# Show $n!<\left(\frac{n}{e}\right)^{n+1}$

What I want to show is $$\left(\frac{n}{e}\right)^{n}

I managed to deal with the first part of inequalities using the Taylor expansion: $$e^n=\frac{1}{0!}n^0+\frac{1}{1!}n^1+\frac{1}{2!}n^2+\dots + \frac{1}{n!}n^n+\dots>\frac{1}{n!}n^n$$

But the second part was not that easy to me. I could do it using induction, however I would like to know if there's a way to do it without induction just like the first inequality.

And I found on the net, without justification, these inequalities are related to the inequalities $$\left(1+\frac{1}{n}\right)^n

but couldn't find a link by myself. (I did the first inequality without using theese inequalities)

Likewise, the first inequality is easy to show using the Taylor expansion: $$1+\frac{1}{n}<1+\frac{1}{n}+\frac{1}{2!n^2}+\frac{1}{3!n^3}+\dots=e^{\frac{1}{n}}$$

But the second one seems not doable with this technic.

Can anyone give me some hint on those two second part inequalities? And if possible, what is the link between them?

• This is not true for all $n$ ! And why would you avoid induction ?
– user65203
Commented Apr 9, 2020 at 7:47

I knew the following, i hope you could take advantadge from this as well

Since $$e^{n} = \sum\limits_{k=0}^{\infty} \frac{n^{k}}{k!}$$ it's a non negative series containing the factor $$\frac{n^{n}}{n!}$$ for sure we obtain that $$e^{n} \geq \frac{n^{n}}{n!}$$

Which give us a first inequality $$n^{n} \geq n! \geq n^{n} \cdot e^{-n}$$

Fortunately we can find better ones,one of whom is given by the Mean Value Theorem with $$f(x) = x\cdot log(x) \hspace{0.2cm}$$:

$$f(k+1)-f(k) = (k+1)\cdot \log(k+1) - k\log(k)$$

$$(k+1)\log(k+1) - k\cdot \log(k) \geq (\log(k) +1)(k+1-k)$$

Summing for per $$k=1,2,\cdots,n-1$$ (Which give us a telescopic sum) we get :

$$n\cdot \log(n) \geq \sum\limits_{k=1}^{n-1}\log(k)+(n-1)$$ Adding $$log(n)$$ on both sides we reach : $$(n+1)\log(n) \geq \log(n!) + (n-1)$$

Taking the exponential $$n! \leq n^{n+1} \cdot e^{-n+1}$$ And we have just improved the first inequality with $$\left(\frac{n}{e}\right)^n \leq n! \leq e^2 \left(\frac{n}{e}\right)^{n+1}$$

• This is a very nice trick! Commented Apr 9, 2020 at 10:42

Hint: You could use Stirling's approximation, $$n!\sim \sqrt{2\pi n}(n/e)^n$$, if you can show $$\sqrt{2\pi n}\lt n/e$$. But this is clearly true for large enough $$n$$.

We can use mathematical induction to prove the second part. Assuming the case when $$k\leq n$$ is right,we only need to prove the following inequality to prove the case when $$k=n+1.$$ $$n+1<\frac{(\frac{n+1}{e})^{n+2}}{(\frac{n}{e})^{n+1}}.$$ $$n+1<(\frac{n+1}{e})(\frac{n+1}{n})^{n+1}$$ $$e<(1+\frac{1}{n})^{n+1}$$ And this is just the inequality you find on the internet.