Absolute and conditional convergence
Problem 7.246 · medium
Does \( \displaystyle \sum_{n=1}^{\infty} \frac{\left(-1\right)^{n + 1}}{n^{\frac{3}{2}}} \) converge absolutely, converge conditionally, or diverge?
- \[ \lim_{n \to \infty} \frac{1}{n^{\frac{3}{2}}} = 0 \]bₙ → 0.✓ Proved
- bₙ is eventually decreasing, so the alternating series test gives convergence.
- \[ \lim_{n \to \infty} 1 = 1 \]Limit comparison of |aₙ| with 1/n^3/2: a p-series that converges.✓ Proved
Answer \( \text{converges absolutely} \)
Lines: 2 proved, 1 not checked. The answer was also checked a second way, without looking at the solution. The reviewers disagree about how one step is explained; every verdict is in the receipt.
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 2 | Not checked | — | a sentence; read, not computed |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | SymPy's Sum.is_convergent on the series and on its absolute values |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: fail (error) — The solution claims to use limit comparison with 1/n^(3/2) but the equation shows the limit of 1, which is the limit of |a_n|/|a_n|, not a comparison with a p-series term. Furthermore, the limit comparison test requires comparing |a_n| to a known series b_n; the equation provided does not show the ratio with a p-series term, making the justification for absolute convergence invalid as written.
Every verdict on record (4)
qwen3.6:27b-mlx: fail (error) 2026-10-04 — The solution claims to use limit comparison with 1/n^(3/2) but the equation shows the limit of 1, which is the limit of |a_n|/|a_n|, not a comparison with a p-series term. Furthermore, the limit comparison test requires comparing |a_n| to a known series b_n; the equation provided does not show the ratio with a p-series term, making the justification for absolute convergence invalid as written.gpt-oss:20b: pass 2026-10-04qwen3.6:27b-mlx: fail (error) 2026-10-04 — The limit comparison test is applied incorrectly: the limit of |a_n| divided by 1/n^(3/2) is 1, not 1. The solution claims the limit is 1 but labels it as a comparison with 1/n^(3/2) while showing Limit(1, n, oo) = 1, which is nonsensical in this context. Furthermore, the limit comparison test requires comparing to a known convergent series; while p-series with p=3/2 converges, the setup of the limit calculation is garbled and mathematically incoherent as written.gpt-oss:20b: pass 2026-10-04
Proved: SymPy reduced the difference between the two sides to zero. Checked independently: a separate
method, named above, confirmed it. Checked numerically: the two sides agree at every sampled point, which is evidence,
not proof. Reviewed: a model or a person read it; that is all a sentence can have. Solution by
generator:structured/alternating_series, checked 2026-10-04 with SymPy 1.14.0.