Find $c$ such that $\text{P}(\limsup\limits_{n\to \infty} X_n/\sqrt{\log n}=c)= 1$ Need to find $c$ such that $\text{P}(\limsup\limits_{n\to \infty} X_n/\sqrt{\log n}=c)= 1$, where $X_n$ are a sequence of independent random variables such that $X_k\sim\mathcal{N}(0,1) $ I need to use the fact that
$\sqrt{2\pi}\xi\exp\left({\xi^2\over 2}\right)\text{P}(X\geq \xi)\to 1$ as $\xi \to \infty$
This implies that $\exists \alpha,\beta>0$ such that $\forall k,n\geq1$
$$\frac{\alpha}{\sqrt{2\pi}\gamma n^{\gamma^2} \sqrt{2 \log n}} \leq \text{P}(X\geq\gamma\sqrt{2 \log n})\leq \frac{\beta}{\sqrt{2\pi}\gamma n^{\gamma^2}\sqrt{2\log n}}$$
Just by setting $a=\gamma\sqrt{2\log n}$ in the previous relation:
$$\text{P}(X\geq\gamma \sqrt{2\log n})\to \frac{1}{\sqrt{2\pi}\gamma n^{\gamma ^2}\sqrt{2\log n}}$$

Is this true in general for limits? $$\lim_{n\to\infty}A_n=A\Rightarrow\exists\alpha,\beta>0:\forall n\geq 1, \alpha A \leq A_n\leq\beta A$$

 A: Hint: One needs to find $c$ such that, for every $a\lt c\lt b$,
$$
\mathbb P\left(\limsup\limits_{n\to\infty}X_n/\sqrt{\log n}\leqslant a\right)=0,
\qquad
\mathbb P\left(\limsup\limits_{n\to\infty}X_n/\sqrt{\log n}\leqslant b\right)=1.
$$
Both limits stem from Borel-Cantelli lemma and from the estimations you recalled, stated more weakly as
$$
\mathbb P(X\geqslant\gamma\sqrt{2\log n})=n^{-\gamma^2+\varepsilon_n(\gamma)},\qquad\lim\limits_{n\to\infty}\varepsilon_n(\gamma)=0.
$$
In particular, if $\gamma\gt1$, there exists some $\delta\gt1$ and a finite $N$ such that, for every $n\geqslant N$,
$$
\mathbb P(X\geqslant\gamma\sqrt{2\log n})\leqslant n^{-\delta},
$$
hence the series $\sum\limits_n\mathbb P(X\geqslant\gamma\sqrt{2\log n})$ converges, while, if $\gamma\lt1$, there exists some $\delta\lt1$ and a finite $N$ such that, for every $n\geqslant N$,
$$
\mathbb P(X\geqslant\gamma\sqrt{2\log n})\geqslant n^{-\delta},
$$
hence the series $\sum\limits_n\mathbb P(X\geqslant\gamma\sqrt{2\log n})$ diverges. Note finally that in each case, $\sum\limits_n\mathbb P(X\geqslant\gamma\sqrt{2\log n})$ is also $\sum\limits_n\mathbb P(X_n\geqslant\gamma\sqrt{2\log n})$. Can you carry on from here?
Edit:

Is this true in general for limits? $$\lim_{n\to\infty}A_n=A\Rightarrow\exists\alpha,\beta>0:\forall n\geq 1, \alpha A \leq A_n\leq\beta A$$

This is logically unrelated to the rest of the post but a short answer is as follows. If $A=0$, no hope. If $A\gt0$ and $A_n\leqslant0$ for some $n$, no hope either. If $A\gt0$ and $A_n\gt0$ for every $n$, then $A_n/A\to1$ and $A_n/A\gt0$ for every $n$ hence indeed $(A_n/A)_{n\geqslant1}$ is bounded below by a positive constant and above by a finite constant, this the statement you suggest holds. Likewise if $A\lt0$ and $A_n\lt0$ for every $n\geqslant1$.
