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Trying to do Big O proofs and I'm stuck on this proof.

Need to prove if O(nlog(2,n)) is in O(n^2)

After playing around with it I get

log(2,n)/n <= c but I'm not too sure what to do after or how to conclude the proof

Johnathon Pomenrantz
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  • Your question makes no sense. Do you want to prove that $n\log_2(n)\in\mathcal O(n^2)$ instead? Or is $\mathcal O(n\log_2(n))\subseteq \mathcal O(n^2)$ what you want to show? – user127.0.0.1 Feb 10 '14 at 02:52

1 Answers1

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Yes. Since $${n\log(n)\over n^2 } = {\log(n)\over n} \to 0, $$ as $n\to\infty$. In fact it's $o(n^2)$. Since ${\log(n)\over n}\to 0$, this sequence is bounded, so be reassured.

ncmathsadist
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