Can one deduce $\lim_{n\rightarrow\infty}P(X_n>1)=0$ from $P(X_n>1+\frac{x}{2\log n})\leq e^{-x}\quad\forall x$ I want to control the tail probability of some random variable $(X_n)_n$. (In particular, $X_n = \max_{1\leq i\leq n} N_i$, where $N_i$ are independent standard normal)
I come up with the following two upper bounds:
(1)$$P(\frac{X_n}{\sqrt{2\log n}}>1+\frac{x}{2\log n})\leq e^{-x}\quad\forall x>0.$$
Can I deduce $\lim_{n\rightarrow\infty}P(\frac{X_n}{\sqrt{2\log n}}>1)=0$ from it?
I only know that for every $\epsilon_1,\epsilon_2>0$, there exists $N_1$ such that for every $n>N_1$, we have
$$P(\frac{X_n}{\sqrt{2\log n}}>1+\epsilon_1)\leq \epsilon_2.$$ I have no idea how to do next.
(2)$$P(\frac{X_n}{\sqrt{2\log n}}\leq \alpha)\leq (1-n^{-\alpha+o(1)})^n\quad 0<\alpha< 1 .$$
Can I deduce $\lim_{n\rightarrow\infty}P(\frac{X_n}{\sqrt{2\log n}}< 1)=0$ from it?
I only know $\lim_{n\rightarrow\infty}(1-n^{-\alpha+o(1)})^n=0$. Don't know how to deal with $\alpha$ inside the probability.
(3) Suppose we have $$P(\frac{X_n}{\sqrt{2\log n}}>1+\frac{x}{2\log n})\leq e^{-x}\quad\forall x>0.$$ and $$P(\frac{X_n}{\sqrt{2\log n}}\leq \alpha)\leq (1-n^{-\alpha+o(1)})^n\quad 0<\alpha< 1 .$$
Can we deduce $\frac{X_n}{\sqrt{2\log n}}\rightarrow 1$ in distribution?

First, we determine the probability function of $X_n$, we have:
$$\begin{align}
P(X_n<a)&=P(\max_{i=1,..,n}N_i<a)\\
&=P^n(N_i<a)\\
&=\Phi^n(a)
\end{align}$$
With $\Phi(\cdot)$ the cumulative distribution function of $\mathcal{N}(0,1)$.
Return back to the main problem, it suffices to prove that for all $\epsilon>0$, there exists $N$ such that for all $n>N$, we have  $P(\frac{X_n}{\sqrt{2\log n}}>1)<\epsilon$.
We have
$$\begin{align}
P(\frac{X_n}{\sqrt{2\log n}}>1)&=P\left(\frac{X_n}{\sqrt{2\log n}}>1+\frac{x}{2\log n}\right)+P\left(1<\frac{X_n}{\sqrt{2\log n}}<1+\frac{x}{2\log n}\right) \\
&=P\left(\frac{X_n}{\sqrt{2\log n}}>1+\frac{x}{2\log n}\right)+P\left(\sqrt{2\log n}<X_n<\sqrt{2\log n}+\frac{x}{\sqrt{2\log n}}\right) \tag{1}\\
\end{align}$$
We know already how to bound the first term of $(1)$, so let's study the second term, we have
$$\begin{align}
L&:=P\left(\sqrt{2\log n}<X_n<\sqrt{2\log n}+\frac{x}{\sqrt{2\log n}}\right) \\
&=\Phi^n\left(\sqrt{2\log n}+\frac{x}{\sqrt{2\log n}}  \right)  -\Phi^n\left(\sqrt{2\log n}  \right) \\
\end{align}$$
Applying the mean value theorem with $(a,b) = \left(\sqrt{2\log n}+\frac{x}{\sqrt{2\log n}},\sqrt{2\log n} \right)$ and $f=\Phi^n(\cdot)$, there exists $c$ such that $\sqrt{2\log n} <c<\sqrt{2\log n}+\frac{x}{\sqrt{2\log n}}$ and
$$L =\frac{x}{\sqrt{2\log n}}\left( \Phi^n(c) \right)' = \frac{x}{\sqrt{2\log n}}\cdot n \cdot \Phi^{n-1}(c)\cdot\phi(c)< \frac{x}{\sqrt{2\log n}}\cdot n \cdot \phi(c)\tag{2}$$
with the normal density function $\phi(x) =\frac{\exp{-\frac{x^2}{2}}}{\sqrt{2\pi}}$.
As $\phi(\cdot)$ is decreasing for $x>0$, then
$$\phi(c)<\phi(\sqrt{2\log n}) = \frac{\exp{\left(-\frac{2\log n}{2} \right) }}{\sqrt{2\pi}} = \frac{1}{n\sqrt{2\pi}}\tag{3}$$
From $(2),(3)$, we deduce that
$$L<\frac{x}{2\sqrt{\pi \log n}}\tag{4}$$
From $(1),(4)$, we have
$$P(\frac{X_n}{\sqrt{2\log n}}>1)< e^{-x}+\frac{x}{2\sqrt{\pi \log n}} \tag{5}$$
For all $\epsilon >0$, set
$$x= - \ln \left(\frac{\epsilon}{2} \right)$$ and $$N>\exp\left(-\frac{1}{\pi\epsilon^2} \ln \left( \frac{\epsilon}{2}  \right)  \right)$$
Then, for all $n>N$, the two terms on RHS of $(5)$ are both smaller than $\frac{\epsilon}{2}$. Then, for all $n>N$,
$$P\left(\frac{X_n}{\sqrt{2\log n}}>1\right)< \epsilon$$
We can conclude that
$$\lim_{n\to +\infty}P\left(\frac{X_n}{\sqrt{2\log n}}>1\right) = 0$$
PS:  $\frac{X_n}{\sqrt{2\log n}} \xrightarrow{n\infty}{1}$ in distribution if and only if $\forall x$
$$\lim_{n\to +\infty}P\left(\frac{X_n}{\sqrt{2\log n}}<x\right)  = \mathbb{I}_{\{ x \ge 1 \}}$$
So, your last result is not sufficient to prove that $\frac{X_n}{\sqrt{2\log n}}$ converge to $1$ in distribution.
