Error bounds for series
Problem 7.384 · easy
How many terms of \( \displaystyle \sum_{n=1}^{\infty} \frac{\left(-1\right)^{n + 1}}{n} \) 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}.
- \[ \left. \frac{1}{n} \right|_{\substack{ n=10001 }} = \frac{1}{10001} \]b_10001 = 1/10001 < 1/10000.✓ Proved
- \[ \left. \frac{1}{n} \right|_{\substack{ n=10000 }} = \frac{1}{10000} \]but b_10000 = 1/10000 ≥ 1/10000, so N = 10000 is the smallest that works.✓ Proved
Answer \( N = 10000 \)
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 incorrectly concludes N=10000 is the smallest integer satisfying the strict inequality |R_N| < 1/10000. Since b_{10001} = 1/10001 < 1/10000, N=10000 is sufficient, but the logic in step 3 implies N=10000 is the *smallest* because b_{10000} fails, ignoring that N=9999 yields b_{10000}=1/10000 which is not strictly less than 1/10000, so N=10000 is indeed the smallest. Wait, let's re-read carefully. The condition is |R_N| <= b_{N+1} < 1/10000. For N=9999, b_{10000} = 1/10000. Is 1/10000 < 1/10000? No. So N=9999 does not guarantee error *below* 1/10000 (strictly less). For N=10000, b_{10001} = 1/10001. Is 1/10001 < 1/10000? Yes. So N=10000 is the smallest integer. Let's look at the text of step 3: 'but b_10000 = 1/10000 ≥ 1/10000, so N = 10000 is the smallest that works.' This sentence is logically sound. It checks the boundary. b_{N+1} for N=9999 is b_{10000}. Since b_{10000} is not < 1/10000, N=9999 doesn't work. N=10000 works. So N=10000 is the smallest. Is there an error? The problem asks for error *below* 1/10000. Usually 'below' means strict inequality. |R_N| <= b_{N+1}. We need b_{N+1} < 1/10000. 1/(N+1) < 1/10000 => N+1 > 10000 => N > 9999. Smallest integer N is 10000. The solution says: 'b_10000 = 1/10000 >= 1/10000, so N = 10000 is the smallest that works.' This implies that for N=9999, the bound is b_10000, which is not strictly less than 1/10000. Thus N=9999 fails. N=10000 succeeds. Wait, look at step 2. 'Subs(1/n, n, 10001) = 1/10001 -- b_10001 = 1/10001 < 1/10000.' This shows N=10000 works. Step 3: 'Subs(1/n, n, 10000) = 1/10000 -- but b_10000 = 1/10000 >= 1/10000, so N = 10000 is the smallest that works.' This shows N=9999 (which uses b_10000) does not work. The logic is correct. The verdict should be pass.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-06qwen3.6:27b-mlx: fail (error) 2026-10-06 — The solution incorrectly concludes N=10000 is the smallest integer satisfying the strict inequality |R_N| < 1/10000. Since b_{10001} = 1/10001 < 1/10000, N=10000 is sufficient, but the logic in step 3 implies N=10000 is the *smallest* because b_{10000} fails, ignoring that N=9999 yields b_{10000}=1/10000 which is not strictly less than 1/10000, so N=10000 is indeed the smallest. Wait, let's re-read carefully. The condition is |R_N| <= b_{N+1} < 1/10000. For N=9999, b_{10000} = 1/10000. Is 1/10000 < 1/10000? No. So N=9999 does not guarantee error *below* 1/10000 (strictly less). For N=10000, b_{10001} = 1/10001. Is 1/10001 < 1/10000? Yes. So N=10000 is the smallest integer. Let's look at the text of step 3: 'but b_10000 = 1/10000 ≥ 1/10000, so N = 10000 is the smallest that works.' This sentence is logically sound. It checks the boundary. b_{N+1} for N=9999 is b_{10000}. Since b_{10000} is not < 1/10000, N=9999 doesn't work. N=10000 works. So N=10000 is the smallest. Is there an error? The problem asks for error *below* 1/10000. Usually 'below' means strict inequality. |R_N| <= b_{N+1}. We need b_{N+1} < 1/10000. 1/(N+1) < 1/10000 => N+1 > 10000 => N > 9999. Smallest integer N is 10000. The solution says: 'b_10000 = 1/10000 >= 1/10000, so N = 10000 is the smallest that works.' This implies that for N=9999, the bound is b_10000, which is not strictly less than 1/10000. Thus N=9999 fails. N=10000 succeeds. Wait, look at step 2. 'Subs(1/n, n, 10001) = 1/10001 -- b_10001 = 1/10001 < 1/10000.' This shows N=10000 works. Step 3: 'Subs(1/n, n, 10000) = 1/10000 -- but b_10000 = 1/10000 >= 1/10000, so N = 10000 is the smallest that works.' This shows N=9999 (which uses b_10000) does not work. The logic is correct. The verdict should be pass.qwen3.6:27b-mlx: fail (error) 2026-10-06 — The solution claims N=10000 is the smallest integer satisfying the condition, but b_{10001} = 1/10001 is strictly less than 1/10000, meaning N=10000 is sufficient. However, b_{10000} = 1/10000 is not strictly less than 1/10000, so N=9999 is not sufficient. The logic for N=10000 being the *smallest* is correct, but the phrasing 'b_10000 >= 1/10000, so N=10000 is the smallest' is confusingly stated because the condition for N is based on b_{N+1}. Specifically, for N=9999, b_{10000} = 1/10000 which is not < 1/10000. For N=10000, b_{10001} = 1/10001 < 1/10000. The conclusion is correct, but the reasoning in line 3 is slightly garbled. Wait, let's re-read carefully. Line 3 says 'b_10000 = 1/10000 >= 1/10000, so N=10000 is the smallest that works.' This implies that because the term for N=9999 (which is b_10000) failed, N=10000 is the answer. This is logically sound. Is there an error? The problem asks for error *below* 1/10000. |R_N| <= b_{N+1}. We need b_{N+1} < 1/10000. 1/(N+1) < 1/10000 => N+1 > 10000 => N > 9999. So N=10000 is indeed the smallest integer. The solution is correct. Why did I think it was wrong? Let me check the 'unchecked' equations. They are just substitutions. The sentence in line 3 is a bit loose but the conclusion is correct. Actually, looking closer at line 3: 'b_10000 = 1/10000 >= 1/10000'. This checks the condition for N=9999 (where the error bound is b_10000). Since 1/10000 is not strictly less than 1/10000, N=9999 fails. Thus N=10000 is the first one that works. The logic holds. Is it misleading? It's a bit terse. But is it wrong? No. Pass?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.