Series convergence tests
Problem 7.155 · easy
Does \( \displaystyle \sum_{n=1}^{\infty} \frac{5^{n}}{n^{2}} \) converge or diverge?
- Use the ratio test.Reviewed
- \[ \frac{5^{- n} 5^{n + 1} n^{2}}{\left(n + 1\right)^{2}} = \frac{5 n^{2}}{\left(n + 1\right)^{2}} \]a_(n+1)/a_n.✓ Proved
- \[ \lim_{n \to \infty}\left(\frac{5 n^{2}}{\left(n + 1\right)^{2}}\right) = 5 \]The limit is 5, > 1: diverges.✓ Proved
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 | 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 |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | SymPy's own convergence decision (Sum.is_convergent) agrees |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: pass — The solution correctly applies the Ratio Test. The limit calculation is correct, and the conclusion that the series diverges because the limit is greater than 1 is valid.
Every verdict on record (4)
qwen3.6:27b-mlx: pass 2026-09-27 — The solution correctly applies the Ratio Test. The limit calculation is correct, and the conclusion that the series diverges because the limit is greater than 1 is valid.gpt-oss:20b: pass 2026-09-27qwen3.6:27b-mlx: pass 2026-09-27 — The solution correctly applies the Ratio Test. The limit calculation is correct, and the conclusion that the series diverges because the limit is greater than 1 is sound.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/series_convergence, checked 2026-09-27 with SymPy 1.14.0.