Earlier this week, I was trying to correctly write up why the gamma function $\Gamma(s)$ converges for $s > 0$ and I realised that I did not know how to show that the exponential function grows faster than any polynomial function. It’s only just a simple application of the Taylor expansion of the exponential but I’ll write it up since I’m sure a lot of people never saw the proof.

First, what I mean by that is that for any $n \in \textbf{N},$ there exists an $N \in \textbf{N}$ such that

$x \geq N \Rightarrow e^x > \displaystyle\sum_{i=0}^{n} |a_ix^i|$

for any $a_i \in \textbf{R}, i = 1, \ldots, n.$ To show this, we will show that

$\displaystyle\lim_{x \to \infty} \frac{e^x}{\sum |a_ix^i|} = \infty.$

This suffices because then for $x$ large enough this ratio will be greater than $1$ and therefore we will have $e^x > \sum |a_ix^i|.$ In fact, we could only show that this limit is greater than 1.

First, note that

$\displaystyle\frac{\sum_{i=0}^n a_ix^i}{a_nx^n} = \left( 1 + \displaystyle\frac{a_{n-1}}{a_nx} + \cdots + \frac{a_0}{a_nx^n} \right) \to 1$

as $x \to \infty.$ This is often written $\sum a_ix^i \sim_{x \to \infty} a_nx^n.$ Thus, for any $\epsilon > 0,$ when $x$ is large enough we have

$\left|\displaystyle\sum_{i=0}^n a_ix^i - a_nx^n\right| < \epsilon.$

Therefore, we only need to show that

$\displaystyle\lim_{x \to \infty} \frac{e^x}{|a_n|x^n} = \infty.$

Since this limit goes to infinity iff $\frac{e^x}{x^n}$ goes, we will drop the $|a_n|$ in the denomiator.

Expanding $e^x$ in its Taylor series, we get

$\displaystyle\frac{e^x}{x^n} = \displaystyle\sum_{k=0}^{\infty} \frac{x^k}{x^nk!}$

$= \displaystyle\frac{1 + x + x^2/2! + x^3/3! + \cdots + x^n/n! + x^{n+1}/(n+1)! + \cdots}{x^n}$

$= \displaystyle\frac{1}{x^n} + \frac{1}{x^{n-1}} + \frac{1}{x^{n-2}2!} + \cdots + \frac{1}{n!} + \frac{x}{(n+1)!} + \cdots$

$> \displaystyle\frac{x}{(n+1)!}$

which tends to $\infty$ as $x \to \infty.$