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Basically what i am trying to prove is this : $f(n),g(n) \neq 0\quad , n>0 \ \ \ \ and f(n)=Ω(g(n)) \ \ \ , \ then \frac{1}{f(n)}=O(\frac{1}{g(n)}) $

I guess that if we take the definition of $f(n) = Ω (g(n))$ , which means that $f(n) \ge cg(n) $ for every $ n \ge n0$ and then flip the relation so that it becomes $\frac{1}{f(n)} \le \frac{1}{c} \frac{1}{g(n)}$ then we can say that the statement is true , however i am not sure if this is correct or not . Can anyone help me prove it correctly or disprove it ? Thank you .

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    May 2, 2023 at 20:26

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$$\forall n>n_0:f(n)\ge c\,g(n)>0\implies\frac1{f(n)}\le \frac 1c\frac1{g(n)}$$ is true.

Just a remark: $n_0$ must be taken larger than any root of $f$ and $g$. A pathological counter-example is

$$1=\Omega\left(\sin\left(\frac{n\pi}2\right)\right).$$

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