Error bounds for series
Problem 7.391 · easy
How many terms of \( \displaystyle \sum_{n=1}^{\infty} \frac{\left(-1\right)^{n + 1}}{n} \) guarantee an error below \( \displaystyle \frac{1}{100} \)? 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}.
- \[ \left. \frac{1}{n} \right|_{\substack{ n=101 }} = \frac{1}{101} \]b_101 = 1/101 < 1/100.✓ Proved
- \[ \left. \frac{1}{n} \right|_{\substack{ n=100 }} = \frac{1}{100} \]but b_100 = 1/100 ≥ 1/100, 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: pass — The solution correctly applies the alternating series estimate, but sentence 3 mistakenly refers to b_{100} instead of b_{101}. The conclusion that N=100 is the smallest integer satisfying the error bound remains correct.qwen3.6:27b-mlx: fail (error) — The solution incorrectly concludes N=100 is the smallest integer satisfying the condition. Since b_{101} = 1/101 < 1/100, the error after N=100 terms is already below 1/100. The condition b_{N+1} < 1/100 is satisfied for N=100, but the solution's logic implies N=100 is the boundary where it *just* works, whereas N=99 yields b_{100}=1/100 which is not strictly less than 1/100. Wait, let's re-read carefully. The problem asks for error *below* 1/100. |R_N| <= b_{N+1}. We need b_{N+1} < 1/100. 1/(N+1) < 1/100 => N+1 > 100 => N > 99. So N=100 is indeed the smallest integer. However, the sentence in step 3 says 'b_100 = 1/100 >= 1/100, so N=100 is the smallest that works'. This logic is flawed because it compares b_100 (which corresponds to N=99) to determine N=100. It should say 'Since b_100 is not < 1/100, N=99 does not work. Since b_101 < 1/100, N=100 works.' The current phrasing is confusing and potentially misleading about which term corresponds to which N.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-06 — The solution correctly applies the alternating series estimate, but sentence 3 mistakenly refers to b_{100} instead of b_{101}. The conclusion that N=100 is the smallest integer satisfying the error bound remains correct.qwen3.6:27b-mlx: fail (error) 2026-10-06 — The solution incorrectly concludes N=100 is the smallest integer satisfying the condition. Since b_{101} = 1/101 < 1/100, the error after N=100 terms is already below 1/100. The condition b_{N+1} < 1/100 is satisfied for N=100, but the solution's logic implies N=100 is the boundary where it *just* works, whereas N=99 yields b_{100}=1/100 which is not strictly less than 1/100. Wait, let's re-read carefully. The problem asks for error *below* 1/100. |R_N| <= b_{N+1}. We need b_{N+1} < 1/100. 1/(N+1) < 1/100 => N+1 > 100 => N > 99. So N=100 is indeed the smallest integer. However, the sentence in step 3 says 'b_100 = 1/100 >= 1/100, so N=100 is the smallest that works'. This logic is flawed because it compares b_100 (which corresponds to N=99) to determine N=100. It should say 'Since b_100 is not < 1/100, N=99 does not work. Since b_101 < 1/100, N=100 works.' The current phrasing is confusing and potentially misleading about which term corresponds to which N.qwen3.6:27b-mlx: fail (error) 2026-10-06 — The solution claims N=100 is the smallest integer satisfying the condition, but for N=99, the error bound is b_{100} = 1/100, which is not strictly less than 1/100. However, for N=100, the error bound is b_{101} = 1/101, which is strictly less than 1/100. The logic in step 3 is confused: it compares b_100 to the tolerance to justify N=100, but b_100 corresponds to the error after N=99 terms. The correct reasoning is that we need b_{N+1} < 1/100, so 1/(N+1) < 1/100 implies N+1 > 100, so N > 99, meaning the smallest integer N is 100. The sentence 'b_100 = 1/100 >= 1/100, so N=100 is the smallest that works' is logically flawed because it mixes up the index of the term with the number of terms summed.gpt-oss:20b: pass 2026-10-06
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 2026-10-06 with SymPy 1.14.0.