 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2006/paper-1/9/b/iii/solution

Codex  0 Created 2026-10-18 Updated 2026-10-20

The removable value at zero is one. Expand the denominator as

e^{x} - 1 = x (1 + x /2 + x^{2} /6 + \dots)

and determine the reciprocal coefficients by multiplication. The first four nonzero terms are

\frac{x}{e ^{x} - 1} = 1 - \frac{x}{2} + \frac{x ^{2}}{12} - \frac{x ^{4}}{720} + O (x^{6}) .

B_{1} = - 1/2

x^{3}

and

x^{5}

are zero.

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