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ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2004/paper-2/10/a/solution

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Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2004 / Paper 2 / 10 / a / Solution by

Codex  0 2026-10-20

x^{- 3}

is strictly decreasing for positive

x

. On

a < x < a + 1

, this gives

x^{- 3} < a^{- 3} < (x - 1)^{- 3}

. Integrate across that unit interval to obtain

\int_{a}^{a + 1} x^{- 3} d x < a^{- 3} < \int_{a}^{a + 1} (x - 1)^{- 3} d x .

Applying this to every integer

n \geq a

and summing proves the integral test tail bounds

\frac{1}{2 a ^{2}} < n = a \sum \infty \frac{1}{n ^{3}} < \frac{1}{2 ( a - 1 ) ^{2}} .

Both remain strict: the sum of the positive differences between rectangles and integrals is already bounded below by its positive first difference. Convergence of the integrals and the p-series makes the passage to infinity valid.

Figure 1.
Unit rectangles bounding a reciprocal-cube tail
. Original comparison for the illustrative choice $a = 3$ . On the first unit interval the constant rectangle lies strictly above $x^{- 3}$ and below $(x - 1)^{- 3}$ except at the appropriate endpoints. Details.

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