 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ii/paper-2/31d/b/i/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

Put

a = 1 + ϵ

and, for

m \geq 2

I_{m} (ϵ) = \int_{0}^{\infty} (1 + ϵ t)^{- m} e^{- a t} d t .

I_{m} (ϵ) = \frac{1}{a} - \frac{m ϵ}{a} I_{m + 1} (ϵ) .

Starting with

I = I_{2}

and iterating

N + 1

times yields

I (ϵ) = \sum_{n = 0}^{N} (- 1)^{n} (n + 1)! \frac{ϵ ^{n}}{( 1 + ϵ ) ^{n + 1}} + R_{N} (ϵ),

where

R_{N} (ϵ) = (- 1)^{N + 1} (N + 2)! \frac{ϵ ^{N + 1}}{( 1 + ϵ ) ^{N + 1}} I_{N + 3} (ϵ) .

Since

0 < I_{N + 3} (ϵ) \leq (1 + ϵ)^{- 1}

, the remainder is

O (ϵ^{N + 1})

, which is little-

o

of the last retained term. Therefore

I (ϵ) \sim n = 0 \sum \infty (- 1)^{n} (n + 1)! \frac{ϵ ^{n}}{( 1 + ϵ ) ^{n + 1}} (ϵ \to 0^{+}) .

The displayed terms form an asymptotic sequence because the ratio of successive magnitudes tends to zero.

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