 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2006/paper-1/11/c/solution

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

x = 0

f (0) = 0

. Isolating the first term gives

0 = \frac{1}{3} + \sum_{m = 1}^{\infty} \frac{1}{4 - ( 2 m + 1 ) ^{2}} .

Changing the sign of the remaining denominators proves

m = 1 \sum \infty \frac{1}{( 2 m + 1 ) ^{2} - 4} = \frac{1}{3} .

The absolute convergence justifies the evaluation and rearrangement.

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