 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2008/paper-1/13/a/solution

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

For

∣ t ∣ < 1

(1 + t)^{- 1} = \sum_{n \geq 0} (- t)^{n}

. Its uniform convergence on every closed interval strictly inside

(- 1, 1)

permits termwise integration from zero to

x

. Therefore

lo g (1 + x) = \sum_{n = 1}^{\infty} \frac{( - 1 ) ^{n + 1} x ^{n}}{n}, lo g (1 - x) = - \sum_{n = 1}^{\infty} \frac{x ^{n}}{n} .

Subtracting cancels every even power and doubles every odd power, giving the Taylor series

lo g \frac{1 + x}{1 - x} = 2 j = 0 \sum \infty \frac{x ^{2 j + 1}}{2 j + 1}, ∣ x ∣ < 1.

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