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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?

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  • $\begingroup$ Factor out by $e^{-x}$ and use geometric sum. $\endgroup$ Commented Feb 19, 2019 at 17:57
  • $\begingroup$ The double limit depends on how $n$ and $x$ tend to $\infty$. In particular, if $x=n$, the limit won't exist. $\endgroup$
    – Peter
    Commented Feb 19, 2019 at 17:57

1 Answer 1

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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

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

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  • $\begingroup$ We have to consider $x\rightarrow \infty$ , so $|x|<1$ does not hold. $\endgroup$
    – Peter
    Commented Feb 19, 2019 at 18:00
  • $\begingroup$ @Peter The limit does not exist for $|x|\ge 1$. $\endgroup$
    – Mark Viola
    Commented Feb 19, 2019 at 18:01
  • $\begingroup$ I just noticed that two different limits are mentioned. $\endgroup$
    – Peter
    Commented Feb 19, 2019 at 18:06
  • $\begingroup$ @Peter I've edited the content to provide a hopefully clearer explanation. $\endgroup$
    – Mark Viola
    Commented Feb 19, 2019 at 18:10
  • $\begingroup$ Would the downvoter care to comment? $\endgroup$
    – Mark Viola
    Commented Feb 19, 2019 at 18:10

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