 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ii/paper-2/32e/a/i/solution

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

For

n \geq 1

, put

r_{n} = ϕ_{n} / ϕ_{n - 1}

. Since

(ϕ_{n})

r_{n} \to 0

. We can rewrite

ψ_{n} = \frac{ϕ _{n}}{1 + r _{n}},

\frac{ψ _{n}}{ϕ _{n}} = \frac{1}{1 + r _{n}} ⟶ 1.

Together with

ψ_{0} = ϕ_{0}

, this proves

ψ_{n} \sim ϕ_{n} for every n \geq 0.

Moreover,

\frac{ψ _{n + 1}}{ψ _{n}} = \frac{ψ _{n + 1}}{ϕ _{n + 1}} \frac{ϕ _{n + 1}}{ϕ _{n}} \frac{ϕ _{n}}{ψ _{n}} ⟶ 1 \cdot 0 \cdot 1 = 0.

Hence $(ψ_{n})$ is an asymptotic sequence. This is the harmonic-mean refinement of an asymptotic sequence.

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