 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2013/ia/paper-1/4f/a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-07

Let

s_{N} = \sum_{n = 1}^{N} (- 1)^{n - 1} b_{n}

. Pairing consecutive terms shows

s_{2 m} \geq 0

and

s_{2 m + 2} - s_{2 m} = b_{2 m + 1} - b_{2 m + 2} \geq 0, s_{2 m + 3} - s_{2 m + 1} = - b_{2 m + 2} + b_{2 m + 3} \leq 0.

The even partial sums increase and the odd partial sums decrease. They satisfy

s_{2 m} \leq s_{2 m + 1} \leq b_{1}

, while

s_{2 m + 1} - s_{2 m} = b_{2 m + 1} ⟶ 0.

Both are monotone bounded sequences and hence have limits, and the displayed difference makes their limits equal. Every partial sum belongs to one subsequence, so the alternating series converges. This is a proof of the alternating series test; the monotonicity and zero-limit hypotheses play separate roles.

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