Establishing convergence in probability from a related convergence in distribution Is it true that $\sqrt n (\hat{\theta}-\theta) \ \rightarrow_d \ N(0,\sigma^2)$ 
implies
$\text{plim} \ \hat{\theta} = \theta $?  If so, how can I prove this?    

Attempted proof: My proof is like this:
Since $\hat{\theta} \rightarrow_{asy.} N(\theta, \frac{\sigma^2}{n})$ I have $\lim_{n \to \infty} E(\hat{\theta})= \theta  \ $ and 
$\lim_{n \to \infty} Var(\hat{\theta})=0 $.  Therefore, from convergence in quadratic mean, $\text{plim} \ \hat{\theta} = \theta $.  Is this proof correct?
 A: We are given that
$$\sqrt n (\hat{\theta}-\theta) \to_d  N(0,\sigma^2) \tag{1} $$
Make the additional assumptions that
1) For the finite distribution of $\sqrt n (\hat{\theta}-\theta)$, the $2+\delta,\; \delta >0$ absolute moment exists and is finite.
2) The sequences of 1st and 2nd moments of $\{\sqrt n (\hat{\theta}-\theta)\}$ converge each to a constant.
Then these constants are the corresponding moments of the limiting distribution.
In particular, this means that
$$\lim \text {Var}[\sqrt{n}(\hat \theta -\theta)] = \sigma^2 \tag{2}$$
At the same time, we have
$$\lim \text {Var}[\sqrt{n}(\hat \theta -\theta)] = \lim \mathbb E[\sqrt{n}(\hat \theta -\theta)]^2 - \lim\left(\mathbb E[\sqrt{n}(\hat \theta -\theta)]\right)^2$$
$$= \lim \mathbb E[\sqrt{n}(\hat \theta -\theta)]^2 - \left(\lim\mathbb E[\sqrt{n}(\hat \theta -\theta)]\right)^2 = \lim \mathbb E[\sqrt{n}(\hat \theta -\theta)]^2 - 0 \tag{3}$$
Combining $(2)$ and $(3)$ we have
$$\lim \mathbb E[\sqrt{n}(\hat \theta -\theta)]^2 = \sigma^2 \implies \lim n\mathbb E(\hat \theta -\theta)^2 = \sigma^2 < \infty \implies \mathbb E(\hat \theta -\theta)^2 = O(1/n) $$
$$\implies \mathbb E(\hat \theta -\theta)^2 = o(1) \implies \mathbb E(\hat \theta -\theta)^2 \to 0 $$
The last result is convergence in quadratic mean - and convergence in quadratic mean of a random variable to a constant, implies that this constant is also its probability limit (see here for an exposition why).
So $\hat \theta_n \to_p \theta$.
While in general it does not hold that convergence in distribution implies convergence in probability, we see how, for a subset of cases and under some additional conditions, we can go from the limiting distribution to the probability limit.
