Geometric series
Problem 7.153 · medium
Does \( \displaystyle \sum_{n=1}^{\infty} - 2 \left(\frac{2}{3}\right)^{n - 1} \) converge? If so, find its sum.
- \[ -2 \]The first term, a.✓ Proved
- \[ \frac{2}{3} \]The common ratio, r.✓ Proved
- |r| = 2/3 < 1, so the series converges to a/(1 − r).Reviewed
- \[ -6 \]a/(1 − r).✓ Proved
Answer \( -6 \)
✓ Nihil obstat Lines: 3 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 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 2 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 3 | Reviewed | reviewers: gpt-oss:20b, qwen3.6:27b-mlx | a sentence; read, not computed |
| 4 | ✓ 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 | the 400th partial sum agrees |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: pass — The solution correctly identifies the series as a geometric series, verifies the convergence condition |r| < 1, and applies the correct sum formula.
Every verdict on record (4)
qwen3.6:27b-mlx: pass 2026-09-27 — The solution correctly identifies the series as a geometric series, verifies the convergence condition |r| < 1, and applies the correct sum formula.gpt-oss:20b: pass 2026-09-27qwen3.6:27b-mlx: pass 2026-09-27 — The solution correctly identifies the first term and common ratio, verifies the convergence condition for a geometric series, and applies the sum formula correctly.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/geometric_series, checked 2026-09-27 with SymPy 1.14.0.