Inequality proof including $n!$

Question

$n^{n}e^{-n+1} \le n! \le n^{n}e^{-n+1} n$,

$n \in \mathbb{N}$

I'm struggling solving the inequality above, I have tried AM-GM, Bernoulli but I guess now that the proof is based maybe on induction.The squeeze theorem can make maybe also sense. I appreciate any help.

$ n^{n}e^{-n+1} \le n! \le n^{n}e^{-n+1} n$
which is equal to

$\lim_{n\to\infty} e^{-n+1} \le\lim_{n\to\infty} \frac{n^n}{n!} \le \lim_{n\to\infty} e^{-n+1}n $ where $\lim_{n\to\infty} \frac{n^n}{n!} = 0$ is that correct? — Herrpeter, Jun 18 '20 at 16:48
@Herrpeter: bracketing with limits does not guarantee the inequalities. — , Jun 18 '20 at 16:53
@Herrpeter You misplayed the numerator and the denominator. The middle should be $\frac{n!}{n^n}$ — Zerox, Jun 18 '20 at 17:40
@Herrpeter FYI, the same inequality was previously asked to be proven, but specifically using integration, in Prove with integration the inequality $e(\frac{n}{e})^n < n! < n \times e(\frac{n}{e})^n$. — John Omielan, Jun 18 '20 at 17:51

score 3 · Accepted Answer · answered Jun 18 '20 at 17:07

3

$1. \log$ and $\exp$ are monotonoe increasing functions, so if $\log x < \log y \to x <y$, same for $\exp$.

$2.n! = e^{\sum_{k=1}^{n}\log k}$

$3. \int_{1}^{n} \log x dx = n \log n -n +1 $

Can you do the rest?

answered Jun 18 '20 at 17:07

Alex

19,262

Inequality proof including $n!$

1 Answers1

Linked

Related