Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

Why?

share|improve this question
7  
$$\log n! =\sum_{k=1}^n \log k \leq \sum_{k=1}^n \log n$$ –  Pedro Tamaroff May 4 '12 at 16:02
6  
The two statements you give are consistent. In fact both are true. –  Qiaochu Yuan May 4 '12 at 16:03
    
$\log(n!)$ is actually $\Theta(n \log n )$. See math.stackexchange.com/questions/46892/… –  lhf May 4 '12 at 16:12
    
@lhf Indeed, $\dfrac 1 2 n \log n$ and $n \log n$ suffice. –  Pedro Tamaroff May 4 '12 at 16:14
2  
The obvious inequalities $(n/2)^{n/2} \leq n! \leq n^n$ suffice to prove $\Theta(n \log n)$. –  sdcvvc May 4 '12 at 16:18

2 Answers 2

up vote 4 down vote accepted

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

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

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

share|improve this answer
    
Thanks! Using Stirling's Approximation like that does the trick. –  David Faux May 23 '12 at 21:25

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...

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.