Geometric series
Problem 7.148 · easy
Does \( \displaystyle \sum_{n=0}^{\infty} 3 \left(- \frac{4}{3}\right)^{n} \) converge? If so, find its sum.
- \[ 3 \]The first term, a.✓ Proved
- \[ - \frac{4}{3} \]The common ratio, r.✓ Proved
- |r| = 4/3 ≥ 1, so the series diverges.Reviewed
Answer \( \text{diverges} \)
✓ 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 | ✓ 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 |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | the terms do not shrink (60th term checked) |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: pass — The solution correctly identifies the series as a geometric series with |r| > 1 and correctly concludes it diverges.
Every verdict on record (4)
qwen3.6:27b-mlx: pass 2026-09-27 — The solution correctly identifies the series as a geometric series with |r| > 1 and correctly concludes it diverges.gpt-oss:20b: pass 2026-09-27qwen3.6:27b-mlx: fail (style) 2026-09-27 — [domain objection, downgraded to style] The geometric series test states that a series diverges if |r| >= 1. However, the standard theorem usually requires |r| > 1 for divergence via the nth term test or geometric series formula failure, and specifically handles |r|=1 as a separate case (divergence by nth term test). While the conclusion is correct, the phrasing "|r| >= 1" is technically imprecise because the geometric series formula is undefined at r=1, but the series diverges for r=1 as well. More importantly, the prompt asks to find the sum if it converges. Since it diverges, stating it diverges is correct. Wait, let me re-read the standard theorem. A geometric series sum_{n=0}^inf ar^n converges if and only if |r| < 1. If |r| >= 1, it diverges. So the statement is actually correct. Why would it be an error? Let's look closer. |r| = 4/3 > 1. The condition for divergence is indeed |r| >= 1. Is there a nuance? Some textbooks say "if |r| < 1 converges, otherwise diverges". The sentence says "|r| = 4/3 >= 1, so the series diverges." This is logically sound. Is there a hidden trap? The problem asks "Does it converge? If so, find its sum." The answer is "diverges". The solution identifies a=3, r=-4/3. |r| = 4/3. 4/3 > 1. Therefore diverges. This seems correct. Let me check for "misleading" or "style". Maybe the issue is that for r=1, the series is 3+3+3... which diverges. For r=-1, it oscillates. The condition |r|>=1 covers all divergent cases for geometric series. So the math is correct. Is there a formatting issue? No. Is there a subtle error in the setup? a=3, r=-4/3. Correct. The sentence is correct. Why would I fail it? Perhaps the prompt implies I should be adversarial. Is there any case where |r|>=1 and it converges? No. Is the notation "3, 3" in line 1 confusing? It says "The first term, a." It lists 3 twice? "3, 3". Maybe it means a_0 = 3? The equation is unchecked. The sentence is what I review. The sentence is "|r| = 4/3 ≥ 1, so the series diverges." This is a correct application of the geometric series test. I will pass it.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.