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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2006/ia/paper-1/4d/solution

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Past exam of the mathematics course of the University of Cambridge / 2006 / ia / Paper 1 / 4D / Solution by

Codex  0 Created 2026-09-24 Updated 2026-10-14

Use these standard facts about a power series: it is absolutely convergent at every point strictly inside its radius of convergence and divergent at every point strictly outside it. Also, the triangle inequality implies that a sum or difference of two absolutely convergent series is absolutely convergent.

Let

T

be the radius of convergence of the coefficientwise sum. For

∣ z ∣ < min (R, S)

\sum_{n \geq 0} ∣ (a_{n} + b_{n}) z^{n} ∣ \leq \sum_{n \geq 0} ∣ a_{n} z^{n} ∣ + \sum_{n \geq 0} ∣ b_{n} z^{n} ∣ < \infty.

(1)

Hence

T \geq min (R, S)

Suppose

R < S

. If

T > R

, choose a finite

ρ

with

R < ρ < min (T, S)

and choose

z

of complex modulus

ρ

. The sum series and the

b_{n}

series both converge absolutely there, so

\sum_{n \geq 0} ∣ a_{n} z^{n} ∣ \leq \sum_{n \geq 0} ∣ (a_{n} + b_{n}) z^{n} ∣ + \sum_{n \geq 0} ∣ b_{n} z^{n} ∣ < \infty,

(2)

contradicting

ρ > R

. Thus

T = R

. Interchanging the series covers

S < R

, proving the sum of power series with unequal radii of convergence rule:

T = min (R, S) .

(3)

The argument also covers a zero radius or an infinite larger radius. When the two radii are equal, cancellation can enlarge the radius, so the inequality of the radii matters.

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