Say we have two sequences of scalar random variables $X_n\geq0$ and $\widehat{\theta}_n>0$, where
$X_n = o_p$(1) and we know $X_n = \mathcal{O}_p(n^{-1})$, i.e., $X_n$ converges (in probability) to zero at rate $n^{-1}$.
$\widehat{\theta}_n$ is a $\sqrt{n}$-consistent estimator of $\theta>0$ : $\widehat\theta_n - \theta = \mathcal{O}_p(n^{-1/2})$, i.e., $\widehat\theta_n$ converges (in probability) to $\theta$ at rate $n^{-1/2}$.
Q: Is it true for the ratio $R_n := X_n / \widehat\theta_n$ that,
$R_n = o_p(1)$, i.e, $R_n$ converges to zero (in probability)?
$R_n = \mathcal{O}_p(n^{-1})$, i.e., the rate is $n^{-1}$?
I think that 1. follows along the argument in this answer,
$$\frac{X_n}{n^{-1}} = \mathcal{O}_p(1) \, ; \, \frac{n^{-1}}{\widehat\theta_n} = o_p(1) \quad \Rightarrow \quad R_n = \frac{X_n}{n^{-1}}\cdot\frac{n^{-1}} {\widehat\theta_n} = \mathcal{O}_p(1)\cdot o_p(1) = o_p(1).$$
But 2. is just a guess from looking at computer simulation results and I don't know how to obtain the rate.
Any help is highly appreciated.
We can write $$ \frac{X_n}{\widehat{\theta}_n} = \frac{X_n}{\theta} \cdot \frac{\theta}{\widehat{\theta}_n} = \frac{X_n}{\theta} + \frac{X_n}{\theta}\left( \frac{\theta}{\widehat{\theta}_n} - 1 \right) . $$ Note that $\frac{X_n}{\theta} = O_P(n^{-1}$) and $\frac{\theta}{\widehat{\theta}_n} - 1 = o_P(1)$. (Since $\widehat{\theta}_n - \theta = o_P(1)$, a bit of limit manipulation using the fact that $\theta$ is a real number can be used to show this.)
Therefore, using $O_P(n^{-1}) o_P(1) = O_P(n^{-1})$, the above expression satisfies $$ \frac{X_n}{\widehat{\theta}_n} = O_P(n^{-1}) + O_P(n^{-1}) \, o_P(1) = O_P(n^{-1}) . $$