How to prove that $n\log(n) \in \mathcal{O}((n)^k)$?

Question

I need to prove for $k>1$ that is:

$n\log(n) \in \mathcal{O}(n^k)$

A also need to mention what happens in case that $k = 1$. So actually, to prove $n\log(n) \in \mathcal{O}(n)$

which I did so:

For big-Oh Notation $n\log(n) \in \mathcal{O}(n)$ exists an constant $c$ and $n_0$ such that for $n\gt n_0$ :

$n\log(n) \le c * \mathcal{n}$ // $\div\mathcal{n}$ gives:

$log(n) \le c$ // this is NOT TRUE weil there is no such a constant $c$ that for all $n\gt n_0$

But I have no idea how to prove $(n^k)$. Please help :)

$n\log n$ is not in $O(n)$, so no wonder you cannot prove it is. (It is, however, in $O(n^k)$ for every fixed $k>1$). Note that this is equivalent to proving that $\log n$ is in $O(n^a)$ for any fixed $a>0$. — Clement C., Mar 07 '21 at 11:09
thank you for the answer. But I still don't know how to prove it correctly :( — No Ziffer, Mar 07 '21 at 11:26

score 0 · Accepted Answer · answered Mar 07 '21 at 11:31

It suffices to prove that $\lim_{n \to \infty} \frac{n\log_2n}{n^k}$ exists. We have: $$\lim_{n \to \infty} \frac{n\log_2n}{n^k} =\lim_{n \to \infty} \frac{\log_2n}{n^{k-1}} = \lim_{n \to \infty} \frac{\frac{1}{ln(2)n}}{(k-1)n^{k-2}} = \lim_{n \to \infty} \frac{1}{ln(2)(k-1)n^{k-1}} = 0 $$ So the limit exists. Note that I have used the L'Hopital rule in the third equality and $k>1$.

How to prove that $n\log(n) \in \mathcal{O}((n)^k)$?

1 Answers1