In _All of Statistics_, chapter 11 (pg. 183), Larry Wasserman states in his description of the Wald Test:
> We are testing the null hypothesis $ \hat{\theta} = \theta_0 $ versus the alternative hypothesis $ \theta \neq \theta_0 $. 

He then says we should assume that $ \hat{\theta} $ is asymptotically normal, i.e. converges in distribution to a standard normal:
$$
 \frac{\sqrt{n}(\hat{\theta} - \theta_0)}{\hat{\text{se}}} \rightsquigarrow N(0, 1).
$$

(_Wasserman uses $ \rightsquigarrow $ to denote converging in distribution._)

My impression is that this is a reasonable assumption because of the Central Limit Theorem.

But then, when proving the following theorem:
> _Asymptotically the Wald test has size $ \alpha $, that is, 
$$ \mathbb{P}_{\theta_0}\left(\lvert Z \rvert > z_{\alpha/2}\right) \rightarrow \alpha $$ as $ n \rightarrow \infty $,_ 

He says:
> Under $ \theta = \theta_0 $, $ (\hat{\theta} - \theta_0)/\hat{se} \rightsquigarrow N(0, 1). $

I don't see how $ (\hat{\theta} - \theta_0)/\hat{se} \rightsquigarrow N(0, 1) $ follows from the fact that 
$$
 \frac{\sqrt{n}(\hat{\theta} - \theta_0)}{\hat{\text{se}}} \rightsquigarrow N(0, 1).
$$

Doesn't this imply that
$$
\lim_{n \rightarrow \infty}  P\left(\frac{\sqrt{n}(\hat{\theta} - \theta_0)}{\hat{\text{se}}} \leq z\right) = \lim_{n \rightarrow \infty} P\left((\hat{\theta} - \theta_0)/\hat{se} \leq z\right),
$$
which is not true?

Is this a mistake or am I missing something?