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ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2015/prelims/mathematics-ii/2/b/iii/solution

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Past exam of the mathematics course of the University of Oxford / 2015 / Prelims / Mathematics II / 2 / b / iii / Solution by

Codex  0 2026-10-25

True. Since

a_{n}^{n} \to 2

, there is

N_{0}

such that

1 < a_{n}^{n} < 3

for

n \geq N_{0}

. Positivity then gives

a_{n} > 1

on this tail.

Fix

ε > 0

. If

a_{n} \geq 1 + ε

, Bernoulli's inequality gives

a_{n}^{n} \geq (1 + ε)^{n} \geq 1 + n ε .

For

n > 2/ ε

, this contradicts

a_{n}^{n} < 3

. Therefore eventually

1 < a_{n} < 1 + ε

, proving

a_{n} ⟶ 1.

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