$\forall n\in \mathbb N$, $\exists$ a $b(n)$ s.t. $x^n < b(n)\exp(\frac{x^2}{3})$, for $x > 0$

Question

$\forall$I want to show that for any $n\in \mathbb N$, there exists a $b(n) \in \mathbb R$ such that $x^n < b(n)\exp(\frac{x^2}{3})$, for $x > 0$.

To find this $b(n)$, I would suppose that this $b(n)$ exists, then we would have $$\ln \frac{x^n}{b(n)} \le \frac{1}{3}x^2.$$

Let $f(x) = \frac{1}{3}x^2 - \ln \frac{x^n}{b(n)}.$ Then, $f'(x) = \frac{2}{3}x - \frac{n}{x} = 0 \implies x = \sqrt{3n/2}$.

$f$ is decreasing on $(0,\sqrt{3n/2})$ and is increasing on $(\sqrt{3n/2}, +\infty)$ and $f(\sqrt{3n/2}) >0$.

$b(n)$ is canceled out when we calculate $f'(x)$, so it seems like it does not matter what its value is, but in reality it matters.

What is wrong with this verification? How do we show the existence of the $b(n)$?

Answer 1

Based on what you have, it is sufficient to find a value for $b(n)$ such that $ f( \sqrt{3n/2}) > 0$.
This implies that $b(n) > \sqrt{3n/2}^n / e^{(n/2)}$ will work.