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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?

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Multiply both sides by $x^n$. It holds for $x > 0$. –  Qiaochu Yuan Aug 3 '11 at 2:16
    
To add to what Qiaochu said, since every term in the series is positive we can say $e^x$ is bigger than every individual term in the series. Restrict your focus to $x^{n+1}/(n+1)!$, and you should see what to do from there. –  anon Aug 3 '11 at 2:29
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Despite its appearance in Hardy and Wright, this isn't Number Theory, so I've retagged it real-analysis. –  Gerry Myerson Aug 3 '11 at 2:56

3 Answers 3

up vote 6 down vote accepted

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.

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I think that the heart of the difficulty is understanding what’s meant by the statement that one function tends to infinity faster than another.

Let $f(x) = 2x$ and $g(x) = 3x$; the ratio $\frac{f(x)}{g(x)} = \frac{2x}{3x} = \frac{2}{3}$ regardless of $x$, so we can say that $f(x)$ and $g(x)$ tend to infinity at the same rate as $x\to\infty$. This is simply what’s meant by ‘tend to infinity at the same rate’, at least in the ideal case. Now let $f(x) = 2x+1$ and $g(x) = 3x+1$; it’s no longer true that $\frac{f(x)}{g(x)}$ is a constant, but it is true that $\frac{f(x)}{g(x)} \to \frac{2}{3}$ as $x\to\infty$, so we can still reasonably say that $f(x)$ and $g(x)$ tend to infinity at the same rate. Finally, let $f(x) = 2x + x\sin x$ and $g(x) = 3x$; now $\frac{f(x)}{g(x)}$ doesn’t even have a limit as $x\to\infty$, but $x \le f(x) \le 3x$, so $\frac{f(x)}{g(x)}$ is always between $\frac{1}{3}$ and $1$. In this case we can at least say that $f(x)$ and $g(x)$ go to infinity at roughly the same rate: neither consistently outpaces the other. (See this article on Big O Notation, especially the Big $\Theta$ notation.

Extending this idea, if the ratio $\frac{f(x)}{g(x)}$ itself goes to infinity as $x\to\infty$, we say that $f(x)$ goes to infinity faster than $g(x)$: as $x$ increases, $f(x)$ gets further and further ahead of $g(x)$, in the sense that it becomes a bigger and bigger multiple of $g(x)$.

Now apply this to the functions $f(x) = e^x$ and $g(x) = x^n$ for some fixed $n$: $\frac{e^x}{x^n} = x^{-n}e^x > \frac{x}{(n+1)!}$, and certainly $\lim\limits_{x\to\infty}\frac{x}{(n+1)!} = \infty$ for any fixed $n$, so $\lim\limits_{x\to\infty}\frac{e^x}{x^n} = \infty$, i.e., $e^x$ tends to infinity faster than $x^n$.

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Imagine that $n$ is a fixed number. That is very important for the argument. If you like, imagine that $n=20000$. On second reading, imagine that $n=10000000$.

For any positive integer $N$, let $F(N)=N(n+1)!$. Then, by the inequality that you quote, if $x \gt F(N)$, then $$x^{-n}e^x \gt\frac{x}{(n+1)!}\gt N.$$ So given any $N$, after a while (more precisely, if $x \gt F(N)$), we will have $x^{-n}e^x>N$. That is precisely what is meant by "$e^x$ tends to infinity more rapidly" than $x^n$. Informally, for any fixed $n$, $e^x$ is ultimately much larger than $x^n$, in the sense that when we divide $e^x$ by $x^n$, or equivalently multiply $e^x$ by $x^{-n}$, the result is still, after a while, huge. In the long run, for any fixed $n$, $e^x$ beats $x^n$, by a lot.

I would be slightly more comfortable rewording the argument as follows. Fix $n$. We want to show that $\dfrac{x^n}{e^x}$ approaches $0$ as $x \to \infty$. This is another way of saying that in the long run, $e^x$ beats $x^n$.

The same basic inequality shows that if $x\gt F(N)$ then $$\frac{x^n}{e^x}<\frac{1}{N}.$$ So however close we want $\dfrac{x^n}{e^x}$ to be to $0$, by choosing $x$ large enough, we can make it that close (or closer). In the long run, $e^x$ is much bigger than $x^n$.

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