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 Incoming links: Integral bounds for a logarithmic factorial

Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2005 / Paper 1 / 7 / a / Solution Created 2026-10-18 Updated 2026-10-20

The logarithm turns the product defining the factorial into

ln (N!) = \sum_{k = 2}^{N} ln k

. Since

ln x

is increasing, each term bounds integrals on its adjacent unit intervals:

\int_{k - 1}^{k} ln x d x \leq ln k \leq \int_{k}^{k + 1} ln x d x .

Summing from

k = 2

to

N

proves the integral bounds for a logarithmic factorial:

N ln N - N + 1 \leq ln (N!) \leq (N + 1) ln (N + 1) - N + 1 - 2 ln 2.

For

N = 1

, both sides and the empty sum are zero. For

N \geq 2

, the rectangle diagram gives the same inequalities geometrically. These bounds establish the leading

N ln N - N

growth, while Stirling's formula gives finer corrections.

Figure 1.
Unit rectangles bound the logarithm of a factorial between two integrals
. Original example with $N = 6$ . Left: each right-endpoint rectangle lies above $ln x$ on $[k - 1, k]$ . Right: the same rectangle heights shifted to $[k, k + 1]$ lie below $ln x$ . Both rectangle totals equal $ln (6!)$ . Details.

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