Proving a sequence of random variables converges under certain conditions 
Let $\{X_{n}\}$ be a sequence of independent standard normal random
  variables with $S_{n} := \{X_{n} > \sqrt{\alpha \ln n}\}$ . Also set
  $S = \limsup_{n\to\infty} S_{n}$. 
(a) Prove if $\alpha > 2$ then $P(S) = 0$.
(b) Prove if $\alpha = 2$ then $P(S) = 1$.
(c) Prove the sequence $X_{n}/\sqrt{\ln(n)}$ converges to $0$ in probability but
  not almost surely.

I am studying for my exam but I am not so sure about this problem. I will really appreciate any help. I have tried to utilize many tools such as Borel-Cantelli Lemma to write the sets as the intersection of union. I think Borel-Cantelli Lemma is correct here but I have had no help for (a) and (b).
In (c) I am given a hint to use the result $(x + 1/x)^{-1} e^{-x^2/2} \leq \int_{x}^{\infty} e^{-y^2/2}\mathop{dy} \leq e^{-x^2/2}/x$. 
But I do not quite  see how to show prove it converges in probability but not almost surely. I tried doing something like $\lim_{n\to\infty} Pr(|X_{n}| > \epsilon)$ and I tried to apply Markov or Chebyshev  inequality with no luck.  Of course I need to use the fact that the $X_i$'s are standard normal random variables here. This must be how they get the exponential terms. But I have had no luck.
I will greatly appreciate your assistance in the problem.
 A: The inequality which you are given should be rather used to prove a), b) and not c).
For a): It suffices to show that $$\mathbb{P}(X_n > \sqrt{\alpha \ln n} \, \, \text{infinitely often})=0.$$ By Borel-Cantelli, this is true if $$\sum_{n \geq 1} \mathbb{P}(X_n> \sqrt{\alpha \ln n})< \infty. \tag{1}$$ Using the upper bound from your hint, we have $$\mathbb{P}(X_n> \sqrt{\alpha \ln n}) \leq \sqrt{2\pi} \frac{1}{\sqrt{\alpha \ln n}} \underbrace{\exp \left(- \frac{\alpha \ln(n)}{2} \right)}_{n^{-\alpha/2}}$$ and this gives the convergence of the series in $(1)$.
For b) It suffices to show that $$\mathbb{P}(X_n \geq \sqrt{2 \ln n} \, \, \text{infinitely often})=1.$$ By Borel-Cantelli, this is true if $$\sum_{n \geq 1} \mathbb{P}(X_n> \sqrt{\alpha \ln n})= \infty. \tag{2}$$Use the lower bound from your hint to prove this.
For c): It is immediate from b) that $X_n/\sqrt{\ln n}$ cannot converge almost surely to zero (because otherwise the probability in b) would be zero). To prove convergence in probability, just apply Markov's inequality and use that $\mathbb{E}(X_n^2)=1$:
$$\mathbb{P}\left( \left| \frac{X_n}{\sqrt{\ln n}} \right| > \epsilon \right) \leq \frac{1}{\epsilon^2 \ln n} \mathbb{E}(X_n^2) = \frac{1}{\epsilon^2 \ln n} \xrightarrow[]{n \to \infty} 0.$$
A: Hints: You have to use the stated  inequalities for all three parts.
We have $P(S_n) \leq e^{-\alpha ln n /2}\frac 1 {\sqrt {\alpha \ln n}}$. Note that $e^{-\alpha ln n /2}=\frac 1 {n^{\alpha /2}}$. If $\alpha >2$ then $\sum P(S_n) <\infty$ so $P(S)=0$ by Borel -Cantelli Lemma. If $\alpha =2$ then the left of the given inequality shows that $\sum P(S_n)=\infty$ so $P(S)=1$ by the second part of Borel -Cantelli Lemma.
Now $\frac {X_n} {\sqrt n} \to 0$ in probability because $P(\frac {X_n} {\sqrt n} >\epsilon) \leq (\frac 1 {\sqrt {n\epsilon}}) E|X_n|$. (You will have to use symmetry of the distribution to consider $P(-\frac {X_n} {\sqrt n} >\epsilon)$]. If $\frac {X_n} {\sqrt n}$ converges almost surely then $\sum P(\frac {X_n} {\sqrt n} >1))<\infty$ by Borel -Cantelli Lemma. Use the left side inequality to get a  contradiction. [This is just the case $\alpha =1$ of b)].
