 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ia/paper-1/4e/solution

Codex  0 Created 2026-09-23 Updated 2026-10-03

The comparison test says that if

0 \leq b_{n} \leq c_{n}

eventually and

\sum c_{n}

converges, then

\sum b_{n}

converges. If

\sum a_{n} z_{0}^{n}

converges, its terms are bounded:

∣ a_{n} z_{0}^{n} ∣ \leq M

. For

∣ z_{1} ∣ < ∣ z_{0} ∣

∣ a_{n} z_{1}^{n} ∣ \leq M \frac{z _{1}}{z _{0}}^{n},

so comparison with a geometric series proves absolute convergence.

R

is the number for which the power series converges absolutely for

∣ z ∣ < R

and diverges for

∣ z ∣ > R

. Given

r_{i} < R_{i}

, both

\sum ∣ a_{n} ∣ r_{1}^{n}

and

\sum ∣ b_{n} ∣ r_{2}^{n}

converge, so their terms are bounded, say

∣ a_{n} ∣ r_{1}^{n} \leq M

. Then

\sum ∣ a_{n} b_{n} ∣ (r_{1} r_{2})^{n} \leq M \sum ∣ b_{n} ∣ r_{2}^{n} < \infty.

Thus $R \geq r_{1} r_{2}$ ; letting $r_{i} ↑ R_{i}$ gives $R \geq R_{1} R_{2}$ .

 New to topics? Read the docs here!