In which condition can I say that $X_n=O_p(\alpha_n)$ implies $X_n^{-1}=O_p(1/\alpha_n)$, since it holds for $X_n\sim N(0,1)$.(In this case, $X_n=O_p(1)$ and $\frac{1}{X_n}$ is a.s. finite and hence $\frac{1}{X_n}=O_p(1)$.)
From the definition of big O notaion, $X_n=O_p(\alpha_n)$ means, for $\forall \epsilon>0,\exists M_\epsilon,N_\epsilon$, that $P(|X_n|\geq M_\epsilon\alpha_n)<\epsilon$. To prove $\frac{1}{X_n}=O_p(\frac{1}{\alpha_n})$, I must find some $C_\epsilon>0$ to make the following probability small enough,
\begin{align} P(|\frac{1}{X_n}|\geq C_\epsilon/\alpha_n)&=P(|X_n|\leq \frac{\alpha_n}{C_\epsilon})\\ &=1-P(|X_n|> \frac{\alpha_n}{C_\epsilon})\\ &=? \end{align} Is something wrong with my proof?