Suppose that $(\hat{\theta}_n -\theta)/b_n \stackrel{d}{\to} N(0,1)$. Does this imply $b_n/se(\hat{\theta}_n){\to}1 $?
$b_n$ here is any sequence, but the example I have in mind is where $b_n$ is $1/\sqrt{n I(\theta)}$, where $I$ is the Fisher information of the MLE $\hat{\theta}_n$.
Basically, this is asking does an asymptotic approximation to the standard error for the MLE $b_n$ asymptotically equal to the actual standard error $se(\hat{\theta}_n)$.
Intuitively it must be true since $\hat{\theta}_n -\theta$ is approximately distributed $N(0,b_n^2)$, while on the other hand it is also approximately distributed $N(0,se(\hat{\theta}_n)^2)$. The converses is true by Slutsky's theorem. My question would be answered by the claim here, but this claim is stated without proof.
How can this be proved?
