Error bounds for series
Problem 7.385 · easy
How many terms of \( \displaystyle \sum_{n=1}^{\infty} \frac{\left(-1\right)^{n + 1}}{n} \) 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}.
- \[ \left. \frac{1}{n} \right|_{\substack{ n=1001 }} = \frac{1}{1001} \]b_1001 = 1/1001 < 1/1000.✓ Proved
- \[ \left. \frac{1}{n} \right|_{\substack{ n=1000 }} = \frac{1}{1000} \]but b_1000 = 1/1000 ≥ 1/1000, so N = 1000 is the smallest that works.✓ Proved
Answer \( N = 1000 \)
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=1000 is the smallest integer satisfying the condition, but b_{1000} = 1/1000 is not strictly less than 1/1000. The smallest N such that b_{N+1} < 1/1000 is N=1000 only if the inequality is non-strict, but standard error bounds require strict inequality for 'below'. Even if non-strict were allowed, N=999 gives b_{1000}=1/1000 which is not < 1/1000. Actually, for |R_N| < 1/1000, we need 1/(N+1) < 1/1000 => N+1 > 1000 => N > 999. So N=1000 is correct. Wait, let me re-read. b_{N+1} < 1/1000. If N=1000, b_{1001} = 1/1001 < 1/1000. This works. If N=999, b_{1000} = 1/1000 which is NOT < 1/1000. So N=1000 is indeed the smallest. The text says 'b_1000 = 1/1000 >= 1/1000, so N=1000 is the smallest'. This logic is slightly garbled but the conclusion is correct. However, the prompt asks for error BELOW 1/1000. 1/1000 is not below 1/1000. So N=999 fails. N=1000 works. The solution is actually correct in its conclusion. Why did I think it was wrong? Let's look closer. Step 3 says 'b_1000 ... so N=1000 is the smallest'. It implies that since b_1000 didn't work (as an upper bound for N=999), N=1000 is the next candidate. This is correct reasoning. Is there a style issue? 'Subs(1/n, n, 1001)' is not standard notation but marked as equation/proved. The sentence logic is sound. Wait, is the verdict pass? Let's check the constraints. 'error below 1/1000'. |R_N| <= b_{N+1}. We need b_{N+1} < 1/1000. 1/(N+1) < 1/1000 => N+1 > 1000 => N > 999. Smallest integer N is 1000. The solution concludes N=1000. The reasoning in step 3 is a bit loose ('b_1000 ... so N=1000 is the smallest') but technically correct because it identifies that the previous term didn't satisfy the strict inequality. I will mark it pass.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-06qwen3.6:27b-mlx: fail (error) 2026-10-06 — The solution claims N=1000 is the smallest integer satisfying the condition, but b_{1000} = 1/1000 is not strictly less than 1/1000. The smallest N such that b_{N+1} < 1/1000 is N=1000 only if the inequality is non-strict, but standard error bounds require strict inequality for 'below'. Even if non-strict were allowed, N=999 gives b_{1000}=1/1000 which is not < 1/1000. Actually, for |R_N| < 1/1000, we need 1/(N+1) < 1/1000 => N+1 > 1000 => N > 999. So N=1000 is correct. Wait, let me re-read. b_{N+1} < 1/1000. If N=1000, b_{1001} = 1/1001 < 1/1000. This works. If N=999, b_{1000} = 1/1000 which is NOT < 1/1000. So N=1000 is indeed the smallest. The text says 'b_1000 = 1/1000 >= 1/1000, so N=1000 is the smallest'. This logic is slightly garbled but the conclusion is correct. However, the prompt asks for error BELOW 1/1000. 1/1000 is not below 1/1000. So N=999 fails. N=1000 works. The solution is actually correct in its conclusion. Why did I think it was wrong? Let's look closer. Step 3 says 'b_1000 ... so N=1000 is the smallest'. It implies that since b_1000 didn't work (as an upper bound for N=999), N=1000 is the next candidate. This is correct reasoning. Is there a style issue? 'Subs(1/n, n, 1001)' is not standard notation but marked as equation/proved. The sentence logic is sound. Wait, is the verdict pass? Let's check the constraints. 'error below 1/1000'. |R_N| <= b_{N+1}. We need b_{N+1} < 1/1000. 1/(N+1) < 1/1000 => N+1 > 1000 => N > 999. Smallest integer N is 1000. The solution concludes N=1000. The reasoning in step 3 is a bit loose ('b_1000 ... so N=1000 is the smallest') but technically correct because it identifies that the previous term didn't satisfy the strict inequality. I will mark it pass.qwen3.6:27b-mlx: fail (error) 2026-10-06 — The solution incorrectly concludes that N=1000 is the smallest integer satisfying the condition. Since b_{1000} = 1/1000 is not strictly less than 1/1000, N=1000 does not guarantee an error below 1/1000; the correct answer is N=1001.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.