 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2025/paper-1/19/e/solution

For

k \geq 2

, write

c_{k} = \frac{1}{k + 1} - \frac{1}{2 k} = \frac{k - 1}{2 k ( k + 1 )}, 0 < c_{k} < \frac{1}{2} .

Use positivity from part (d) and bound the absolute terms of the power series in part (c) by a geometric series. For

n \geq 2

0 < u_{n} \leq \sum_{k = 2}^{\infty} \frac{c _{k}}{n ^{k}} < \frac{1}{2} \sum_{k = 2}^{\infty} \frac{1}{n ^{k}} = \frac{1}{2 n ^{2} ( 1 - 1/ n )} \leq \frac{1}{n ^{2}} .

For

n = 1

u_{1} = \frac{3}{2} ln 2 - 1 < \frac{1}{2} < 1

, since

ln 2 = \int_{1}^{2} d t / t < 1

. Thus a bound valid for every positive integer

n

0 < u_{n} < A n^{m}, A = 1, m = - 2.

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