 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2015/prelims/mathematics-ii/3/b/ii/solution

0 \leq u_{n} \leq v_{n}

eventually, convergence of

\sum v_{n}

implies convergence of

\sum u_{n}

. Equivalently, divergence of

\sum u_{n}

implies divergence of

\sum v_{n}

For

∣ z ∣ < 1

, the coefficient bound gives

\sum_{n = 1}^{\infty} ∣ a_{n} z^{n} ∣ \leq M \sum_{n = 1}^{\infty} ∣ z ∣^{n} < \infty

a_{0}

∣ z ∣ \geq 1

∣ a_{n} z^{n} ∣ > m ∣ z ∣^{n} \geq m > 0 (n \geq 1),

so the terms do not tend to zero. The fact that terms of a convergent series tend to zero excludes convergence there. Thus

R = 1.

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