Convergence to a distribution implies convergence of a logarithm? Let $X_n$ be a sequence of almost surely positive real-valued random variables s.t. $$\sqrt{n} \, \left( X_n -a \right) \to_D \mathcal{N} ( 0, 1)$$
where $\to_D$ denotes convergence in distribution and $a>0$. Now I'm interested in what happens to$$\sqrt{n} \, \log \left( \frac{X_n}{a} \right)$$as $n \to \infty$.
My thoughts were the following: By expanding $\log(x)$ to a series at x=1 we get $$\sqrt{n} \, \log \left( \frac{X_n}{a} \right)= \sqrt{n} \, \frac{X_n-a}{a} + \sqrt{n} \,\mathcal{O} \left( \left( X_n-a \right)^2 \right),$$
and because the higher order terms tend to zero in probability
$$\sqrt{n} \, \log \left( \frac{X_n}{a} \right) \to_D \mathcal{N} \left( 0, \frac{1}{a^2} \right).$$

THIS WAS DUE TO A SIMPLE ERROR IN MY CODE
However, numerical simulations seem to suggest that actually
$$\sqrt{n} \, \log \left( \frac{X_n}{a} \right) \to_P 0,$$where $\to_P$ denotes convergence in probability. I'd appreciate any comments on why my reasoning may not be valid. Even better if someone has an idea of how to do this correctly.

Not a homework.

Indeed $\sqrt{n}\log(X_n/a)$ converges in distribution to a Gaussian $\mathcal{N}(0,1/a^2)$. One way to prove is to use the identity: 
$$ \frac{x}{1+x} \leq \log(1+x) \leq x $$ 
which holds for all $x>-1$ (i.e., whenever $\log(1+x)$ is nicely defined). 

So now define $G_n = \sqrt{n}(X_n-a)$.  Then $\log(X_n/a) = \log(1 + \frac{G_n}{a\sqrt{n}})$ and so the above identity gives: 
$$ \frac{\frac{G_n}{a\sqrt{n}}}{1+\frac{G_n}{a\sqrt{n}}} \leq \log(X_n/a) \leq \frac{G_n}{a\sqrt{n}} $$
Multiplying by $\sqrt{n}$ gives: 
$$ \frac{G_n}{a}\left(\frac{1}{1+ \frac{G_n}{a\sqrt{n}}}\right) \leq \sqrt{n}\log(X_n/a) \leq \frac{G_n}{a} $$

Now define: 
\begin{align} 
M_n &=\sqrt{n}\log(X_n/a) \\
Z_n &= \frac{1}{1+\frac{G_n}{a\sqrt{n}}}
\end{align} 
Thus, 
$$  \frac{G_nZ_n}{a} \leq M_n \leq \frac{G_n}{a}  $$
Define $N$ as a Gaussian $\mathcal{N}(0,1/a^2)$. 
Note that $G_n/a$ converges to $N$ in distribution, and $Z_n$ converges to 1 in distribution. 
Upper bound:
For all $x \in \mathbb{R}$ we have: 
$$ Pr[M_n\leq x] \geq Pr[G_n/a \leq x]$$
and so 
$$ \liminf_{n\rightarrow\infty} Pr[M_n \leq x] \geq Pr[N \leq x] $$
Lower bound: For simplicity, fix $x>0$ (similar techniques can be used for $x \leq 0$). We have: 
$$ Pr[M_n \leq x] \leq Pr[G_nZ_n/a \leq x] $$
For all $\epsilon>0$: 
$$ \{G_nZ_n/a \leq x\} \subseteq \{G_n/a \leq x + \epsilon\} \cup \{Z_n < x/(x+\epsilon)\} $$
So: 
$$Pr[M_n \leq x] \leq Pr[G_n/a \leq x+ \epsilon] + Pr[Z_n < x/(x+\epsilon)]\rightarrow Pr[N\leq x+\epsilon] $$
This holds for all $\epsilon>0$, and so (assuming $x>0$): 
$$ \limsup_{n\rightarrow\infty}Pr[M_n \leq x] \leq Pr[N\leq x] $$

The upper and lower bounds together imply $\lim_{n\rightarrow\infty} Pr[M_n\leq x] = Pr[N\leq x]$. 
