Big $\Omega$ notation inverses Is $\Omega(1/n^2)$ of the same order as that of $1/\Omega(n^2)$
I am kind of new to it. It seems $f$ is order of $\Omega(1/n^2)$ this implies that $f.n^2 > 0$ in limit this mean $1/f.1/n^2 < \infty$ this mean $1/f $is $O(n^2)$ does this imply $f$ is $1/O(n^2)$ or $f$ is $1/\Omega(n^2)$
 A: Let me bring definition for big-$O$ for non-negative functions:
$$O(g)=\{f\colon \exists C \gt0, \exists N \in \mathbb{N},\forall n \gt N,f(n) \leqslant C \cdot g(n)  \}$$
definition for big-$\Omega$ differs in that we have "$\geqslant$" in place "$\leqslant$".
Now, suppose we have $f \in \Omega\left(\frac{1}{n^2}\right)$. Strictly speaking this doesn't allow to consider $\frac{1}{f}$, because definition puts restriction only on some "remainder" of sequence $f(n)$. We can have any fixed members equal to zero, which makes it impossible. But for functions which have no zero value we can say, that $f \in \Omega\left(\frac{1}{n^2}\right) \Rightarrow \frac{1}{f} \in O(n^2)$.
Same with your last comment/question: having $B\in O(n^2)$ we, generally, cannot consider $\frac{1}{B}$. But even in case $B(n)\ne 0$ for all $n$ and $A\in O(n^2)$ we cannot obtain $\frac{A}{B} \in O(1)$ as it comes from example $B(n)=1, A(n)=n^2$.
A: No. Saying $f(n)$ is $\Omega(1/n^2)$ is a lower bound on $f(n)$ - it is at least a constant times $1/n^2$.
However, saying $f(n)$ is $1/\Omega(n^2)$ is an upper bound on $f(n)$, since it gives a lower bound on $1/f(n)$.
In other words, assuming everything is strictly positive, $f(n)$ is $1/\Omega(n)$ is the same as saying $f(n)$ is $O(1/n^2)$.
