Home›Calculus 2›Comparison test for improper integrals›Problem 4.1095

Comparison test for improper integrals

Problem 4.1095 · medium

Use a comparison to decide whether \( \displaystyle \int_1^\infty \frac{x + 3}{x^{2}} \, dx \) converges or diverges.

(x + 3)/x² = 1/x + 3/x² ≥ 1/x.
Reviewed
\[ - \frac{1}{x} + \frac{x + 3}{x^{2}} = \frac{3}{x^{2}} \]
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: pass
qwen3.6:27b-mlx: pass — The solution correctly applies the Direct Comparison Test. The inequality is valid for the given domain, and the divergence of the comparison integral 1/x is correctly identified.

Every verdict on record (6)

qwen3.6:27b-mlx: pass 2026-10-17 — The solution correctly applies the Direct Comparison Test. The inequality is valid for the given domain, and the divergence of the comparison integral 1/x is correctly identified.
gpt-oss:20b: pass 2026-10-17
qwen3.6:27b-mlx: pass 2026-10-14 — The solution correctly applies the Direct Comparison Test. The inequality f(x) >= 1/x is valid for x >= 1, and since the integral of 1/x diverges, the integral of f(x) must also diverge.
gpt-oss:20b: pass 2026-10-14
qwen3.6:27b-mlx: pass 2026-10-14 — The solution correctly applies the Direct Comparison Test. The inequality (x+3)/x^2 >= 1/x holds for x >= 1, and since the integral of 1/x diverges, the original integral diverges.
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.