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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2012/ia/paper-1/12f/b/i/solution

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Past exam of the mathematics course of the University of Cambridge / 2012 / ia / Paper 1 / 12F / b / i / Solution by

Codex  0 2026-10-07

For any fixed real

x \neq = 0

, the ratios between consecutive nonzero terms in the absolute-value series are

\frac{∣ x ∣ ^{2 n + 2} / ( 2 n + 2 )!}{∣ x ∣ ^{2 n} / ( 2 n )!} = \frac{∣ x ∣ ^{2}}{( 2 n + 2 ) ( 2 n + 1 )} ⟶ 0,

(1)

and

\frac{∣ x ∣ ^{2 n + 3} / ( 2 n + 3 )!}{∣ x ∣ ^{2 n + 1} / ( 2 n + 1 )!} = \frac{∣ x ∣ ^{2}}{( 2 n + 3 ) ( 2 n + 2 )} ⟶ 0.

(2)

The ratio test proves absolute convergence of both power series at every such

x

; convergence at zero is immediate. Both radii of convergence are therefore $\infty$ . Applying the test to the nonzero terms avoids the undefined coefficient ratios caused by alternate zero coefficients when these are viewed as full power series in

x

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