How to prove this approximation of logarithm of factorial

Question

In other words, how to prove this equation

$$\lg{n!} = \Theta(n\lg{n})$$

Answer 1

Hint: $$ n\log \bigg(\frac{n}{e}\bigg) + 1 \le \log n! \le (n + 1)\log \bigg(\frac{{n + 1}}{e}\bigg) + 1. $$ Taken from Wikipedia, see here.

EDIT: Actually it is stated in that link "Hence $\log n!$ is $\Theta (n\log n)$''.

EDIT 2: The derivation of the above result is given in the Wikipedia link (and is very elementary). Based on it, let's show that $$ \log n! = n \log n - n + O(\log n) \;\; {\rm as}\; n \to \infty, $$ which implies, in particular, that $$ \log n! \sim n\log n \;\; {\rm as}\; n \to \infty $$ (that is, $\log n! / (n\log n) \to 1$ as $n \to \infty$), which implies, in particular, that $\log n!$ is $\Theta (n\log n)$. So, first note that $$ n\log n - n + 1 \le \log n! \le (n + 1)\log (n + 1) - (n + 1) + 1. $$ Hence, $$ n\log n - n + 1 \le \log n! \le n\log (n + 1) + \log (n + 1) - n. $$ Next note that, since $\log(1+x) = x + O(x^2)$ as $x \to 0$, $$ \log(n+1) - \log n = \log \bigg(\frac{{n + 1}}{n}\bigg) = \log \bigg(1 + \frac{1}{n}\bigg) = \frac{1}{n} + O\bigg(\frac{1}{{n^2 }}\bigg), $$ and hence $$ \log(n+1) = \log n + \frac{1}{n} + O\bigg(\frac{1}{{n^2 }}\bigg). $$ Now, $$ n\log n - n + 1 \le \log n! \leq n\bigg[\log n + \frac{1}{n} + O\bigg(\frac{1}{{n^2 }}\bigg)\bigg] + \bigg[\log n + \frac{1}{n} + O\bigg(\frac{1}{{n^2 }}\bigg)\bigg] - n, $$ hence $$ n\log n - n + 1 \le \log n! \leq n\log n - n + \log n + O(1), $$ and finally $$ \log n! = n \log n - n + O(\log n). $$

Answer 2

The easy part is $\log(n!) \le n \log n$, which follows from $n! \le n^n$.

The hard part is discussed in How do you prove that $n^n$ is $O(n!^2)$?.

Answer 3

$e^x = \sum_{n \ge 0} \frac{x^n}{n!}$ implies $e^n \ge \frac{n^n}{n!}$, hence $n! \ge \left( \frac{n}{e} \right)^n$, hence $\log n! \ge n \log n - n$. On the other hand,

$$n! \le \left( \frac{1 + 2 + ... + n}{n} \right)^n = \left( \frac{n+1}{2} \right)^n$$

by AM-GM, hence $\log n! \le n \log \left( \frac{n+1}{2} \right)$.

Answer 4

Looking at the graph of $\log$ one immediately verifies the inequalities $$\int_1^n\log x\ dx <\sum_{k=1}^n\log k<\int_2^{n+1}\log x\ dx\ .$$ Evaluating the integrals one gets $$n\log n -n+1< \log(n!)<(n+1)\log(n+1)-n + c$$ with $c=1-\log 4$, and this easily implies ${\log(n!)\over n\log n}\to 1$ $(n\to\infty)$.