Suppose I have the following expression:
$$f(n) = \frac{\bigg(1+O(1/n)\bigg)}{\bigg(1+O(1/n)\bigg)\bigg(1+O(1/n)\bigg)}.$$
Is there a way to manipulate $f$ (i.e. multiply by terms involving $n$) in order to get an expression that looks like $(1+O(1))$?
If $a_n,b_n,c_n\in O(1/n)$, then \begin{align} \frac{1+a_n}{(1+b_n)(1+c_n)}-1 &=\frac{1+a_n-(1+b_n)(1+c_n)}{(1+b_n)(1+c_n)}\\ &=\frac{a_n-b_n-c_n-b_nc_n}{(1+b_n)(1+c_n)}\\ &\leq\frac{|a_n|+|b_n|+|c_n|+|b_n||c_n|}{|1+b_n||1+c_n|}\\ &=O(1/n) \end{align} hence $$f(n)=\frac{1+O(1/n)}{(1+O(1/n))(1+O(1/n))}=1+O(1/n)$$ consequently, $f(n)=1+O(1)$ as well.