 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ii/paper-2/32e/b/solution

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

Recall that positivity and asymptoticity mean

\frac{ϕ _{j + 1}}{ϕ _{j}} \to 0, \frac{ψ _{j + 1}}{ψ _{j}} \to 0

for each fixed

j

. For

0 \leq k \leq n

, compare the

k

th term of

χ_{n + 1}

with the

k

th term of

χ_{n}

\frac{ϕ _{k} ψ _{n + 1 - k}}{ϕ _{k} ψ _{n - k}} = \frac{ψ _{n + 1 - k}}{ψ _{n - k}} \to 0.

For the remaining term, compare with the final term of

χ_{n}

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

Since

χ_{n}

is positive and dominates every comparison term, summing these finitely many bounds gives

\frac{χ _{n + 1} ( x )}{χ _{n} ( x )} \to 0 .

Thus $(χ_{n})$ is an asymptotic sequence.

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