 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ii/paper-4/11f/solution

Codex  0 Created 2026-09-23 Updated 2026-10-03

p_{- 2} = 0, p_{- 1} = 1, q_{- 2} = 1, q_{- 1} = 0,

and recursively

p_{n} = a_{n} p_{n - 1} + p_{n - 2}, q_{n} = a_{n} q_{n - 1} + q_{n - 2} .

Then

p_{n} / q_{n} = [a_{0}, l d o t s, a_{n}]

. The determinant identity

p_{n} q_{n - 1} - p_{n - 1} q_{n} = (- 1)^{n - 1}

follows by induction. Writing the remaining complete quotient as

θ_{n + 1} > 1

gives

θ = \frac{p _{n} θ _{n + 1} + p _{n - 1}}{q _{n} θ _{n + 1} + q _{n - 1}},

θ

lies strictly between consecutive convergents and their order alternates. For odd

n

\frac{p _{n - 1}}{q _{n - 1}} < θ < \frac{p _{n}}{q _{n}} .

Moreover

∣ θ - p_{n} / q_{n} ∣ < 1/ (q_{n} q_{n + 1})

, and

q_{n} \to \infty

, proving convergence.

7 = [2; \overline{1, 1, 1, 4}] .

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