 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2014/ia/paper-1/9d/b/solution

Codex  0 Created 2026-09-24 Updated 2026-10-06

Let

r = e^{i θ} \neq = 1

. The geometric series from any starting index is bounded:

B_{j} = \sum_{n = m}^{j} r^{n} = \frac{r ^{m} ( 1 - r ^{j - m + 1} )}{1 - r}, ∣ B_{j} ∣ \leq K = \frac{2}{∣1 - r ∣} .

\sum_{n = m}^{N} a_{n} r^{n} = a_{N} B_{N} + \sum_{n = m}^{N - 1} (a_{n} - a_{n + 1}) B_{n} .

Since the coefficients are nonincreasing and positive, the modulus is at most

K (a_{N} + a_{m} - a_{N}) = K a_{m} \to 0

. The Cauchy criterion proves convergence for every $θ \in / 2 π Z$ . This is a direct proof of the Dirichlet test in this setting.

For

a_{n} = 1/ n

, the imaginary parts of the convergent partial sums give convergence of the sine series at those angles. If

θ \in 2 π Z

, every sine term is zero. Hence the sine series converges for all real angles.

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