# Prove that $\ln n = o(n^\alpha)$ for all strictly positive real numbers $\alpha$

asymptoticsconvergence-divergencelogarithmsreal-analysissequences-and-series

I am trying to prove that for all $$\alpha \in \mathbb R_+^*$$:
$$\frac{\ln n}{n^\alpha} \longrightarrow{} 0$$
For $$\alpha = 1$$, I proved it using the squeeze theorem after showing (using the derivative of the difference) that $$\ln n \le \sqrt n$$ for almost all $$n$$. I tried to generalise this approach to an arbitrary positive exponent $$\alpha$$. To do that, I tried to prove that the function
$$f: x \mapsto \ln n – x^\alpha$$
is negative for almost all natural numbers. I showed that is was decreating over $$I = [\left(\frac{2}{\alpha}\right)^\frac{2}{\alpha}, +\infty[$$, but I can't manage to find a point $$x_0 \in I$$ such that $$f(x_0) \le 0$$.

What should I do?

If $$\dfrac{\ln n}{n}\to 0,$$ then $$\dfrac{\ln n^\alpha}{n^\alpha} \to 0.$$ Thus
$$\frac{\ln n}{n^\alpha} = \frac{1}{\alpha}\left (\frac{\ln n^\alpha}{n^\alpha}\right)\to \frac{1}{\alpha}\cdot 0 = 0.$$