Comparison test for improper integrals
Problem 4.1104 · medium
Use a comparison to decide whether \( \displaystyle \int_1^\infty \frac{x + 1}{x^{2}} \, dx \) converges or diverges.
- (x + 1)/x² = 1/x + 1/x² ≥ 1/x.Reviewed
- \[ - \frac{1}{x} + \frac{x + 1}{x^{2}} = \frac{1}{x^{2}} \]The difference, which is ≥ 0 for x ≥ 1.✓ Proved
- \[ \int\limits_{1}^{T} \frac{1}{x}\, dx = \ln{\left(T \right)} \]Up to T the comparison integral is this, which grows without bound as T → ∞: it diverges.✓ Proved
- A smaller positive integrand than a convergent one converges; a larger one than a divergent one diverges. So it diverges.Reviewed
Answer \( \text{diverges} \)
✓ Nihil obstat Lines: 2 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. How we verify
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 |
| 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 integral over each decade [10ᵏ, 10ᵏ⁺¹] for k up to 7: its pieces vanish only if it converges |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: pass — The solution correctly applies the Direct Comparison Test. The inequality (x+1)/x^2 >= 1/x is valid for x >= 1, and since the integral of 1/x diverges, the original integral must also diverge.
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-10-17 — The solution correctly applies the Direct Comparison Test. The inequality (x+1)/x^2 >= 1/x is valid for x >= 1, and since the integral of 1/x diverges, the original integral must also diverge.gpt-oss:20b: pass 2026-10-17qwen3.6:27b-mlx: pass 2026-10-14 — The solution correctly applies the Direct Comparison Test. The inequality (x+1)/x^2 >= 1/x is valid for x >= 1, and since the integral of 1/x diverges, the original integral must also diverge.gpt-oss:20b: pass 2026-10-14qwen3.6:27b-mlx: pass 2026-10-14 — The solution correctly applies the Direct Comparison Test. The inequality (x+1)/x^2 >= 1/x is valid for x >= 1, and since the integral of 1/x diverges, the original integral must also diverge. The reasoning is sound.gpt-oss:20b: pass 2026-10-14
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/improper_comparison, checked 2026-10-17 with SymPy 1.14.0.