4
$\begingroup$

I thought that $\log(n!)$ would be $\Omega(n \log n )$, but I read somewhere that $\log(n!) = O(n\log n)$.

Why?

  • 2
    The obvious inequalities $(n/2)^{n/2} \leq n! \leq n^n$ suffice to prove $\Theta(n \log n)$.2012-05-04

2 Answers 2

6

One idea

$\log n! =\sum_{k=1}^n \log k \leq \sum_{k=1}^n \log n=n\log n$

The other approach would be

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

From where

$\log n !\sim n\log n -n+\frac{1}{2} \log \pi n$

$\frac{\log n !}{n \log n}\sim 1-\frac 1 {\log n}+\frac{1}{2} \frac {\log \pi n} {n \log n}$

  • 0
    Thanks! Using Stirling's Approximation like that does the trick.2012-05-23
5

By Stolz Cezaro

$\lim_{n \to \infty} \frac{\ln (n!)}{n \ln n}=lim_{n \to \infty} \frac{\ln ((n+1)!)- \ln(n!)}{(n+1) \ln (n+1)-n \ln n}=lim_{n \to \infty} \frac{\ln ((n+1))}{\ln (n+1)+n [\ln(n+1)- \ln n]} $

$=lim_{n \to \infty} \frac{\ln ((n+1))}{\ln (n+1)+ [\ln(1+ \frac{1}{n})^n]}=1 $

Thus $\ln (n!) \sim n \ln n$

This implies both big O and Omega...