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Here is the problem: Σ (negative 1) ^ n / n, n equals 1 to ∞.

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Decide whether Σ (negative 1) to the power n / n (n from 1 to ∞) converges or diverges.

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Write the series. Σ a(n) for n equals 1, 2, 3, … with a(n) equals (negative 1) to the power n / n. First terms: negative 1, 0.5, negative 0.333333, 0.25, ….

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nth-term test. lim |a(n)| equals 0, so the divergence test is inconclusive; the terms do shrink, so a finer test is needed.

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Alternating series test. The signs alternate and the magnitudes b(n) equals 1 / n decrease (checked on 310 indices up to 10 to the power 6) and tend to 0, so the series converges.

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Limit comparison with a p-series. Absolute values Σ 1 / n: the terms shrink roughly like 1/n (estimated from a(2N)/a(N)), so compare with Σ 1/n: lim a(n) times n equals 1. The limit is a positive finite number, so the series behaves exactly like the p-series Σ 1/n, which diverges because p less than or equal to 1.

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Absolute or conditional?. Σ |a(n)| diverges (limit comparison with Σ 1/n), so the series converges only conditionally.

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Estimate the sum numerically. Partial sums S N were computed for several N and extrapolated to N → ∞ (Richardson extrapolation; alternating terms are averaged first). Two independent extrapolations agree to better than 10 to the power minus 10, giving S approximately negative 0.6931471806. Any closed form shown is a numerical identification, not a proof.

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So the answer is The series converges conditionally (alternating series test); sum approximately negative 0.6931471806 (matches negative ln(2) to 10+ digits).

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A series converges when its partial sums approach a finite number. The terms must tend to 0 for that to happen, but that alone is not enough (the harmonic series diverges), so a comparison, ratio, root, alternating or integral test is used to decide.

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Partial sums (supporting evidence, not a proof): S₁₀ equals negative 0.64563492, S₁₀₀ equals negative 0.68817218, S₁₀₀₀ equals negative 0.69264743, S₁₀₀₀₀ equals negative 0.69309718.