What happens to the limiting property of $\frac{\hat{\theta_n}}{n^{1/2}}$ if n goes to infinity?

Suppose $\hat{\theta_n}$ is a consistent estimator for $\frac{\sigma}{\mu}$. Specifically, it takes the following form:

$$\hat{\theta_n} = \frac{n^{-1}\Sigma (X_i -\bar{X_n})^2}{n^{-1}\Sigma X_i}$$

If $\mu = 0$, what happens to the limiting properties of $\frac{\hat{\theta_n}}{n^{1/2}}$?

Doesn't it still go to infinity because it converges in probability to something that explodes?

Solution 1:

Assuming all variables are IID with mean $0$ and variance $\sigma^2$, in this case $n^{-1/2}\hat{\theta}_n \stackrel{D}{\to} \sigma/Z$ by Slutsky's theorem and continuous mapping, where $Z\sim N(0,1)$. The only flaw I see with the solution in the comments is that the normal variable seems like it should be in the denominator.