 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2017/prelims/mathematics-ii/5/c/ii/solution

The finite geometric series identity gives, for each

∣ x ∣ < 1

f_{n} (x) = \frac{x ( 1 - x ^{n} )}{1 - x} ⟶ f (x) = \frac{x}{1 - x}, f (x) - f_{n} (x) = \frac{x ^{n + 1}}{1 - x} .

For every fixed

n

, the absolute value of this error tends to infinity as

x \to 1^{-}

. Thus

- 1 < x < 1 sup ∣ f (x) - f_{n} (x) ∣ = \infty,

and there is no uniform convergence on

(- 1, 1)

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