Proving $\log n < \sqrt n$

Question

I am trying to prove $\exists n_0 > 0: \forall n > n_0: \log n < \sqrt n$. My attempt uses the series representation of the exponential function, but it does not seem to accomplish the proof:

$$ \log n < \sqrt n \\ \Leftrightarrow n < e^{\sqrt n} = \sum_{k=0}^\infty \frac{(\sqrt n)^k}{k!} = 1 + \sqrt n + \frac n 2 + \sum_{k=3}^\infty \frac{(\sqrt n)^k}{k!} \\ \Leftrightarrow \frac n 2 < 1 + \sqrt n + \sum_{k=3}^\infty \frac{(\sqrt n)^k}{k!} \\ \Leftrightarrow ??? $$

How could I complete this proof? Or would a different strategy e.g. via monotonicity be more convenient?

Answer 1

Hint: The series representation for the exponential will work. Use the fact that if $x$ is positive then $$e^x\gt \frac{x^3}{3!}.$$

Answer 2

In fact, $\ln x < \sqrt x$ for all $x>0.$ Let $f(x) = \sqrt x - \ln x.$ Then $f'(x) = 1/(2\sqrt x) - 1/x.$ This tells us $f'<0$ on $(0,4), f'(4)=0,$ and $f'>0$ on $(4,\infty).$ Thus $f(4)$ is the absolute minimum of $f$ on $[1,\infty).$ It is quite simple to check that $f(4) > 0,$ hence $f> 0$ on $(0,\infty).$