Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 2 / 32E / a / i

Table of contents
- Solution i

Solution

For

n \geq 1

, put

r_{n} = ϕ_{n} / ϕ_{n - 1}

. Since

(ϕ_{n})

is an asymptotic sequence,

r_{n} \to 0

. We can rewrite

ψ_{n} = \frac{ϕ _{n}}{1 + r _{n}},

so

\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.

 Ancestors (11)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version