Show that $\frac{1}{n!} \leq \frac{e^n}{n^n}$

Question

Using the fact that

$$\frac{1}{n!} = \frac{1}{2\pi r^n} \int_0^{2\pi} \! e^{re^{it}}e^{-int} \, \mathrm{d}t$$

and choosing $r$ in an appropriate way, how can I prove that $$\frac{1}{n!} \leq \frac{e^n}{n^n}$$ ?

Thanks for any hint.

Answer 1

HINT: Try $r = n$, and use an inequality concerning the absolute value of an integral.

Answer 2

Let $z=e^{it}$ and $r=n$. Then, we have

$$2\pi \frac{n^n}{n!}= \int_0^{2\pi}e^{ne^{it}}e^{-int}\,dt=\frac1{ i}\oint_{|z|=1}\frac{e^{nz}}{z^{n+1}}\,dz$$

Note that we have

$$\left|\frac1i\oint_{|z|=1}\frac{e^{nz}}{z^{n+1}}\,dz\right|\le \oint_{|z|=1}\frac{e^{n|z|}}{|z|^{n+1}}\,|dz|=e^n$$

Putting it together yields the coveted inequality.

Answer 3

Mark's approach is perfectly fine (so (+1) to him), and I would just point out that you do not really need the residue theorem or Cauchy's integral formula to prove $n!\geq\left(\frac{n}{e}\right)^n$, just basic combinatorics. Indeed, for any $n\geq 2$ we have $$ n = \prod_{k=1}^{n-1}\left(1+\frac{1}{k}\right)\qquad(\text{telescopic product}) $$ hence for any $N\geq 2$ $$ N! = \prod_{n=2}^{N}n = \prod_{n=2}^{N}\prod_{k=1}^{n-1}\left(1+\frac{1}{k}\right)=\prod_{k=1}^{N-1}\left(1+\frac{1}{k}\right)^{N-k}=\frac{N^N}{\prod_{k=1}^{N-1}\left(1+\frac{1}{k}\right)^k}$$ and since the sequence $\left\{\left(1+\frac{1}{k}\right)^k\right\}_{k\geq 1}$ is increasing towards $e$, $N!\geq\left(\frac{N}{e}\right)^N\cdot e$.
Similarly $N!\leq\left(\frac{N}{e}\right)^N\cdot e\sqrt{N}$.