Prove $\log x!$ is $\Omega (xlogx)$

Question

Find a positive real number $C$ and a nonnegative real number $x_o$ such that

$Cx$$\log x$ $\leq$ $\log x!$ for all real numbers $x > x_o$.

I tried to expand $\log x!$ into $\log 1 + \log2 +\log3 +....\log x$. But how do I choose $C$ and $x_o$ so the above inequality hold?

Any hints would be appreciated.

Answer 1

Note that $$e^n = \sum_{k=1}^{\infty} \dfrac{n^k}{k!} \geq \dfrac{n^n}{n!} \implies \log(n!) \geq n \log(n) - n$$ Also, $$\log(n!) = \sum_{k=1}^n \log(k) \leq \sum_{k=1}^n \log(n) = n \log(n)$$ We hence have $$n \log(n)-n \leq \log(n!) \leq n \log(n)$$

Answer 2

For $k\ge 2$, we have $\log k\ge \int_{k-1}^{k}\log t\,dt$. Summing from $k=2$ to $n$, we find that $$\log 2 +\log 3 + \log 4+ \cdots +\log n \ge \int_1^n \log t\,dt=n\log n-n.$$
If $n\ge 9$, then $n\log n-n \gt \frac{1}{2}n\log n$.