Power series from the geometric series
Problem 7.436 · medium
Find a power series for \( \displaystyle f(x) = \frac{3}{2 - x} \) centered at 0, and its interval of convergence.
- \[ \frac{3}{2 - x} \]Write f in the form (something)/(1 − u) with u = x/2.✓ Proved
- 1/(1 − u) = Σ uⁿ for |u| < 1; substitute and multiply through.Reviewed
- \[ \lim_{x \to 0^+}\left(\frac{- \frac{3 x^{3}}{16} - \frac{3 x^{2}}{8} - \frac{3 x}{4} - \frac{3}{2} + \frac{3}{2 - x}}{x^{3}}\right) = 0 \]The first terms 3*x**3/16 + 3*x**2/8 + 3*x/4 + 3/2 agree with f to order x^3.✓ Proved
- |u| < 1 means |x| < 2; the series diverges at both endpoints (the terms do not shrink).Reviewed
Answer \( \sum_{n=0}^{\infty} 3 \cdot 2^{- n - 1} x^{n},\quad |x| < 2 \)
Lines: 2 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
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 2 | Reviewed | reviewers: gpt-oss:20b, qwen3.6:27b-mlx | a sentence; read, not computed |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 4 | 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 series summed numerically at half the radius matches f |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: pass — The solution correctly derives the power series and interval of convergence. The logic is sound, and the check of the first few terms confirms the coefficients.
Every verdict on record (4)
qwen3.6:27b-mlx: pass 2026-10-08 — The solution correctly derives the power series and interval of convergence. The logic is sound, and the check of the first few terms confirms the coefficients.gpt-oss:20b: pass 2026-10-08gpt-oss:20b: fail (misleading) 2026-10-08 — The solution’s check of the first terms shows a sign error (it writes negative terms), which would mislead a student into thinking the series has alternating signs. The correct series has all positive coefficients 3·2^{‑n‑1}x^n. The rest of the reasoning is sound.qwen3.6:27b-mlx: pass 2026-10-08 — The solution correctly derives the power series and interval of convergence. The reasoning for divergence at endpoints is sound.
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/power_series_geometric, checked 2026-10-08 with SymPy 1.14.0.