Series convergence tests
Problem 7.136 · easy
Does \( \displaystyle \sum_{n=1}^{\infty} 2^{- n} n \) converge or diverge?
- Use the ratio test.Reviewed
- \[ \frac{2^{n} 2^{- n - 1} \left(n + 1\right)}{n} = \frac{n + 1}{2 n} \]a_(n+1)/a_n.✓ Proved
- \[ \lim_{n \to \infty}\left(\frac{n + 1}{2 n}\right) = \frac{1}{2} \]The limit is 1/2, < 1: converges.✓ Proved
Answer \( \text{converges} \)
✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation.
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | Reviewed | reviewers: gpt-oss:20b, qwen3.6:27b-mlx | a sentence; read, not computed |
| 2 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 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 own convergence decision (Sum.is_convergent) agrees |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: pass
Every verdict on record (4)
qwen3.6:27b-mlx: pass 2026-09-26gpt-oss:20b: pass 2026-09-26qwen3.6:27b-mlx: pass 2026-09-26gpt-oss:20b: pass 2026-09-26
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/series_convergence, checked 2026-09-26 with SymPy 1.14.0.