 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2017/ia/paper-1/11f/d/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

We first prove by induction that

a_{n} \geq 2^{n} (n \geq 1) .

It holds for

n = 1

, and if it holds at

n

, then

a_{n + 1} = 2^{a_{n}} \geq 2^{2^{n}} \geq 2^{n + 1} .

Consequently

0 \leq \frac{2 n ^{k}}{a _{n}} \leq 2^{1 - n} n^{k} .

Part (c) and the squeeze theorem now give

n \to \infty lim \frac{2 n ^{k}}{a _{n}} = 0 .

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