Prove that $\ln n = o(n^\alpha)$ for all strictly positive real numbers $\alpha$ 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
$$\frac{\ln n}{n^\alpha} = \frac{1}{\alpha}\left (\frac{\ln n^\alpha}{n^\alpha}\right)\to \frac{1}{\alpha}\cdot 0 = 0.$$
