R.v. Normal distribution 
Let $(X_1, X_2 ,\ldots )$ be an independent sequence of normally distributed random variables with mean $0$ and standard deviation $\sigma$. How can I find an increasing sequence $(a_1, a_2, \ldots )$ such that $$\lim sup \frac{X_n}{a_n}=1  \  \ a.s$$
And if $Y_n= \max_{1\leq k < n} X_n$ Is it true that $\lim \frac{Y_n}{a_n}=1  \  \ a.s$?

My attempt:
First I have proof that if $X$ is a normally distributed random variable having mean $0$ and standard deviation $\sigma$, then as $x \rightarrow \infty$, $$ P(\{w: X(w) > x\}) \sim \frac{\sigma}{x\sqrt{2\pi}} \  exp(-x^2 /2\sigma^2 ) \ (\ast)$$
Let $\delta >0$, for $x>\frac{\sigma}{\sqrt{\delta}}$, we have
$$ \frac{1}{\sqrt{2\pi\sigma^2}}\int_x^{\infty} e^{-u^2 /2\sigma^2} du < \frac{1}{\sqrt{2\pi\sigma^2}}\int_x^{\infty} (1+\frac{\sigma^2}{u^2})e^{-u^2 /2\sigma^2} du < \frac{1+\delta}{\sqrt{2\pi\sigma^2}}\int_x^{\infty} e^{-u^2 /2\sigma^2} du$$
The expression between the two inequality signs is equal to the right side of ($\ast$).
I replaced the integrand by a slightly dierent integrand that has a simple antiderivative. I used integration by parts along a few diferent paths, and, then, if, for one of these paths, the new integral is small compared with the original integral, combined it with the original integral.
Ok, My teacher told me that it is correct that argument and for the sequence $(a_1,a_2,\ldots)$ , he suggested me $a_n=\sqrt{2\sigma^2 \log n}$, clearly this is an increasing sequence but...
How can I construct or find this sequence such that $\lim sup \frac{X_n}{a_n}=1  \  \ a.s$ given that $ P(\{w: X(w) > x\}) \sim \frac{\sigma}{x\sqrt{2\pi}} \  exp(-x^2 /2\sigma^2 )$ ?
Could someone help me for this pls. Thanks for your time and help.
 A: You can prove that $P(\limsup_n \frac{X_n}{\sigma \sqrt{2 \log n}} = 1) =1$. It suffices to prove that $$P(\limsup_n \frac{X_n}{\sigma \sqrt{2 \log n}} > 1) = P(\limsup_n \frac{X_n}{\sigma \sqrt{2 \log n}} < 1)=0$$

Let $\displaystyle Y_n = \frac{X_n}{\sigma \sqrt{2 \log n}}$.
Note that $\displaystyle \limsup_n \frac{X_n}{\sigma \sqrt{2 \log n}} > 1 = \{w\in \Omega, \limsup_n \frac{X_n(w)}{\sigma \sqrt{2 \log n}} > 1\}  = \bigcup_N \bigcap_n \bigcup_{k\geq n}\left(Y_k> 1+\frac 1N \right)$.
It suffices to prove that $\forall N, P\left(\bigcap_n \bigcup_{k\geq n}\left(Y_k> 1+\frac 1N \right) \right)=0$.
But $P\left(\bigcap_n \bigcup_{k\geq n}\left(Y_k> 1+\frac 1N \right) \right)=P\left(\limsup_n \left( Y_n> 1+\frac 1N\right) \right)$.
By Borel-Cantelli, it suffices to prove that $\sum_n P\left( Y_n> 1+\frac 1N\right)$ converges. Note that $$P\left( Y_n> 1+\frac 1N\right) = P\left(X_n > \sigma \left( 1 + \frac 1N \right)\sqrt{2\log n}\right)$$
We have the standard estimate $$\left(\frac{\sigma}x-\frac{\sigma^3}{x^3} \right)e^{-\frac{x^2}{2\sigma^2}}\leq P(X_n>x)\leq \frac{\sigma}xe^{-\frac{x^2}{2\sigma^2}}$$
Hence $\displaystyle P\left( Y_n> 1+\frac 1N\right) \leq \frac{1}{n^{\left(1+\frac 1N\right)^2}\sqrt{2\log n}}$ and the series converges.
This proves $P(\limsup_n \frac{X_n}{\sigma \sqrt{2 \log n}} > 1) =0$.

Note that $$ \begin{aligned}[t] \limsup_n \frac{X_n}{\sigma \sqrt{2 \log n}} < 1 &= \{w\in \Omega, \limsup_n \frac{X_n(w)}{\sigma \sqrt{2 \log n}} < 1\}  \\ &\subset \bigcup_n \bigcap_{k\geq n}\left(\frac{X_k}{\sigma \sqrt{2 \log n}} < 1\right)\\ &= \liminf_n \left( \frac{X_n}{\sigma \sqrt{2 \log n}} < 1\right) \end{aligned}$$
Since $\liminf_n \frac{X_n}{\sigma \sqrt{2 \log n}}<1=\left(\limsup_n \frac{X_n}{\sigma \sqrt{2 \log n}}>1 \right)^c $, it suffices to prove that $$P\left( \limsup_n \frac{X_n}{\sigma \sqrt{2 \log n}}>1\right)= 1$$
As expected we make use of the second Borel Cantelli lemma : $$P\left( \frac{X_n}{\sigma \sqrt{2 \log n}}>1\right) =P\left( X_n>\sigma \sqrt{2\log n}\right)\geq \frac{1}{\sqrt{2 \log n}}\left(1-\frac{1}{2\log n} \right)\frac 1n$$
Since $$\frac{1}{\sqrt{2 \log n}}\left(1-\frac{1}{2\log n} \right)\frac 1n \sim \frac{1}{\sqrt{2}n \sqrt{\log n}}$$the sum diverges, proving the claim.
