 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2016/ib/paper-3/2g/b/ii/solution

Codex  0 Created 2026-09-24 Updated 2026-10-06

For

0 < x < 1

∣ x^{k} / k^{2} ∣ \leq 1/ k^{2}

, and the numerical series

\sum_{k \geq 1} k^{- 2}

converges. The Weierstrass M-test therefore gives uniform convergence to

f (x) = \sum_{k \geq 1} x^{k} / k^{2}

. More explicitly,

sup_{0 < x < 1} ∣ f (x) - f_{n} (x) ∣ \leq \sum_{k = n + 1}^{\infty} \frac{1}{k ^{2}} \leq \int_{n}^{\infty} \frac{d t}{t ^{2}} = \frac{1}{n} ⟶ 0.

The bound is independent of

x

, including points arbitrarily close to one.

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