=1}^{\infty}(A_k \cup (\bigcup_{n\geq k}A_n^c\cap A_{n+1}))) \\ lim_{k\rightarrow\infty}\mathbb{P}(\bigcup_{n\geq k}A_n) &= lim_{k\rightarrow\infty}\mathbb{P}(A_k \cup (\bigcup_{n\geq k}A_n^c\cap A_{n+1})) \\ &\leq lim_{k\rightarrow\infty}\mathbb{P}(A_k) + \sum_{n\geq k}\mathbb{P}(A_n^c\cap A_{n+1}) \\ &=? \text{ }0 \end{align*}$

If I use my assumption that $\mathbb{P}(A_n) \rightarrow 0$, I know that the left part of the final line will tend to zero, but can I say that with $\sum_{n=1}^{\infty}\mathbb{P}(A_n^c\cap A_{n+1})<\infty$ implies that the right side tends to zero?

My reasoning is that as n tends to infinity, $P(A_{n+1})=0$ and $A_n^c\cap A_{n+1} \subseteq A_{n+1}$, so its probability goes to 0. But I am not sure.

And if I can't say this, how should I use my second assumption to show that this probability is then zero?

If so, I think the reasoning is good and I am done, but I am not sure.

Thanks!

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    For the series, note that it is the tail of a convergent series. Such a thing converges to $0$ necessarily, because the other, omitted terms from the whole convergent series approach this series (by definition of a convergent series).2012-10-25

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As the sequence $\{\bigcup_{n\geq k}A_n\}_k$ is decreasing, we have to show that $\lim_{k\to +\infty}P\left(\bigcup_{n\geq k}A_n\right)=0$. Using the idea in the OP, we have $P\left(\bigcup_{n\geq k}A_n\right)\leq P(A_k)+\sum_{n\geq k}P(A_{n+1}\setminus A_n).$