How to prove log n! = Ω(nlog n)?

Question

I saw this proof but i didn't understand the bold part

log(n!) = log(1) + log(2) + log(3) + ... + log(n)

Deleting the first half of the terms gives

log(n!) >= log(n/2) + log(n/2+1) + log(n/2+2) + ... + log(n)

could any one explain to me this part?

Answer 1

Well, for any $k \ge 1$ we have $\log ( k) \ge 0$.
And as a sum of positive elements is positive, we have $\sum_{k=1}^{\left\lfloor\frac{n}{2}\right\rfloor -1} \log (k) \ge 0 $. So we have $$ \log( n!) = \sum_{k=1}^n \log(k) = \sum_{k=1}^{\left\lfloor\frac{n}{2}\right\rfloor-1} \log(k) + \sum_{k = \left\lfloor\frac{n}{2}\right\rfloor}^n \log (k) \ge \sum_{k = \left\lfloor\frac{n}{2}\right\rfloor}^n \log (k) = \sum_{k=0}^{\left\lceil \frac{n}{2} \right\rceil } \log (\left\lfloor \frac{n}{2} \right\rfloor +k)$$

Answer 2

For example when $n=6$: $$\begin{aligned}\log(6!)&=\log(1)+\log(2)+\log(3)+\log(4)+\log(5)+\log(6)\\ &\ge\log(3)+\log(4)+\log(5)+\log(6)\\ &=\log\frac{6}{2}+\log\left(\frac{6}{2}+1\right)+\log\left(\frac{6}{2}+2\right)+\log\left(\frac{6}{2}+3\right). \end{aligned}$$