Show that $\operatorname{ln}(n!)=\Theta(n\operatorname{ln}(n))$

Question

Another question about asymptotic approximations.

We are asked to show that $\operatorname{ln}(n!)=\Theta(n\operatorname{ln}(n))$

I'm stuck tho and can use help.

What I did is:

$\operatorname{ln}(n!) = \operatorname{ln}(n)+\operatorname{ln}(n-1)+...+\operatorname{ln}(1) \leq n\operatorname{ln}(n)$ and so $\operatorname{ln}(n!)=O(n\operatorname{ln}(n))$.

That's the easy part.

How do I show that $\operatorname{ln}(n!)=\Omega(n\operatorname{ln}(n))$?

Answer 1

Since the function $x\mapsto \ln x$ is increasing we have $$\ln(n!)=\sum_{k=1}^n\ln k\sim_\infty\int_1^n\ln t dt\sim_\infty n\ln n$$

Added Using the monotonicity of the $\ln $ function we have

$$\int_{k-1}^k\ln tdt\le\ln k\le\int_k^{k+1}\ln t dt\;\;\forall k\ge2$$ then by summing for $k=2$ to $n$ and using some algebra we find easily $$\sum_{k=1}^n\ln k\sim_\infty\int_1^n\ln t dt$$