 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2025/paper-1/18/d/solution

x = π /2

, all odd-indexed cosine terms vanish, while

cos (2 kπ /2) = (- 1)^{k}

. Thus the Fourier series from part (c) gives

e^{π /2} = \frac{e ^{π} - 1}{π} + \frac{2 ( e ^{π} - 1 )}{π} \sum_{k = 1}^{\infty} \frac{( - 1 ) ^{k}}{4 k ^{2} + 1} .

Divide by

2 (e^{π} - 1) / π

and rearrange. The hyperbolic sine identity

e^{π} - 1 = 2 e^{π /2} sinh (π /2)

yields

k = 1 \sum \infty \frac{( - 1 ) ^{k}}{4 k ^{2} + 1} = \frac{π}{4 sinh ( π /2 )} - \frac{1}{2}, m = 4.

The evaluation is legitimate by the convergence established in part (c).

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