The first Borel-Cantelli Lemma states
If $\sum \mathbf{P}(A_n) < \infty$ then $\mathbf{P}(A_n i.o.) = 0$
Question :
If $\mathbf{P}(A_n) \rightarrow 0$ and $\sum \mathbf{P}(A_{n+1} / A_{n}) < \infty$ then $\mathbf{P}(A_n i.o.) = 0$
Now of course $\sum \mathbf{P}(A_{n+1} / A_{n}) < \infty$ does not imply $\sum \mathbf{P}(A_n) < \infty$. For example we can take $\mathbf{P}(A_n) = 1/n$.
So how can we takle such a problem.
My approach: We can find indices $n_k$ s.t. $\sum_{n_k}^{n_{k+1}}\mathbf{P}(A_n) < 1/k^2$ $k = 1,2,3..$ Define an event $B_k = \cup_{n_k}^{n_{k+1}}\mathbf{P}(A_n) $ Then By Borel Cantelli $\mathbf{P}(B_n i.o.) = 0$ Thus as only a finite number of $B_n$s happen we conclude on a finite number of $A_n$s happen.
Any help is appreciated
Set $B_n := A_{n+1} \setminus A_n$. Then by Borel-Cantelli we obtain $$\mathbb{P}(B_n \quad i.o.) = 0 $$ since $\sum_{n \in \mathbb{N}} \mathbb{P}(B_n) < \infty$. This means that for almost all $\omega \in \Omega$ there are only finitely many $n \in \mathbb{N}$ such that $\omega$ is in $A_{n+1}$ but not in $A_n$. Thus we can choose $N$ sufficiently large such that either for all $n > N$ it is $\omega \in A_n$ or for all $n > N$ it is $\omega \not \in A_n$. Intuitively $\omega$ cannot change infinitely often between being inside of $A_n$ or being outside of $A_n$.
It remains to show that the first is impossible, i.e. that for almost all $\omega$ it is not $\omega \in A_n$ for all $n > N$. For that we use the assumption $\lim_{n \rightarrow \infty} \mathbb{P}(A_n) = 0$, which yields $$ \mathbb{P}(\bigcap_{n>N} A_n) \leq \mathbb{P}(A_n) \rightarrow 0 \quad (n \rightarrow \infty).$$ Thus for almost all $\omega \in \Omega$ it is eventually $\omega \not \in A_n$.