Borel Cantelli Problem - Normal Random Variables

Question

Let $\{Z_n\}_{n=1}^\infty$ be a sequence of i.i.d standard random variables. Prove that $P(\limsup_n |Z_n|/\sqrt{2ln(n)}=1)=1$. (Attempt) It is clear that both B.C lemmas will be used. Let $\delta>0$ be arbitrary and choose the sets $A_n=\{|Z_n|/\sqrt{2ln(n)}<1+\delta\}$ and $B_n=\{|Z_n|/\sqrt{2ln(n)}>1-\delta\}$. Once I show that the series $\sum P(A_n)$ is finite and $\sum P(B_n)$ is infinite I am done. However I have no estimate on $P(A_n)$ or $P(B_n)$ as the c.d.f is messy to work with.

Answer 1

I suppose by 'standard' you mean 'standard normal'.

Hint: Consider $\int_c^{\infty} e^{-x^{2}/2} dx$. Use L\Hopital's Rule to show that $\frac {\int_c^{\infty} e^{-x^{2}/2} dx} {\frac 1 c e^{-c^{2}/2}} \to 1$ as $c \to \infty$. This fact makes it fairly easy to show that one of the series under consideration is divergent and the other convergent.