I have the following problem regarding random walks where I need to show that:

> 
> $$P\left(\frac{\sum\_{k=1}^n Y\_k}{n} \not \to
> 0\right)=P\left(\bigcup\_{N=1}^\infty\bigcap\_{m=1}^\infty\bigcup\_{n>m}\left|\frac{\sum\_{k=1}^n Y\_k}{n}\right| \geq \frac1N\right)$$
> 
> 
> 

where the $Y\_i$s are i.i.d. random variables. I'm given a hint that I should use the definition of convergence here, but I don't see at least yet how that would help me. 

> 
> **My question is: How do I prove this equality?**
> 
> 
> 

This is what I have tried so far: 

I denote $X\_n=\sum\_{k=1}^n Y\_k$. I began by thinking that $P\left(\frac{X\_n}{n} \not \to
0\right)$ consists from those events where the average $\frac{X\_n}{n}$ either converges to some $L\neq0$ or does not converge at all. That is, the set of events 

$$\left\{\frac{X\_n}{n} \not \to
0\right\}=\left\{\frac{X\_n}{n}\to L\neq0\right\}\bigcup\left\{\frac{X\_n}{n} \to \pm\infty\right\}\triangleq A.$$

My idea was that if I can show that $A=\left\{\bigcup\_{N=1}^\infty\bigcap\_{m=1}^\infty\bigcup\_{n>m}\left|\frac{X\_n}{n}\right| \geq \frac1N\right\}$ then I would have solved this problem. 

Next I took a closer look at the set $\left\{\bigcap\_{m=1}^\infty\bigcup\_{n>m}\left|\frac{X\_n}{n}\right| \geq \frac1N\right\}$. It seemed to me that: 

$$\left\{\bigcup\_{n>1}\left|\frac{X\_n}{n}\right| \geq \frac1N\right\}\supseteq \left\{\bigcup\_{n>2}\left|\frac{X\_n}{n}\right| \geq \frac1N\right\}\supseteq \left\{\bigcup\_{n>3}\left|\frac{X\_n}{n}\right| \geq \frac1N\right\}\supseteq \cdots,$$

but now I got stucked and don't know yet how to proceed.