 Solution

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

a_{n} = ⌈ α^{- n} ⌉ (n \geq 0) .

Because

0 < α < 1

, we have

α^{- n} \geq 1

and

α^{- n} \leq a_{n} < α^{- n} + 1 \leq 2 α^{- n} .

For

∣ z ∣ < α

\sum_{n = 0}^{\infty} ∣ a_{n} z^{n} ∣ \leq 2 \sum_{n = 0}^{\infty} (\frac{∣ z ∣}{α})^{n} < \infty.

For

∣ z ∣ \geq α

, we instead have

∣ a_{n} z^{n} ∣ \geq (∣ z ∣/ α)^{n} \geq 1

, so the terms do not tend to zero. Thus the power series has radius of convergence exactly $α$ .

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