∫Calc Practice

Radius of convergence

Problem 7.167 · easy

Find the radius of convergence of \( \displaystyle \sum_{n=1}^{\infty} 2^{- n} \left(x + 3\right)^{n} \).
  1. Use the ratio test on the absolute values of the terms.
    Reviewed
  2. \[ \lim_{n \to \infty}\left(\frac{2^{n} 2^{- n - 1} \left(x + 3\right)^{- n} \left(x + 3\right)^{n + 1}}{x + 3}\right) = \frac{1}{2} \]
    |a_(n+1)/a_n| → 1/2·|x − -3|.✓ Proved
  3. The series converges when 1/2|x − -3| < 1, that is |x − -3| < 2.
    Reviewed
Answer \( R = 2 \)

Lines: 1 proved, 2 reviewed. 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
LineStatusChecked byDetail
1Reviewedreviewers: gpt-oss:20b, qwen3.6:27b-mlxa sentence; read, not computed
2✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
3Reviewedreviewers: gpt-oss:20b, qwen3.6:27b-mlxa sentence; read, not computed
answer, a second way✓ Checked independentlysympy 1.14.0 + mpmath 1.3.0terms evaluated just inside and just outside the radius shrink and grow

Reviewers

  • gpt-oss:20b: pass
  • qwen3.6:27b-mlx: pass
Every verdict on record (4)
  • qwen3.6:27b-mlx: pass 2026-09-27
  • gpt-oss:20b: pass 2026-09-27
  • qwen3.6:27b-mlx: fail (error) 2026-09-27 — The limit calculation in step 2 is incorrect; the ratio of consecutive terms simplifies to |x+3|/2, not 1/2. Consequently, the inequality in step 3 is derived from a wrong limit, although the final radius happens to be correct by coincidence.
  • gpt-oss:20b: pass 2026-09-27

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/radius_of_convergence, checked 2026-09-27 with SymPy 1.14.0.