$X_1,...,X_n$ are independent identical random variables of standard normal distribution, $S_n=X_1+...+X_n$, $a_n \uparrow \infty$, try to give the necessary and sufficient conditions of
(1)$\frac{X_n}{a_n}\rightarrow0$ a.e.
(2)$\frac{S_n}{a_n}\rightarrow0$ a.e.
(3)$\frac{\max\{X_1,...,X_n\}}{a_n}\rightarrow0$ a.e.
My ideas so far: (1) since $\frac{X_n}{a_n}\rightarrow0$ is equivalent to $|\frac{X_n}{a_n}|\rightarrow0$, then for each $\epsilon>0$, I tried to proove $p\left(\limsup|\frac{X_n}{a_n}| \geq \epsilon\right)=0$. By Borel 0-1 lemma, this is equivalent to proof $\sum p(|\frac{X_n}{a_n}| \geq \epsilon) < \infty$.
Since $a_n$ is non-negative to some extent,
\begin{equation}
\sum p\left(|\frac{X_n}{a_n}| \geq \epsilon\right) =\sum p(|X_n| \geq \epsilon a_n)=2\int_{\epsilon a_n}^{\infty} \frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}dx=2\sum(1-F(\epsilon a_n))\
\end{equation}
Since for normal distribution $1-F(x)=p(X>x)$~$\frac{1}{2\pi x}e^{-\frac{x^2}{2}}$ as $x \rightarrow \infty$.
Therefore the summation above convergence iff
\begin{equation}
\sum \frac{1}{a_n}e^{-\frac{{\epsilon} a_n^2}{2}} < \infty\
\end{equation}
The proof of (2) and (3) seems more difficult, since $\{\frac{S_n}{a_n}\}$ are not independent any more, so I cannot use Borel 0-1 lemma to give a sufficient and necessary condition like problem(1).
The results are weired, and I am doubt whether my ideas are right?
Thanks in advance for any tips or help in general.