Limits of sequences
Problem 7.293 · easy
Write the first four terms of \( \displaystyle a_n = \frac{2 n + 4}{4 n^{2} + 1} \), and determine whether the sequence converges. If it does, find its limit.
- \[ \left[\begin{matrix}\frac{6}{5}\\\frac{8}{17}\\\frac{10}{37}\\\frac{12}{65}\end{matrix}\right] \]The first four terms.✓ Proved
- The denominator has the higher degree.
- \[ \lim_{n \to \infty}\left(\frac{2 n + 4}{4 n^{2} + 1}\right) = 0 \]The limit.✓ Proved
Answer \( a_1, \dots, a_4 = \frac{6}{5}, \frac{8}{17}, \frac{10}{37}, \frac{12}{65};\ 0 \)
Lines: 2 proved, 1 not checked. 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 | Not checked | — | a sentence; read, not computed |
| 3 | ✓ 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 terms evaluated at n = 10⁶, 10⁶ + 1 and 10⁷ |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: fail (misleading) — The justification 'The denominator has the higher degree' is an informal heuristic, not a rigorous theorem. A student should be taught to divide by the highest power of n or apply the formal limit laws to prove convergence, rather than relying on degree comparison as a standalone proof.
Every verdict on record (4)
qwen3.6:27b-mlx: fail (misleading) 2026-10-05 — The justification 'The denominator has the higher degree' is an informal heuristic, not a rigorous theorem. A student should be taught to divide by the highest power of n or apply the formal limit laws to prove convergence, rather than relying on degree comparison as a standalone proof.gpt-oss:20b: pass 2026-10-05qwen3.6:27b-mlx: pass 2026-10-05 — The solution correctly calculates the first four terms and identifies the limit as 0 based on the degree of the denominator being higher than the numerator. The reasoning is sound and concise.gpt-oss:20b: pass 2026-10-05
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/sequence_limits, checked 2026-10-05 with SymPy 1.14.0.