This is the problem 3.5.1 of Shiryaev's Probability:
Let $\xi_1,\xi_2,\ldots$ be a sequence of independent identically distributed random variables with $E\xi_1=0$ and $E\xi_1^2=1$. Show that $\max\left(\frac{\mid\xi_1\mid}{\sqrt{n}},\ldots,\frac{\mid\xi_n\mid}{\sqrt{n}}\right)\overset{d}{\rightarrow} 0,\ n\to\infty$
I tried to solve it with this:
$\forall\varepsilon>0$, then $\lim_{n\to\infty}P\left\{\max\left(\frac{\mid\xi_1\mid}{\sqrt{n}},\ldots,\frac{\mid\xi_n\mid}{\sqrt{n}}\right)\leq\varepsilon\right\}=\lim_{n\to\infty}[P\{\mid\xi_1\mid\leq\sqrt{n}\varepsilon\}]^n=\lim_{n\to\infty}[1-P\{\mid\xi_1\mid>\sqrt{n}\varepsilon\}]^n=\exp[\lim_{n\to\infty}(-nP\{\mid\xi_1\mid>\sqrt{n}\varepsilon\}]$
How can I prove $\lim_{n\to\infty}(-nP\{\mid\xi_1\mid>\sqrt{n}\varepsilon\}\to 0\ ?$ Thank you!