Error bounds for series
Problem 7.388 · easy
How many terms of \( \displaystyle \sum_{n=1}^{\infty} \frac{\left(-1\right)^{n + 1}}{n^{2}} \) guarantee an error below \( \displaystyle \frac{1}{100} \)? 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}.
- \[ \left. \frac{1}{n^{2}} \right|_{\substack{ n=11 }} = \frac{1}{121} \]b_11 = 1/121 < 1/100.✓ Proved
- \[ \left. \frac{1}{n^{2}} \right|_{\substack{ n=10 }} = \frac{1}{100} \]but b_10 = 1/100 ≥ 1/100, 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: fail (error) — The solution incorrectly concludes that N=10 suffices. For the alternating series estimate we need |R_N| ≤ b_{N+1} < 1/100. Since b_{10}=1/100 is not strictly less than 1/100, N must be at least 11. The conclusion that N=10 is the smallest that works is false.qwen3.6:27b-mlx: fail (error) — The solution incorrectly concludes N=10 is the smallest integer satisfying the strict error bound |R_N| < 1/100. Since b_11 = 1/121 < 1/100, N=10 does satisfy the condition, but the reasoning in step 3 is flawed: it claims N=10 is the smallest because b_10 >= 1/100, ignoring that N=9 yields b_10 = 1/100 which is not strictly less than 1/100, but N=10 yields b_11 < 1/100. Wait, let's re-evaluate. For N=9, error <= b_10 = 1/100. Is 1/100 < 1/100? No. So N=9 fails. For N=10, error <= b_11 = 1/121. Is 1/121 < 1/100? Yes. So N=10 is indeed the smallest. The logic in step 3 says 'b_10 = 1/100 >= 1/100, so N=10 is the smallest that works'. This implies that because b_10 is not small enough, we must go to N=10. This is actually correct logic for finding the smallest N. However, the phrasing 'b_10 ... so N=10' is slightly ambiguous. Let's look closer. Step 2 shows b_11 < 1/100. Step 3 shows b_10 >= 1/100. The conclusion is N=10. This is correct. Why did I think it was an error? Ah, the prompt asks for 'error below 1/100'. Strict inequality. b_10 = 1/100 is not below 1/100. So N=9 (which uses b_10 for error bound) does not guarantee error < 1/100. N=10 (which uses b_11) does. So N=10 is correct. The solution is sound. Wait, is there a subtle issue? 'Subs(n**(-2), n, 10) = 1/100 -- but b_10 = 1/100 >= 1/100, so N = 10 is the smallest that works.' This sentence is logically valid. It checks the boundary. If the question was 'error <= 1/100', N=9 would work. Since it is 'below' (<), N=9 fails. N=10 works. The solution correctly identifies N=10. Is there any misleading part? No. It seems correct. Let me double check the 'adversarial' nature. Is there a trap? Sum starts at n=1. Terms are 1, -1/4, 1/9... b_n = 1/n^2. Alternating series test applies. Error |R_N| <= b_{N+1}. We need b_{N+1} < 1/100. 1/(N+1)^2 < 1/100 => (N+1)^2 > 100 => N+1 > 10 => N > 9. Smallest integer N is 10. The solution derives this by checking N=10 (b_11 < 1/100) and N=9 (implied by checking b_10 >= 1/100). The text says 'b_10 ... so N=10 is the smallest'. This is a bit of a jump in explanation but not wrong. It establishes that N=9 doesn't work (via b_10) and N=10 does (via b_11). I will mark it pass.
Every verdict on record (4)
gpt-oss:20b: fail (error) 2026-10-06 — The solution incorrectly concludes that N=10 suffices. For the alternating series estimate we need |R_N| ≤ b_{N+1} < 1/100. Since b_{10}=1/100 is not strictly less than 1/100, N must be at least 11. The conclusion that N=10 is the smallest that works is false.qwen3.6:27b-mlx: fail (error) 2026-10-06 — The solution incorrectly concludes N=10 is the smallest integer satisfying the strict error bound |R_N| < 1/100. Since b_11 = 1/121 < 1/100, N=10 does satisfy the condition, but the reasoning in step 3 is flawed: it claims N=10 is the smallest because b_10 >= 1/100, ignoring that N=9 yields b_10 = 1/100 which is not strictly less than 1/100, but N=10 yields b_11 < 1/100. Wait, let's re-evaluate. For N=9, error <= b_10 = 1/100. Is 1/100 < 1/100? No. So N=9 fails. For N=10, error <= b_11 = 1/121. Is 1/121 < 1/100? Yes. So N=10 is indeed the smallest. The logic in step 3 says 'b_10 = 1/100 >= 1/100, so N=10 is the smallest that works'. This implies that because b_10 is not small enough, we must go to N=10. This is actually correct logic for finding the smallest N. However, the phrasing 'b_10 ... so N=10' is slightly ambiguous. Let's look closer. Step 2 shows b_11 < 1/100. Step 3 shows b_10 >= 1/100. The conclusion is N=10. This is correct. Why did I think it was an error? Ah, the prompt asks for 'error below 1/100'. Strict inequality. b_10 = 1/100 is not below 1/100. So N=9 (which uses b_10 for error bound) does not guarantee error < 1/100. N=10 (which uses b_11) does. So N=10 is correct. The solution is sound. Wait, is there a subtle issue? 'Subs(n**(-2), n, 10) = 1/100 -- but b_10 = 1/100 >= 1/100, so N = 10 is the smallest that works.' This sentence is logically valid. It checks the boundary. If the question was 'error <= 1/100', N=9 would work. Since it is 'below' (<), N=9 fails. N=10 works. The solution correctly identifies N=10. Is there any misleading part? No. It seems correct. Let me double check the 'adversarial' nature. Is there a trap? Sum starts at n=1. Terms are 1, -1/4, 1/9... b_n = 1/n^2. Alternating series test applies. Error |R_N| <= b_{N+1}. We need b_{N+1} < 1/100. 1/(N+1)^2 < 1/100 => (N+1)^2 > 100 => N+1 > 10 => N > 9. Smallest integer N is 10. The solution derives this by checking N=10 (b_11 < 1/100) and N=9 (implied by checking b_10 >= 1/100). The text says 'b_10 ... so N=10 is the smallest'. This is a bit of a jump in explanation but not wrong. It establishes that N=9 doesn't work (via b_10) and N=10 does (via b_11). I will mark it pass.qwen3.6:27b-mlx: fail (error) 2026-10-06 — The solution incorrectly concludes N=10 is the smallest integer satisfying the condition. Since b_{11} = 1/121 < 1/100, N=9 is sufficient (as |R_9| <= b_{10} = 1/100 is not strictly less, wait, the problem asks for error BELOW 1/100. b_{10} = 1/100 is not < 1/100. b_{11} = 1/121 < 1/100. So N=10 gives error <= b_{11} < 1/100. N=9 gives error <= b_{10} = 1/100 which is NOT < 1/100. So N=10 is actually correct. Let me re-read carefully. Problem: error BELOW 1/100. i.e. |R_N| < 1/100. Estimate: |R_N| <= b_{N+1}. We need b_{N+1} < 1/100. b_{N+1} = 1/(N+1)^2. 1/(N+1)^2 < 1/100 => (N+1)^2 > 100 => N+1 > 10 => N > 9. Smallest integer N is 10. Let's check N=9. |R_9| <= b_{10} = 1/100. Is 1/100 < 1/100? No. Let's check N=10. |R_{10}| <= b_{11} = 1/121. Is 1/121 < 1/100? Yes. So N=10 is indeed the correct answer. The logic in step 3 says 'b_10 = 1/100 >= 1/100, so N=10 is the smallest that works'. This implies that for N=9, the bound is b_{10}=1/100 which fails the strict inequality. For N=10, the bound is b_{11}=1/121 which passes. The conclusion is correct. The reasoning is slightly compressed but correct. Wait, let's look at step 3 again. 'Subs(n**(-2), n, 10) = 1/100 -- but b_10 = 1/100 >= 1/100, so N = 10 is the smallest that works.' This sentence is confusing. It calculates b_10. Then it says because b_10 >= 1/100, N=10 is the answer. This skips the logic that N=10 uses b_{11}. It should say 'Since b_{10} is not < 1/100, N=9 does not guarantee error < 1/100. Since b_{11} < 1/100, N=10 does.' However, is it an ERROR? The conclusion is correct. The intermediate statement 'b_10 = 1/100 >= 1/100' is true. The implication 'so N=10 is the smallest' is true, but the connection is missing. A student might think N=10 corresponds to b_10. Actually, usually 'N terms' means sum up to N. Error is bounded by b_{N+1}. If N=9, bound is b_{10} = 1/100. Not < 1/100. If N=10, bound is b_{11} = 1/121. < 1/100. The text says: 'b_10 = 1/100 >= 1/100, so N = 10 is the smallest that works.' This is misleading/wrong reasoning. It suggests that checking b_10 determines N=10. But b_10 determines if N=9 works. Since b_10 fails, N=9 fails. Then we check b_11 (implicitly) to see if N=10 works. The text doesn't explicitly link b_11 to N=10 in the final sentence, it just jumps to the answer after evaluating b_10. Is it 'error' or 'misleading'? If a student reads 'b_10 >= 1/100 so N=10', they might think the condition is on b_N. The condition is on b_{N+1}. This is a significant conceptual gap in the explanation. It teaches the wrong mapping between N and the term index. Let's call it 'misleading' or 'error'. Given the strict instructions 'find what is wrong', the logical jump is flawed. The statement 'b_10 ... so N=10' is not a valid deduction without mentioning b_11. It's a non-sequitur. I will mark it as error because the justification provided does not logically support the conclusion as stated (it skips the crucial step of identifying that N=10 relies on b_{11}).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.