Show that $6n^2+20n$ is not $\Omega ( n^3)$

Question

I am trying to show that $6n^2+20n$ is not $\Omega( n^3)$

Thoughts: By definition there must exist a $c \in \Bbb R$ such that $6n^2+20n > cn^3$ for all $n \in N$. Any hints on how I can show this to be a contradiction algebraically would be appreciated.

It suffices to show that $n$ is not $\Omega(n^2)$ and $n^2$ is not $\Omega(n^3)$. Is this easier for you? — SometimesBlind, Sep 16 '16 at 02:14

score 1 · Accepted Answer · answered Sep 16 '16 at 02:22

By contradiction:

Suppose $6n^2 + 20n > cn^3$ for all $n \in N$ for some $c>0 , c\in \Bbb R$ Then since $6n^2 + 20n^2 = 26n^2 > 6n^2 + 20n > cn^3$, this implies that $n < \frac {26} {c} $, which is a contradiction since $n$ does not have an upper bound.

Show that $6n^2+20n$ is not $\Omega ( n^3)$

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