Let $b > 0$. I want to know whether
$$ \frac{b- O(\frac{1}{n})}{a^2 - (\frac{b}{n} + O (\frac{1}{n^2}))} \leq \frac{b}{a^2},$$ for $n$ sufficiently large.
The remainder $O(...)$ actually come from the same Taylor expansions, namely (I hope you know what I mean here), $$O(\frac{1}{n^2}) = \frac{1}{n} O(\frac{1}{n})= \frac{1}{3!} f^{ \prime \prime \prime} (1) \frac{1}{n} + \dots$$
Suppose $b>a$. You can compare because you don't know what are $O(\frac1n)$ and $O(\frac1{n^2})$. For example, let $$ O(\frac1n)=\frac1n, O(\frac1{n^2})=\frac1{n^2}. $$ Then the inequality is equivalent to $$1-\frac{1}{bn}\le 1-\frac{b}{a^2n}-\frac1{a^2n^2} $$ or $$ \frac{b^2-a^2}{a^2b}\frac1n\le-\frac1{a^2n^2} $$ which is absurd.