Since $$ \begin{align*} & e^x = 1 + x + \ldots + \frac{x^n}{n!} + \frac{x^{n+1}}{(n+1)!} + \ldots , \\ & x^{-n}e^x \gt \frac{x}{(n+1)!} \rightarrow \infty \end{align*}$$
when $ x \rightarrow \infty $. Hence $ e^x $ tends to infinity more rapidly than any power of $ x $.
"An Introduction to the Theory of Numbers" – G.H. Hardy and E. M. Wright
I understand the inequality but I do not see how this statement holds true or under what context it holds true. I may be misreading the statement and misinterpreting it. Obviously the inequality exists because the Power Series is a sum of terms of $ \frac{x^n}{n!} $ and so $ e^x $ will be greater than any single given term. How does this fact prove that $ e^x $ is tending more quickly towards infinity? And also faster than what exactly? If $ n $ is constant and $ x $ grows to infinity and $ \frac{e^x}{x^n} $ tends to infinity and not zero or one this tells me that $ e^x $ is growing larger than $ x^n $ and achieving a magnitude that is not of the same order . . . but what does the inequality have to do with anything? What is $ \frac{x}{(n+1)!} $ and how does it relate back to the issue at hand?
Best Answer
From what you mention, that \begin{align*} e^x > \frac{x^{n+1}}{(n+1)!} \end{align*} for every $n$, we multiply both sides by $x^{-n}$ to find \begin{align*} x^{-n}e^x > \frac{x^{n+1}}{(n+1)!}x^{-n}. \end{align*} But the right hand side is simply $x/(n+1)!$, so for fixed $n$, the left hand side tends to infinity as $x$ tends to infinity. So what the inequality is saying is no matter what power you pick, say $x^{-n}$, when you multiply $e^x$ by this power, you always get something that tends to infinity.
If there were some power of $x$, say $x^m$ that grew as fast or faster than $e^x$, you would have that $x^{-m}$ would shrink too quickly for $x^{-m}e^x$ to tend to infinity.
Perhaps an easy way to see that would be to just deal with powers of $x$. If we have some integers $n$ and $m$, then if $n > m$, what does $x^{-n}x^m$ do as $x$ tends to infinity? What if $n=m$? Lastly, what if $n < m$?
Now with $e^x$ we always have the inequality $x^{-n}e^x > \frac{x^{n+1}}{(n+1)!}x^{-n}$ for every choice of $n$, so there is no power of $x$ that grows faster.