Does $\sqrt{n} \left( \hat{\theta}_n - \theta \right) \overset{d}{\rightarrow} F_\infty$ imply $\hat{\theta}_n \overset{p}{\rightarrow} \theta$? Suppose 
$$
\sqrt{n} \left( \hat{\theta}_n - \theta \right) \overset{d}{\rightarrow} F_\infty,
$$ 
where $\hat{\theta}_n$ is a sequence of random variables, $\theta$ is a constant and $F_\infty$ is some distribution function. 
Does this imply that 
$$
\hat{\theta}_n \overset{p}{\rightarrow} \theta \; ?
$$
 A: YES.  Remember the definition of convergence in probability, $\hat{\theta}_n \overset{p}{\rightarrow} \theta$ means that for all $\delta > 0$ there is an $N=N_{\delta}$ such that for all $n \ge N$ we have $P(\mid \hat{\theta}_n - \theta \mid > \delta ) \le \delta$.
Now suppose we do NOT have convergence in probability. Then, there is some positive $\delta$ such that for all $n$ along some subsequence of $n$ going to infinity we have 
$$
   P(\mid \hat{\theta}_n - \theta \mid > \delta ) > \delta
$$
but that means that, along that subsequence, 
$$
\mid \sqrt{n} (\hat{\theta}_n-\theta) \mid > \sqrt{n} \delta
$$
which goes to infinity, so there is at least a probability mass $\delta$ that escapes to infinity (along that subsequence). That contradicts convergence in distribution. 
A: Write $\hat{\theta}_n - \theta$ as 
$$\frac{1}{\sqrt{n}}\times \sqrt{n}(\hat{\theta}_n - \theta),$$
then $\hat{\theta}_n - \theta \to_P 0$ is a direct consequence of Slutsky's Theorem.
Using the "$O_P, o_P$" notations in probability, this fact is usually stated as 
$$o_P(1)\cdot O_P(1) = o_P(1).$$
