Is $\frac{\mu_X + O_p(n^{-1})}{\mu_Y + O_p(n^{-1})} = \frac{\mu_X}{\mu_Y} + O_p(n^{-1})$? Let $X_n = \mu_X + O_p(n^{-1})$ and let $Y_n = \mu_Y + O_p(n^{-1})$ where $\mu_X$ and $\mu_Y$ are constants. Then is it possible to write $Z_n = X_n/Y_n$ in a simplified form? E.g. I have
$$
\begin{align}
Z_n = \frac{X_n}{Y_n} &= \frac{\mu_X + O_p(n^{-1})}{\mu_Y + O_p(n^{-1})} \\
&= \frac{\mu_X}{\mu_Y + O_p(n^{-1})} + \frac{O_p(n^{-1})}{\mu_Y + O_p(n^{-1})}.
\end{align}
$$
Can we then drop the terms $O_p$ terms in the denominator and say
$$
\begin{align}
Z_n = \frac{\mu_X}{\mu_Y} + \frac{O_p(n^{-1})}{\mu_Y} = \frac{\mu_X}{\mu_Y} + O_p(n^{-1}).
\end{align}
$$
Edit: Further attempt after whuber suggestion.
$$
\begin{align}
\frac{\mu_X + O_p(n^{-1})}{\mu_y + O_p(n^{-1})} &= \bigg(\frac{\mu_X}{\mu_Y}\bigg) \frac{1 + O_p(n^{-1})/\mu_X}{1 + O_p(n^{-1})/\mu_Y} \\
&= \bigg(\frac{\mu_X}{\mu_Y}\bigg) \frac{1 + O_p(n^{-1})}{1 + O_p(n^{-1})} \\
&= \bigg(\frac{\mu_X}{\mu_Y}\bigg) (1 + O_p(n^{-1}))\frac{1}{1 + O_p(n^{-1})} \\
&= \bigg(\frac{\mu_X}{\mu_Y}\bigg) (1 + O_p(n^{-1}))(1 - O_p(n^{-1}) + O_p(n^{-2}) + O_p(n^{-3}) + \dots) \\
&= \bigg(\frac{\mu_X}{\mu_Y}\bigg) (1 + O_p(n^{-1}))(1 + O_p(n^{-1})) \\
&= \bigg(\frac{\mu_X}{\mu_Y}\bigg) (1 + 2O_p(n^{-1}) + O_p(n^{-2})) \\
&= \bigg(\frac{\mu_X}{\mu_Y}\bigg) (1 + O_p(n^{-1})) \\
&= \frac{\mu_X}{\mu_Y}  + \frac{\mu_X}{\mu_Y} O_p(n^{-1}) \\
&= \frac{\mu_X}{\mu_Y}  + O_p(n^{-1}).
\end{align}
$$
Does that seem ok? Is it valid to use a Taylor/Neumann series on the $O_p$ terms like this? Do I need conditions on the $O_p$ terms regarding the radius of convergence of the expansion?
 A: If you're uncertain, you can always go back from order notation to explicit epsilontics.
$\Delta_{X,n}=\mu_X+O_p(N^{-1})$ means that for every $\epsilon>0$ and every finite $M$, there exists $N$ such that $P((\Delta_{X,n}-\mu_X)>M/n)<\epsilon)$ when $n>N$. Similarly, $\Delta_{Y,n}=\mu_Y+O_p(N^{-1})$ means that for every $\epsilon>0$ and every finite $M$, there exists $N$ such that $P((\Delta_{Y,n}-\mu_Y)>M/n)<\epsilon)$ when $n>N$.
Now, consider
$$\frac{\Delta_{X,n}}{\Delta_{Y,n}}-\frac{\mu_X}{\mu_Y}$$
and consider $n>N$. Write $\delta_{x,n}=\Delta_{X,n}-\mu_X$ and similarly for $Y$.
Since $\delta_{x,n}$ and $\delta_{y,n}$ are just specific sequences of random variables, you don't have any issues about Taylor series or anything, and you can run the argument you want to run (as long as $\mu_Y\neq 0$) and conclude that, yes, it's equal to
$$\frac{\mu_X}{\mu_Y}+O_p(N^{-1})$$
The condition that $\mu_Y\neq 0$ is important -- it's needed for the first equality in your derivation, and the result is not (in general) true without it.  If $\mu_Y=0$ but $\mu_X\neq 0$ then the LHS is $\Omega_p(N)$ and the RHS is infinite or undefined.   If $\mu_X=\mu_Y=0$ the LHS can be pretty much anything and the RHS is undefined.
