Limit of $ \frac{e^k k!}{k^k} $

Question

According to WolframAlpha we should have

$$ \lim_{k \to \infty} \frac{a^k k!}{k^k} = \infty $$

whenever $a \geq e$. I know how to show that the limit is 0 when $a < e$ and that the limit is $\infty$ when $a > e$, but I struggle with the case $a = e$. Setting

$$ f(k) = \frac{e^k k!}{k^k} $$

we have that

$$ \frac{f(k+1)}{f(k)} = e\left(1 - \frac{1}{k + 1}\right)^k $$

converges to 1, but this is not sufficient.

Duplicate: $\lim\limits_{n \to\infty}\dfrac{e^nn!}{n^n}$. (Found using Approach0.xyz) — Workaholic, Nov 03 '16 at 12:08

score 6 · Accepted Answer · answered Sep 20 '16 at 15:01

6

Hint

You can use Stirling formula:

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

answered Sep 20 '16 at 15:01

E. Joseph

14,843

Indeed, it is immediately clear from this formula. Now, do you have a good reference to a proof of this? – Bib-lost Sep 20 '16 at 15:04
@Bib-lost some proofs are given on the wikipedia page. – Ben Grossmann Sep 20 '16 at 15:04
Only in french... But like @Omnomnomnom said, you will find many on the Internet. – E. Joseph Sep 20 '16 at 15:05
French is okay for me. I am working through the Wikipedia proofs but they all require something I have never heard of. – Bib-lost Sep 20 '16 at 15:14
2

@Bib-lost Then I like this one. – E. Joseph Sep 20 '16 at 15:15
@E.Joseph so do I! – Bib-lost Sep 20 '16 at 15:23

score 5 · Answer 2 · answered Sep 20 '16 at 15:32

Starting from your $$ \dfrac{f(k+1)}{f(k)} = e \left( 1 - \dfrac{1}{k+1}\right)^k $$ take logarithms: $$ \log f(k+1) - \log f(k) = 1 + k \log \left(1 - \dfrac{1}{k+1}\right) $$ Now if $1/2 < a < 1$, there is $\epsilon > 0$ such that $\log(1-t) \ge -t - a t^2$ for $0 < t< \epsilon$, and thus for sufficiently large $k$,
$$\eqalign{ \log f(k+1) - \log f(k) &\ge 1 + k \left(- \dfrac{1}{k+1} - \dfrac{a}{(k+1)^2} \right)\cr &= \dfrac{(1-a)k+1}{(k+1)^2}\cr }$$ Now use the fact that $\displaystyle \sum_{k=1}^\infty \dfrac{k}{(k+1)^2} = \infty$.

score 4 · Answer 3 · edited Sep 20 '16 at 15:33

4

You don't need Stirling's formula, you can use easier estimates, e.g. this one:

We have:

$$ \log n! = \sum_{x=1}^n \log x $$

so, looking at the graph of $\log x$ we can easy bound it below and above by integrals:

$$ \int_1^n \log x \, dx \leq \sum_{x=1}^n \log x \leq \int_0^n \log (x+1) \, dx $$

which gives us the estimate

$$ n\log\left(\frac{n}{e}\right)+1 \leq \log n! \leq (n+1)\log\left( \frac{n+1}{e} \right) + 1 $$

so

$$ e\left(\frac ne\right)^n \leq n! \leq e\left(\frac{n+1}e\right)^{n+1} $$

This is clearly weaker estimate than Stirling's formula, but will work well enough for your problem.

edited Sep 20 '16 at 15:33

Ben Grossmann

225,327

answered Sep 20 '16 at 15:31

xyzzyz

7,612

I agree with all of your calculations, but how does this solve my initial problem? It only shows that the limit is at least $e$ ... – Bib-lost Sep 20 '16 at 16:02

Limit of $ \frac{e^k k!}{k^k} $

3 Answers3