# if f(n),g(n) =! 0 , for every n > 0 , and f(n) = Ω(g(n)) , then does this mean that 1/f(n) = O(1/g(n))

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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– D.W.
May 2 at 20:26

$$\forall n>n_0:f(n)\ge c\,g(n)>0\implies\frac1{f(n)}\le \frac 1c\frac1{g(n)}$$ is true.
Just a remark: $$n_0$$ must be taken larger than any root of $$f$$ and $$g$$. A pathological counter-example is
$$1=\Omega\left(\sin\left(\frac{n\pi}2\right)\right).$$