How can I formally show that a root n consistent estimator is weakly consistent? How can I formally show that a root n consistent estimator is weakly consistent?
Heuristically its clear that if $\sqrt{n}(\theta_n-\theta_0)$ is bounded in probability then since $\sqrt{n} \to \infty$ as $n \to  \infty$ clearly $\theta_n-\theta_0 \in o_p(1)$ that is it goes to zero in probability but every time I try to write it formally I get stuck. Could someone help me out?
 A: I'm going to write your estimator with a hat on it: $\hat{\theta}_n$. The tricks involve playing around with subsets. If $A \subset B$, then $P(A) \le P(B)$. Observe that
\begin{align*}
P(|\hat{\theta}_n - \theta_0| > \epsilon) &= P(|\hat{\theta}_n - \theta_0| > \epsilon , \sqrt{n}|\hat{\theta}_n - \theta_0| \le M) + P(|\hat{\theta}_n - \theta_0| > \epsilon , \sqrt{n}|\hat{\theta}_n - \theta_0| > M) \\
&\le  P(|\hat{\theta}_n - \theta_0| > \epsilon , \sqrt{n}|\hat{\theta}_n - \theta_0| \le M) + P(\sqrt{n}|\hat{\theta}_n - \theta_0| > M) \\
&\le  P(M/\sqrt{n} > \epsilon , \sqrt{n}|\hat{\theta}_n - \theta_0| \le M) + P(\sqrt{n}|\hat{\theta}_n - \theta_0| > M) \\
&\le  P(M/\sqrt{n} > \epsilon ) + P(\sqrt{n}|\hat{\theta}_n - \theta_0| > M).
\end{align*}
So, more formally, pick $\epsilon,\delta >0$. Then pick $M$ such that 
$$
P(\sqrt{n}|\hat{\theta}_n - \theta_0| > M) \le \delta,
$$
if $n > N_1$, which you can do because it's bounded in probability. Then if $n > \text{max}(N_1, M^2/\epsilon^2)$, you have your result:
$$
P(|\hat{\theta}_n - \theta_0| > \epsilon) \le \delta.
$$
Note that I am not showing how $\sqrt{n}$- consistency implies boundedness in probability, so I hope that this counts as a hint.
