Does an exponential decay faster than a polynomial, in the limit of an infinite power?

Question

We know that $$ \lim\limits_{x \rightarrow \infty} \mathrm{e}^{-x}\, x^n = 0$$ for any $n$. But I assume that usually, this is stated with the understanding that $n$ is finite. But what happens when we take the limit $$ \lim\limits_{n \rightarrow \infty} \lim\limits_{x \rightarrow \infty} \mathrm{e}^{-x}\, x^n = 0\,?$$ The context is that I have an infinite sum of the form $$ \lim\limits_{n \rightarrow \infty} \sum_{i=0}^n \mathrm{e}^{-x}\, x^i,$$ and I want to study its behavior as $x \rightarrow \infty$. In summary,

Does $$ \lim\limits_{x \rightarrow \infty} \sum_{i=0}^\infty \mathrm{e}^{-x}\, x^i,$$ converge?

This question seems to indicate that the answer might be yes, but I wonder if taking $n \rightarrow \infty$ messes anything up?

The double limit depends on how $n$ and $x$ tend to $\infty$. In particular, if $x=n$, the limit won't exist. — Peter, Feb 19 '19 at 17:57

Mark Viola · Accepted Answer · 2019-02-19T19:09:33.237

2

The issue is one of interchanging the order of limits. Note that we have

$$\begin{align} \lim_{n\to\infty}\lim_{x\to\infty}\sum_{i=0}^n e^{-x}x^i&=\lim_{n\to\infty} \sum_{i=0}^n \lim_{x\to\infty}\left(e^{-x}x^i\right)\\\\ &=\lim_{n\to\infty} \sum_{i=0}^n (0)\\\\ &=0 \end{align}$$

Here, we first hold $n$ fixed and let $x\to\infty$. The result of the inner limit is $0$ for any $n$. Then, letting $n\to\infty$ produces $0$ as the result.

However, if the order of the limits is interchanged, then we have

$$\begin{align} \lim_{x\to\infty}\lim_{n\to\infty} \sum_{i=0}^n e^{-x}x^i&=\lim_{x\to\infty}\lim_{n\to\infty} e^{-x}\left(\frac{x^{n+1}-1}{x-1}\right)\\\\ \end{align}$$

which diverges since $\lim_{n\to\infty}x^n=\infty$ for $x>1$. In this case, we first hold $x>1$ fixed and take the limit as $n\to\infty$. The resultant limit diverges and renders the outer limit as $x\to\infty$ meaningless.

Aside, we ask what is the limit, if it exists, of $e^{-x}x^x$ as $x\to\infty$? We find that

$$\begin{align} \lim_{x\to\infty}e^{-x}x^x&=\lim_{x\to\infty}e^{-x} e^{x\log(x)}\\\\ &=\lim_{x\to\infty}e^{x\log(x/e)} \\\\ &=\infty \end{align}$$

edited Feb 19 '19 at 19:09

answered Feb 19 '19 at 17:58

Mark Viola

184,670

We have to consider $x\rightarrow \infty$ , so $|x|<1$ does not hold. – Peter Feb 19 '19 at 18:00
@Peter The limit does not exist for $|x|\ge 1$. – Mark Viola Feb 19 '19 at 18:01
I just noticed that two different limits are mentioned. – Peter Feb 19 '19 at 18:06
@Peter I've edited the content to provide a hopefully clearer explanation. – Mark Viola Feb 19 '19 at 18:10
Would the downvoter care to comment? – Mark Viola Feb 19 '19 at 18:10
Does this imply that if, hypothetically, we get convergence for both possibilities of the order in which limits are applied, then we can safely say that the double limit converges? For example, if I was now considering $\lim\limits_{x\rightarrow\infty} \sum_{i=0}^{\infty}h(x)\mathrm{e}^{-x}x^i$ such that I know that the sum converges for any finite x. – josh_eime Feb 19 '19 at 19:13
Not necessarily, but The Moore-Osgood Theorem provides sufficient conditions when the limits can be interchanged. – Mark Viola Feb 19 '19 at 23:44

Does an exponential decay faster than a polynomial, in the limit of an infinite power?

1 Answers1