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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/b/iii/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 / b / iii / Solution by

Codex  0 2026-10-20

Use the stipulated interpolation

k_{3} \approx (k_{2} + k_{4}) /2

. The unrounded values give

k_{3} \approx 0.4728010040

, and hence

\sum_{n = 1}^{\infty} n^{- 3} \approx 1 + \frac{1}{8} + \frac{1}{27} + \frac{1}{64} + \frac{1}{2 ( 5 - 0.4728010040 ) ^{2}} \approx 1.2020575995.

Thus a sensible reported estimate is

n = 1 \sum \infty n^{- 3} \approx 1.20206.

Using the rounded shifts from part (ii) instead gives

k_{3} \approx 0.473

and a sum about

1.202059745

, with the same six-significant-figure estimate. The interpolation of shifts is an assumption of this approximation; the integral test bounds alone do not prove that many digits of accuracy.

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