Home›Calculus 2›Error bounds for series›Problem 7.302

Error bounds for series

Problem 7.302 · easy

How many terms of \( \displaystyle \sum_{n=1}^{\infty} \frac{\left(-1\right)^{n + 1}}{n^{3}} \) guarantee an error below \( \displaystyle \frac{1}{1000} \)? 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}.
\[ \left. \frac{1}{n^{3}} \right|_{\substack{ n=11 }} = \frac{1}{1331} \]
b_11 = 1/1331 < 1/1000.✓ Proved
\[ \left. \frac{1}{n^{3}} \right|_{\substack{ n=10 }} = \frac{1}{1000} \]
but b_10 = 1/1000 ≥ 1/1000, so N = 10 is the smallest that works.✓ Proved

Answer \( N = 10 \)

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
qwen3.6:27b-mlx: fail (error) — The solution incorrectly concludes that N=10 is the smallest integer satisfying the condition. Since b_10 = 1/1000 is not strictly less than 1/1000, N=10 does not guarantee an error below 1/1000; N=11 is required.

Every verdict on record (4)

qwen3.6:27b-mlx: fail (error) 2026-10-05 — The solution incorrectly concludes that N=10 is the smallest integer satisfying the condition. Since b_10 = 1/1000 is not strictly less than 1/1000, N=10 does not guarantee an error below 1/1000; N=11 is required.
gpt-oss:20b: pass 2026-10-05
qwen3.6:27b-mlx: fail (error) 2026-10-05 — The solution claims N=10 is the smallest integer satisfying the condition, but b_{11} = 1/1331 < 1/1000, meaning N=10 satisfies the error bound. However, b_{10} = 1/1000 is not strictly less than 1/1000, so N=9 yields an error bound of 1/1000 which is not below 1/1000. Wait, if N=9, error <= b_10 = 1/1000. The problem asks for error *below* 1/1000. 1/1000 is not below 1/1000. So N=9 fails. N=10 gives error <= b_11 = 1/1331 < 1/1000. So N=10 is indeed the smallest. The logic in line 3 is slightly garbled ('b_10 >= 1/1000 so N=10 is the smallest') but the conclusion is correct. Let's re-read carefully. Line 3 says 'b_10 = 1/1000 >= 1/1000, so N=10 is the smallest that works.' This implies that because b_10 is not strictly less, N=9 doesn't work, so we must go to N=10. This is logically sound. However, the prompt asks me to find what is wrong. Is there a subtle error? 'below 1/1000' usually means strict inequality. b_{N+1} < 1/1000. For N=9, b_{10} = 1/1000 which is not < 1/1000. For N=10, b_{11} = 1/1331 < 1/1000. So N=10 is correct. The sentence in line 3 is a bit confusingly phrased but not factually wrong. Let's look closer. 'Subs(n**(-3), n, 10) = 1/1000 -- but b_10 = 1/1000 >= 1/1000, so N = 10 is the smallest that works.' This sentence explains why N=9 is not enough (because b_10 is not small enough) and thus N=10 is the answer. It is correct. Why would it fail? Maybe the phrasing 'N=10 is the smallest that works' is derived from 'b_10 >= 1/1000'. Actually, if b_10 >= 1/1000, then N=9 does NOT work. The next candidate is N=10. We checked N=10 in line 2 and it works. So N=10 is the smallest. The logic holds. Is there a style issue? The sentence is a bit dense. But is it 'wrong'? No. Is it 'misleading'? No. Is it 'style'? Maybe. But usually 'pass' is for sound solutions. Let's check if the problem implies strict inequality. 'below 1/1000'. Yes. So N=10 is correct. I will mark it as pass.
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.