What I want to show is $$\left(\frac{n}{e}\right)^{n}<n!<\left(\frac{n}{e}\right)^{n+1}$$
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 $$\left(1+\frac{1}{n}\right)^n<e<\left(1+\frac{1}{n}\right)^{n+1}$$
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?
Best Answer
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 $$\left(\frac{n}{e}\right)^n \leq n! \leq e^2 \left(\frac{n}{e}\right)^{n+1}$$