Suppose that $Z_1,Z_2,...$ are random variables with $Z_n\sim\mathrm{Exp}(1)$. (We do not assume that these random variables are independent.) Show that $Z_n/\big(\ln^2(n)\big)$ converges to $0$ almost surely.
To be honest I am not really sure where to start with this problem , I believe I should start with the lim(sup) of the given function summed up. But after that I have no idea.
Help would be appreciated, thanks.
It suffices to show that $P\Big(\bigcap_{m\geq 1} \bigcup_{N\geq 1}\bigcap_{n\geq 1} \big[ Z_n/\big(\ln^2(n)\big)\leq \frac 1m\big]\Big) = 1$.
In turn, it suffices to show that $\forall m\in \mathbb N\setminus\{1\}$, $P\Big( \bigcup_{N\geq 1}\bigcap_{n\geq 1} \big[ Z_n/\big(\ln^2(n)\big)\leq \frac 1m\big]\Big) = 1$, which is the same as $P\Big(\liminf_n [Z_n/\big(\ln^2(n)\big)\leq \frac 1m]\Big)=1$, or $P\Big(\limsup_n [Z_n/\big(\ln^2(n)\big)>\frac 1m]\Big)=0$.
The last equality follows from Borel-Cantelli. I leave the details to you, but it suffices to show that for fixed $m$, the series $\sum_{n\geq 1} \exp(-\frac{\ln^2(n)}{m})$ is convergent. Note that for $n$ large enough, $\frac{\ln(n)}m \geq 2$ and compare with $\sum_{n\geq 1} \frac 1{n^2}$.