 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ia/paper-1/12f/a/iii/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Apply the root test to the terms

a_{n} z^{n^{2}}

∣ a_{n} z^{n^{2}} ∣^{1/ n} = ∣ a_{n} ∣^{1/ n} ∣ z ∣^{n} .

This tends to zero when

∣ z ∣ < 1

and to infinity when

∣ z ∣ > 1

. On

∣ z ∣ = 1

, the series converges absolutely because

\sum ∣ a_{n} ∣

converges: the point

1

lies strictly inside the original radius two. Thus the lacunary power series has radius

1

and in fact converges on its entire boundary circle.

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