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

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?

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

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}

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