for large enough $t_0 \ge 1$ we have $t \ge t_0 \implies \ln t \le \sqrt[2p]{t}$

Question

Why "for large enough $t_0 \ge 1$ we have $t \ge t_0 \implies \ln t \le \sqrt[2p]{t}$" is there an algebraic proof for this?

Answer 1

Using L'Hopital's rule we get $$\lim_{t \to \infty} \frac{\ln t}{\sqrt[2p]{t}} = \lim_{t\to\infty} \frac{\ln t}{t^{\frac1{2p}}} = \lim_{t\to\infty} \frac{t^{-1}}{\frac1{2p}t^{\frac{1}{2p}-1}}= \lim_{t \to\infty} \frac{2p}{t^{\frac1{2p}}} = 0$$ since $2p > 0$.

Therefore there exists $t_0 \ge 1$ such that $t \ge t_0$ implies $\frac{\ln t}{\sqrt[2p]{t}} \le 1$, or $\ln t \le \sqrt[2p]{t}$.

Answer 2

Hint:

The statement you are quoting can be rewritten as:

There exists some $t_0$ such that we have $$t\geq t_0\implies \frac{\ln t}{t^q} < 0$$ (where $q=\frac{1}{2p}$).

very closely connected to the fact that, for all $s>0$, we have

$$\lim_{t\to\infty}\frac{t^s}{e^t} = 0.$$

Answer 3

Compare the logs: this inequality is equivalent to $$\ln(\ln t)<\frac1{2p}\ln t,$$ and setting $u=\ln t$, use that $\lim_{u\to+\infty}\dfrac{\ln u}{u}=0$, so for any $p$, there exists a $u_0$ such that $$\frac{\ln u}u<\frac1{2p}\iff \ln u<\frac1{2p}u\quad\forall u>u_0.$$ Translating this in terms of $t$, we obtain $$\ln(\ln t)<\frac1{2p}\ln t=\ln\Bigl(t^{\tfrac1{2p}}\Bigr)\iff\ln t<t^{\tfrac1{2p}}$$ for all $t>\mathrm e^{u_0}$.