I have a sample $(X_1, X_2, \ldots, X_n)$ from the density,
$$f_\theta(x) = \frac{\theta}{2} \exp(-\theta |x|), \quad -\infty < x<\infty$$
for some $\theta>0$. Let $V_n$ be the MLE for $\frac1\theta$, find the limiting distribution for $ \sqrt{n}(V_n - \frac1\theta)$ as $n \rightarrow \infty$.
I began by finding the MLE.
$$L(\theta\mid \underline{x}) =\left( \frac{\theta}{2} \right)^n \exp\left(-\theta \sum^n_{i =1} |X_i|\right) $$
$$l = \log(L(\theta \mid \underline{x})) = n \log(\theta) - n \log(2) -\theta\sum^n_{i=1} |x_i|$$
$$\frac{\partial l }{\partial \theta} =\frac{n}{\theta} - \sum^n_{i=1} |X_i| \overset{\text{set}}{=}0$$
$$\Rightarrow \widehat{\frac{1}{\theta}}_\text{MLE} = \frac{\sum ^n_{i=1} |x_i|}{n}$$
Now we can let $Y_i$ = $|X_i|$, then $Y_i \sim \operatorname{Expo}(\frac{1}{\theta})$ and $\sum^n_{i=1} Y_i \sim \Gamma(n, \frac{1}{\theta})$.
Thus $\overline{y} \sim \Gamma(n, \frac{n}{\theta})$.
Now I began working on the limiting distribution.
$$P(\sqrt{n}(V_n - \frac{1}{\theta})<v)$$
$$P\left(V_n <\frac{v}{\sqrt{n}} + \frac{1}{\theta}\right)$$
$$= \int^{\sqrt{n} + \frac{1}{\theta}} _{0} \Gamma(n)^{-n} \left ( \frac{\theta}{n} \right )^n V_n^{n-1} \exp \left ( \frac{-V_n n}{\theta} \right) \, dV_n$$
It got worse from there. I think there must be a trick that I am missing.