# How to prove that $n^b \neq O(n^a)$, if b > a > 1

How can we prove that $$n^b \neq O(n^a)$$, if b > a > 1

Based on Big-O definition:

$$n^b \neq O(n^a) \iff |n^b| \le c|n^a|$$

I know it's funny but I am stuck here and can't figure it out

$$u_n = O(v_n) \iff (\exists c \in \mathbb{R}^{*})\,\, (\exists N \in \mathbb{N}) \,\, n > N \implies u_n < c \, v_n$$
And that mean if $$(v_n)_{n\in\mathbb{N}}$$ not null from certain range the limit $$\displaystyle\lim_{n \rightarrow +\infty} \dfrac{u_n}{v_n}$$ is bounded.
If $$b > a > 1$$ we have $$\displaystyle\lim_{n \rightarrow +\infty} \dfrac{n^b}{n^a} = \lim_{n \rightarrow +\infty} n^{b-a} = +\infty$$ not bounded!
Then if $$b > a > 1$$ : $$n^b \neq O(n^a)$$
• The constante $c$ is fixed. If $b>a>1$, $c$ verify $n^b < c \, n^a$ cannot exists. – LAGRIDA Feb 11 at 19:38