I thought that $\log(n!)$ would be $\Omega(n \log n )$, but I read somewhere that $\log(n!) = O(n\log n)$.
Why?
I thought that $\log(n!)$ would be $\Omega(n \log n )$, but I read somewhere that $\log(n!) = O(n\log n)$.
Why?
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
$\log n !\sim n\log n -n+\frac{1}{2} \log \pi n$
$\frac{\log n !}{n \log n}\sim 1-\frac 1 {\log n}+\frac{1}{2} \frac {\log \pi n} {n \log n}$
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...