I have the following problem.
We have $Y_n$ independent random variables defined on $(\Omega, F, \Bbb{P})$ s.t. $$\Bbb{P}(Y_n=1)=p~~~\Bbb{P}(Y_n=0)=1-p$$ for $p\in [0,1]$. We define $A_0=0$ and $A_n=\sum_{k=0}^n Y_k$. Now I assume $0<p<1$ I need to use the Borell-cantelli lemma to sho that almost surly there is no index $N$ such that for $n\geq N$ the sequence $(A_n)$ is increasing.
Now I first need to rewrite the statement using $\limsup$. As I understood it I need to show that $\Bbb{P}(\limsup_{n\rightarrow \infty} A_{n+1}=A_n)=1$
But since I'm a bit unsure using $\limsup$ I wanted to ask if this is correct, so if I rewrote the scentence correclty with mathematical language or not. If not could you maybe explain this a bit because as I said I'm always a bit confused using $\limsup$ in stochastics.
Thanks for your help
Note $$\begin{aligned}P(\{\omega:A_n(\omega)\textrm{ is increasing after some }N\})&=P(\{\omega:\exists N \textrm{ s.t. }Y_n(\omega)=1,\,\forall n\geq N\})=\\ &=P(Y_n=0\textrm{ finitely often)}=\\ &=1-P(Y_n=0\textrm{ i.o.})\end{aligned}$$ Now $(Y_n)_{n \in\mathbb{N}}$ are IID and then by BC II $$\sum_{n \in \mathbb{N}}P(Y_n=0)=\infty \implies P(Y_n=0\textrm{ i.o.})=1$$ and therefore our set has measure zero.