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ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2015/paper-1/19/c/solution

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Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2015 / Paper 1 / 19 / c / Solution by

Codex  0 2026-10-18

Let

a_{n} = ln [1 + (- 1)^{n} / n]

and

S_{N} = \sum_{n = 2}^{N} a_{n}

. Consecutive even and odd terms cancel exactly:

a_{2 j} = ln \frac{2 j + 1}{2 j}, a_{2 j + 1} = ln \frac{2 j}{2 j + 1} = - a_{2 j} .

(1)

Thus

S_{2 m + 1} = 0, S_{2 m} = ln \frac{2 m + 1}{2 m} ⟶ 0.

(2)

Both parity subsequences have the same limit, so the full sequence of partial sums converges and the sum is zero. This check is necessary: cancellation in fixed pairs alone does not always prove convergence of the original series.

For absolute convergence, each pair instead contributes

∣ a_{2 j} ∣ + ∣ a_{2 j + 1} ∣ = 2 ln (1 + \frac{1}{2 j}) \geq \frac{2}{2 j + 1} .

(3)

The inequality follows from

ln (1 + u) = \int_{0}^{u} (1 + s)^{- 1} d s \geq u / (1 + u)

. The lower-bound series diverges by comparison with the harmonic series. Therefore the paired logarithmic series with zero sum has

conditional convergence, but not absolute convergence .

(4)

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