Stirling’s Formula

Theorem 2.1. For all n ≥ 1, n log n− n < log(n!) < n log n, so log(n!) ∼ n log n. Proof. The inequality log(n!) < n log n is a consequence of the trivial inequality n! < nn. Here are three methods of showing n log n− n < log(n!). Method 1: A Riemann sum approximation for ∫ n 1 log x dx using right endpoints is log 2+ · · ·+ log n = log(n!), which overestimates, so log(n!) > ∫ n 1 log x dx = n log n− n+ 1. Method 2: The power series expansion of en is ∑ k≥0 n k/k!. Comparing en to the nth term in the series gives us en > nn/n!, so n! > nn/en. Therefore log(n!) > n log n− n. Method 3: For all k ≥ 1, e > (1 + 1/k)k. Multiplying this over k = 1, 2, . . . , n− 1, we get en−1 > nn−1/(n− 1)! = nn/n!, so n! > enn/en. Thus log(n!) > n log n− n+ 1. Dividing through the inequality n log n − n < log(n!) < n log n by n log n, we obtain 1− 1/ log n < log(n!)/(n log n) < 1, so log(n!) ∼ n log n.