solve $\ln(n!) = \Theta(n\ln(n))$ without stirling approximation

Question

My homework was proving this equation which is simple using Stirling approximation. I was wondering if there is any other method to prove it - without Stirling - I can prove $\ln(n!) = O(n\ln(n))$ like this: $$\ln(n!)=\ln(n\cdot(n-1)\cdots2\cdot1)=\ln(n)+\ln(n-1)+\cdots+\ln(2)+\ln(1)≤n\ln(n)$$ which is obvious.

But I can't prove that $\ln(n!) = \Omega(n\ln(n))$.

$\log n! \geqslant \sum_{k=n/2}^n \log k \geqslant n/2\log(n/2)$. — Daniel Fischer, Oct 10 '13 at 13:26
thanks. but does log(n!) >= n/2log(n/2) mean Omega(nlogn)? I'm very new to these subjects... — Sida, Oct 10 '13 at 13:33
For $n \geqslant 4$, you have $n/2 \geqslant \sqrt{n}$, so $\log (n/2) \geqslant \frac12 \log n$, thus for $n\geqslant 4$, you have $\log (n!) \geqslant \frac14 n\log n$. — Daniel Fischer, Oct 10 '13 at 13:36
The one you mentioned @ThomasAndrews, means log(n!) = O(nlogn); not log(n!) = Omega(nlogn), am I right? — Sida, Oct 10 '13 at 13:44
See http://math.stackexchange.com/questions/46892/how-do-you-prove-that-nn-is-on2. — lhf, Oct 10 '13 at 13:48
Yeah, I deleted my comment because I hadn't read the whole question. @PardisPashakhanloo — Thomas Andrews, Oct 10 '13 at 13:48
@DanielFischer so, how do you get [log(n!) >= 1/4 nlogn] in the end? — Sida, Oct 10 '13 at 13:50

score 9 · Accepted Answer · answered Oct 11 '13 at 11:30

Use a multiplicative variant of Gauss's trick: $$ (n!)^2 = (1 \cdot n) (2 \cdot (n-1)) (3 \cdot (n-2)) \cdots ((n-2) \cdot 3) ((n-1) \cdot 2) (n \cdot 1) \ge n^n $$ This implies that $\ln(n!) \ge \dfrac12 n \ln n$.

The other direction is easy, as you mention, because $n! \le n^n$ and so $\ln(n!) \le n \ln n$.

So $\dfrac12 n \ln n \le \ln(n!) \le n \ln n$ and $\ln(n!) = \Theta(n\ln(n))$.

solve $\ln(n!) = \Theta(n\ln(n))$ without stirling approximation

1 Answers1

Linked