Show $\sqrt{n}[\log(X_{(1)}) - \log(\alpha)] \overset{p}{\to} 0$ This is a follow-up question to $W_n = \frac{1}{n}\sum\log(X_i) - \log(X_{(1)})$ with Delta method. 
Note: $\log = \ln$. Note also that this is from a past qualifying exam, so I am looking for a solution which can be done under time constraints.
Please only use the following in your solutions: definitions of convergence in probability and distribution, Delta method, Central Limit Theorem, Slutsky's Theorem, Weak Law of Large Numbers, Continuous Mapping Theorem. I am not familiar with big-$O$ notation or being "bounded in probability." For those of you familiar with the literature, this is at the level of Casella and Berger, so any measure-theoretic approaches should not be used.
Suppose $X_1, \dots, X_n \sim \text{Pareto}(\alpha, \beta)$ with $n > \dfrac{2}{\beta}$ are independent. The Pareto$(\alpha, \beta)$ pdf is $$f(x) = \beta\alpha^{\beta}x^{-(\beta +1)}I(x > \alpha)\text{, } \alpha, \beta > 0\text{.}$$

If $X_{(1)}$ is the first order statistic, I wish to show
  $$\sqrt{n}[\log(X_{(1)}) - \log(\alpha)] \overset{p}{\to} 0\text{.}$$

What can I tell you about $X_{(1)}$ at this point of the exam?


*

*$X_{(1)} \overset{p}{\to} \alpha$.

*$n(X_{(1)} - \alpha)$ converges in distribution to an exponential distribution with mean $\alpha/\beta$.

*$F_{X_{(1)}}(x) = \begin{cases}
0, & x \leq \alpha \\
1-\left(\dfrac{\alpha}{x}\right)^{\beta n}, & x > \alpha\text{.}
\end{cases}$

*$\mathbb{E}[X_{(1)}] = \dfrac{\alpha\beta n}{\beta n - 1}$ 

*$\mathbb{E}[X_{(1)}^2] = \dfrac{\alpha^2\beta n}{\beta n - 2}$


One possible approach: let $G$ be the CDF of $\sqrt{n}[X_{(1)}-\alpha]$. Then the support is over $(0, \infty)$ and for $x$ in the support of $G$,
$$\begin{align}
G(x) &= F_{X_{(1)}}\left(\dfrac{x}{\sqrt{n}}+\alpha\right) \\
&= 1 - \left[ \dfrac{x/\sqrt{n}+\alpha}{\alpha}\right]^{-\beta n} \\
&= 1 - \left[ \dfrac{x/\alpha}{\sqrt{n}}+1\right]^{-\beta n} \\
\end{align}$$
From here, we could probably apply the Delta method if I could figure out how to compute the limit as $n \to \infty$. The $\sqrt{n}$ is problematic.
Given the above, I know that $\log(X_{(1)}) \overset{p}{\to} \log(\alpha)$, but I'm not sure if anything can be done with this. Alternative approaches are welcome as well.

Not sure if this is mentioned in the previous post and if you accept this or not. Note that 
$$Y_n = \sqrt{n}(\ln X_{(1)} - \ln\alpha) \sim \text{Exp}(\sqrt{n}\beta)$$
and thus CDF of $Y_n$ is
$$ F_{Y_{n}}(y) = \Pr\{\sqrt{n}(\ln X_{(1)} - \ln\alpha)\leq y\} =
\begin{cases}  1 - e^{-\sqrt{n}\beta y} & \text{when} & y > 0 \\
0 & \text{when} & y \leq 0
\end{cases}$$
So obviously you can show that $Y_n$ converge to $0$ in distribution, and thus in turns you show that it converges in probability.
