Error bounds for series
Problem 7.305 · easy
How many terms of \( \displaystyle \sum_{n=1}^{\infty} \frac{\left(-1\right)^{n + 1}}{n^{2}} \) guarantee an error below \( \displaystyle \frac{1}{10000} \)? Use the alternating series estimate \( \displaystyle |R_N| \le b_{N+1} \).
- The error after N terms is at most the first omitted term, b_{N+1}.
- \[ \left. \frac{1}{n^{2}} \right|_{\substack{ n=101 }} = \frac{1}{10201} \]b_101 = 1/10201 < 1/10000.✓ Proved
- \[ \left. \frac{1}{n^{2}} \right|_{\substack{ n=100 }} = \frac{1}{10000} \]but b_100 = 1/10000 ≥ 1/10000, so N = 100 is the smallest that works.✓ Proved
Answer \( N = 100 \)
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 | Not checked | — | 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 | the bound recomputed in floating point at N and N − 1 |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: fail (error) — The solution claims N=100 is the smallest integer satisfying the condition, but for N=99, the error bound is b_{100} = 1/10000, which is not strictly less than 1/10000. However, for N=100, the error bound is b_{101} = 1/10201 < 1/10000. The logic in step 3 is confusing and potentially misleading regarding the strict inequality required for 'below'. More critically, if the requirement is strictly 'below' (<), then N=99 fails because 1/10000 is not < 1/10000. N=100 works. The text says 'N=100 is the smallest that works', which is correct, but the reasoning 'b_100 >= 1/10000 so N=100 is the smallest' is logically garbled. It should say 'Since b_100 is not < 1/10000, N=99 does not work. Since b_101 < 1/10000, N=100 works.' The current phrasing implies a direct comparison that doesn't clearly establish minimality.
Every verdict on record (4)
qwen3.6:27b-mlx: fail (error) 2026-10-05 — The solution claims N=100 is the smallest integer satisfying the condition, but for N=99, the error bound is b_{100} = 1/10000, which is not strictly less than 1/10000. However, for N=100, the error bound is b_{101} = 1/10201 < 1/10000. The logic in step 3 is confusing and potentially misleading regarding the strict inequality required for 'below'. More critically, if the requirement is strictly 'below' (<), then N=99 fails because 1/10000 is not < 1/10000. N=100 works. The text says 'N=100 is the smallest that works', which is correct, but the reasoning 'b_100 >= 1/10000 so N=100 is the smallest' is logically garbled. It should say 'Since b_100 is not < 1/10000, N=99 does not work. Since b_101 < 1/10000, N=100 works.' The current phrasing implies a direct comparison that doesn't clearly establish minimality.gpt-oss:20b: pass 2026-10-05qwen3.6:27b-mlx: fail (error) 2026-10-05 — The solution incorrectly concludes that N=100 is the smallest N satisfying the condition. Since b_{100} = 1/10000 is not strictly less than 1/10000, N=100 does not guarantee an error below 1/10000; N=101 is required.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/series_error_bounds, checked 2026-10-05 with SymPy 1.14.0.