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I tried using the same trick as $\lim\limits_{n\to\infty}\frac{n!}{n^n}$, where you compare the terms one to one.

$(\frac{1}{n})(\frac{2}{n})(\frac{3}{n})...(\frac{n}{n})\cdot e^n$ = $(\frac{e}{n})(\frac{2e}{n})(\frac{3e}{n})\cdots(\frac{ne}{n})$

I can't really figure out what to do after this. Some of the terms are less than one and some are greater than one. It would be great if someone could lead me in the right direction of what to do here.

Mateus Rocha
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63677
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    I'm not sure how rigorous or allowed this is, but I think that, since $n!$ is asymptotically equivalent to $\sqrt{2 \pi n} \cdot n^ne^{-n}$ by Stirling's approximation, we can substitute that expression in. Then

    $$\lim_{n \to \infty} \frac{n! e^n}{n^n} = \lim_{n \to \infty} \sqrt{2\pi n} \cdot \frac{n^n}{e^n} \cdot \frac{e^n}{n^n}$$

    Some clear cancellation happens, leaving the limit for $\sqrt{2 \pi n}$, which is obviously infinite. But I'm not sure if there are conditions we need to verify first before we can do this.

     – PrincessEev Oct 10 '19 at 02:58
  • @EeveeTrainer That's perfectly fine. – Angina Seng Oct 10 '19 at 03:00

4 Answers4

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Apply Stirling's approximation so that as $n\to\infty$

$$n! \sim \left(\frac{n}{e}\right)^{n}\sqrt{2\pi n}$$

therefore

$$\frac{n!\cdot e^n}{n^n}\sim \sqrt{2\pi n}$$

from which $\sqrt{2\pi n}$ increases monotonically as $n\to\infty$

$$\lim\limits_{n\to\infty}\sqrt{2\pi n}=\infty $$

thus

$$\lim\limits_{n\to\infty}\frac{n!\cdot e^n}{n^n}=\infty$$

Axion004
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Your are going in right direction See as n goes to infinity n>>> e so the first term becomes very less.now for the last term n would cancel out itself so last term is e

So limit equals to 0 because maximum value that is multiplied is e and every other value is very close to zero

Who am I
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  • But there are something like $n - \frac{n}{e}$ terms bigger than 1 here, so that part of the product gets bigger. We can show by Stirling's approximation the limit diverges, so this can't be right. – Cade Reinberger Oct 10 '19 at 03:24
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Use Sterling formula and you have $$\frac {n! e^n}{n^n}\approx \sqrt {2\pi n}(n/e)^n (e/n)^n = \sqrt{2\pi n} \to \infty$$

63677
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Mohammad Riazi-Kermani
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I think another way to go about this would be to use the trick you were initially planning to and combine it with the taylor expansion of $e^n$. Then you will have:

$$ \lim _{n \rightarrow \infty} \frac{n ! \cdot e^{n}}{n^{n}} = \lim _{n \rightarrow \infty} \frac{n !}{n^{n}} \cdot (1+n+\frac{n^{2}}{2 !}+\frac{n^{3}}{3 !}+\frac{n^{4}}{4 !}+\frac{n^{5}}{5 !}+\dots)$$

and you can open the brackets and apply the trick you mentioned.

Elen Khachatryan
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