 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ia/paper-2/3f/a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

Since the natural logarithm is increasing, comparison with an integral gives

\int_{1}^{n} lo g x d x \leq \sum_{k = 1}^{n} lo g k \leq n lo g n .

The lower bound is

n lo g n - n + 1

. Dividing all three expressions by

n lo g n

and applying the squeeze theorem gives

lo g (n!) \sim n lo g n .

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