 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ia/paper-1/4f/ii/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

The inequality

(1 + 1/ n)^{n} \leq e

gives, for

∣ x ∣ < 1

\sum_{n = 1}^{\infty} ∣ x ∣^{n} (1 + \frac{1}{n})^{n} \leq e \sum_{n = 1}^{\infty} ∣ x ∣^{n} < \infty.

For

∣ x ∣ > 1

, the terms have magnitude at least

∣ x ∣^{n}

and therefore do not tend to zero, so the series diverges by the term test for divergence. Hence

R = 1 .

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