Proof that $\log n! = n \log n - n + O(\log n)$

Question

This is from Cameron's Combinatorics text (Theorem 3.6.1). His given proof is a one liner

$n \log n - n + 1 \leq \log n! \leq n \log n- n + (\log(n + 1) + 2 - \log 2)$

He doesn't explain himself. Where exactly is he pulling these inequalities from?

Answer 1

He is comparing $\log n!=\sum_{k=1}^n\log k$ with $\int_1^n\log x\,dx$. Certainly $$I=\int_1^n\log x\,dx=n\log n-n+1\le\sum_{k=2}^n\log k=\log n!$$ since $\log x\le\log k$ on the interval $[k-1,k]$.

Answer 2

This is related to Stirling's approximation for the factorial.

$\ln n! = \sum_{i=1}^{n} \ln(n) \ge \int_{1}^{n} \ln(x) \ dx = n\ln n - n + 1$

Thus, $\ln n! \ge n\ln n - n + 1$

See here: https://en.wikipedia.org/wiki/Stirling%27s_approximation

Answer 3

Playing around.

$\begin{array}\\ \ln(n!) &=\sum_{k=1}^n \ln(k)\\ &=\sum_{k=1}^n \int_1^k \dfrac{dt}{t}\\ &=\sum_{k=1}^n \sum_{j=1}^{k-1}\int_{j}^{j+1} \dfrac{dt}{t}\\ &=\sum_{j=1}^{n-1}\sum_{k=j+1}^n \int_{j}^{j+1} \dfrac{dt}{t}\\ &=\sum_{j=1}^{n-1}(n-j) \int_{j}^{j+1} \dfrac{dt}{t}\\ &=\sum_{j=1}^{n-1}n\int_{j}^{j+1} \dfrac{dt}{t}-\sum_{j=1}^{n-1}j \int_{j}^{j+1} \dfrac{dt}{t}\\ &=n\int_{1}^{n} \dfrac{dt}{t}-\sum_{j=1}^{n-1}j\ln(1+1/j)\\ &=n\ln(n)-\sum_{j=1}^{n-1}j\ln(1+1/j)\\ \end{array} $

Therefore, if $f(n) = n\ln(n)-\ln(n!)$, since $x-x^2/2 < \ln(1+x) < x$,

$\begin{array}\\ f(n) &=\sum_{j=1}^{n-1}j\ln(1+1/j)\\ &<\sum_{j=1}^{n-1}j(1/j)\\ &=n-1\\ \text{and}\\ f(n) &=\sum_{j=1}^{n-1}j\ln(1+1/j)\\ &>\sum_{j=1}^{n-1}j(\frac1{j}-\frac1{2j^2})\\ &=n-1-\sum_{j=1}^{n-1}\frac1{2j}\\ &>n-1-\frac12\ln(n)\\ \end{array} $