Error bounds for series
Problem 7.389 · easy
How many terms of \( \displaystyle \sum_{n=1}^{\infty} \frac{\left(-1\right)^{n + 1}}{n^{3}} \) guarantee an error below \( \displaystyle \frac{1}{10000} \)? 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}.
- \[ \left. \frac{1}{n^{3}} \right|_{\substack{ n=22 }} = \frac{1}{10648} \]b_22 = 1/10648 < 1/10000.✓ Proved
- \[ \left. \frac{1}{n^{3}} \right|_{\substack{ n=21 }} = \frac{1}{9261} \]but b_21 = 1/9261 ≥ 1/10000, so N = 21 is the smallest that works.✓ Proved
Answer \( N = 21 \)
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=21 is the smallest N that works, but the check for N=21 shows b_{22} < 1/10000, meaning N=21 is sufficient. However, it fails to check if a smaller N (like N=20) is sufficient. Since b_{21} >= 1/10000, N=20 is not sufficient (error bound is b_{21}). Wait, the error bound for N terms is b_{N+1}. For N=20, error <= b_{21} = 1/9261 > 1/10000. For N=21, error <= b_{22} = 1/10648 < 1/10000. So N=21 is indeed the smallest. The logic in step 3 is slightly garbled: it says 'b_21 >= 1/10000, so N=21 is the smallest'. This implies that because the term *after* N=20 is too big, N=20 doesn't work. And since the term *after* N=21 is small enough, N=21 works. The conclusion is correct, but the phrasing 'b_21 ... so N=21' is confusingly linking the term index to the N index. Let's look closer. Step 3 says 'b_21 >= 1/10000, so N=21 is the smallest that works'. This is logically incomplete. It should say 'Since b_21 > 1/10000, N=20 does not guarantee the error. Since b_22 < 1/10000, N=21 does.' The current sentence skips the N=20 case explicitly and jumps to the conclusion. However, is it an error? It's a bit of a leap. Let's re-read carefully. 'b_21 = 1/9261 >= 1/10000, so N = 21 is the smallest that works.' This is a non-sequitur. The fact that b_21 is large doesn't prove N=21 is the *smallest*. It proves N=20 is *not* sufficient. One must also show N=21 *is* sufficient (which step 2 did). The sentence combines the negative check for N=20 and the positive conclusion for N=21 without clear logical connectors. It's misleading because it suggests the magnitude of b_21 directly determines N=21's status, rather than N=20's failure. Actually, looking at standard alternating series estimation, we need b_{N+1} < epsilon. We found b_{22} < epsilon, so N=21 works. We found b_{21} > epsilon, so N=20 does not work. Thus N=21 is the smallest. The solution presents this correctly in steps 2 and 3, but the sentence in 3 is poorly phrased. Is it an error? No, the math is right. Is it misleading? Yes, a student might think 'b_N < epsilon' is the condition, or get confused by the index shift. But wait, the prompt asks for 'error below 1/10000'. Step 2 shows N=21 works. Step 3 shows N=20 fails (implicitly, by showing b_21 is too big). The sentence 'so N=21 is the smallest' is the conclusion. The reasoning is sound, just compressed. I will mark it as 'style' or 'pass'? The prompt says 'find what is wrong'. If it's correct, pass. The logic is: 1. Check N=21: b_22 < tol. OK. 2. Check N=20: b_21 > tol. Fail. Therefore 21 is smallest. The text says 'b_21 >= ... so N=21 is smallest'. This is a valid logical step if one assumes monotonicity and that we are checking consecutive integers. It's not strictly 'wrong'. However, usually 'error' is reserved for mathematical mistakes. This is a clarity issue. Let's look for a harder error. Did I miss one? No. The answer N=21 is correct. The steps are correct. The sentence is just a bit loose. I'll call it 'style' if I must fail, but 'pass' is better if it's not misleading. Is it misleading? 'b_21 >= 1/10000, so N=21 is the smallest'. A student might think 'Oh, so if b_N >= tol, then N is the answer?' No, that would be wrong. They might think 'If b_{N+1} >= tol, then N is not the answer'. The sentence attributes the property of b_21 to the conclusion about N=21. It's ambiguous. I will mark it as 'misleading' because it obscures the index relationship (b_{N+1} vs b_N).
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-06qwen3.6:27b-mlx: fail (error) 2026-10-06 — The solution claims N=21 is the smallest N that works, but the check for N=21 shows b_{22} < 1/10000, meaning N=21 is sufficient. However, it fails to check if a smaller N (like N=20) is sufficient. Since b_{21} >= 1/10000, N=20 is not sufficient (error bound is b_{21}). Wait, the error bound for N terms is b_{N+1}. For N=20, error <= b_{21} = 1/9261 > 1/10000. For N=21, error <= b_{22} = 1/10648 < 1/10000. So N=21 is indeed the smallest. The logic in step 3 is slightly garbled: it says 'b_21 >= 1/10000, so N=21 is the smallest'. This implies that because the term *after* N=20 is too big, N=20 doesn't work. And since the term *after* N=21 is small enough, N=21 works. The conclusion is correct, but the phrasing 'b_21 ... so N=21' is confusingly linking the term index to the N index. Let's look closer. Step 3 says 'b_21 >= 1/10000, so N=21 is the smallest that works'. This is logically incomplete. It should say 'Since b_21 > 1/10000, N=20 does not guarantee the error. Since b_22 < 1/10000, N=21 does.' The current sentence skips the N=20 case explicitly and jumps to the conclusion. However, is it an error? It's a bit of a leap. Let's re-read carefully. 'b_21 = 1/9261 >= 1/10000, so N = 21 is the smallest that works.' This is a non-sequitur. The fact that b_21 is large doesn't prove N=21 is the *smallest*. It proves N=20 is *not* sufficient. One must also show N=21 *is* sufficient (which step 2 did). The sentence combines the negative check for N=20 and the positive conclusion for N=21 without clear logical connectors. It's misleading because it suggests the magnitude of b_21 directly determines N=21's status, rather than N=20's failure. Actually, looking at standard alternating series estimation, we need b_{N+1} < epsilon. We found b_{22} < epsilon, so N=21 works. We found b_{21} > epsilon, so N=20 does not work. Thus N=21 is the smallest. The solution presents this correctly in steps 2 and 3, but the sentence in 3 is poorly phrased. Is it an error? No, the math is right. Is it misleading? Yes, a student might think 'b_N < epsilon' is the condition, or get confused by the index shift. But wait, the prompt asks for 'error below 1/10000'. Step 2 shows N=21 works. Step 3 shows N=20 fails (implicitly, by showing b_21 is too big). The sentence 'so N=21 is the smallest' is the conclusion. The reasoning is sound, just compressed. I will mark it as 'style' or 'pass'? The prompt says 'find what is wrong'. If it's correct, pass. The logic is: 1. Check N=21: b_22 < tol. OK. 2. Check N=20: b_21 > tol. Fail. Therefore 21 is smallest. The text says 'b_21 >= ... so N=21 is smallest'. This is a valid logical step if one assumes monotonicity and that we are checking consecutive integers. It's not strictly 'wrong'. However, usually 'error' is reserved for mathematical mistakes. This is a clarity issue. Let's look for a harder error. Did I miss one? No. The answer N=21 is correct. The steps are correct. The sentence is just a bit loose. I'll call it 'style' if I must fail, but 'pass' is better if it's not misleading. Is it misleading? 'b_21 >= 1/10000, so N=21 is the smallest'. A student might think 'Oh, so if b_N >= tol, then N is the answer?' No, that would be wrong. They might think 'If b_{N+1} >= tol, then N is not the answer'. The sentence attributes the property of b_21 to the conclusion about N=21. It's ambiguous. I will mark it as 'misleading' because it obscures the index relationship (b_{N+1} vs b_N).qwen3.6:27b-mlx: fail (error) 2026-10-06 — The solution claims N=21 is the smallest N that works, but for N=21, the error bound is b_{22} = 1/10648, which is less than 1/10000. For N=20, the error bound is b_{21} = 1/9261, which is greater than 1/10000. Thus, N=21 is indeed the correct answer, but the reasoning in step 3 is flawed: it checks b_{21} against the tolerance to justify N=21, whereas it should check b_{N+1} (i.e., b_{22}) for N=21 and b_{21} for N=20. Specifically, the statement 'b_{21} >= 1/10000, so N=21 is the smallest that works' is logically disconnected; one must show that for N=20, the error bound b_{21} fails, and for N=21, the error bound b_{22} succeeds. The current text implies checking b_{N} instead of b_{N+1} for the condition, or is just poorly phrased. Actually, looking closer: Step 3 says 'b_{21} >= 1/10000, so N=21 is the smallest that works'. This is non-sequitur. The condition for N=20 to fail is b_{21} > 1/10000. The condition for N=21 to work is b_{22} < 1/10000. Step 2 establishes N=21 works. Step 3 establishes N=20 fails (since b_{21} > 1/10000). The conclusion 'so N=21 is the smallest' follows from combining 2 and 3, but the sentence structure in 3 is confusing/wrong because it cites b_{21} as the reason for N=21 being the answer, rather than b_{22}. However, strictly speaking, if N=20 fails, then 21 is the smallest integer > 20. The logic is: N=20 fails because b_{21} > tol. N=21 works because b_{22} < tol. Therefore 21 is the smallest. The text in 3 says 'b_{21} >= ... so N=21 is the smallest'. This is technically correct logic (if N=20 doesn't work, 21 is the next candidate, and we already showed 21 works). But it's very poorly phrased. Is it an error? It's misleading. A student might think they check b_N <= tol. Let's look at the standard Alternating Series Estimation Theorem: |R_N| <= b_{N+1}. We need b_{N+1} < 1/10000. For N=21, b_{22} < 1/10000. For N=20, b_{21} > 1/10000. So N=21 is the first one. The solution says: b_{22} < tol (so 21 works). b_{21} >= tol (so 20 doesn't work). Conclusion: 21 is smallest. This is correct logic. Why did I mark it fail? Let me re-read carefully. 'b_{21} = 1/9261 >= 1/10000, so N = 21 is the smallest that works.' This sentence is logically valid if interpreted as 'Since N=20 fails (due to b_{21}), and N=21 works (from prev step), N=21 is the smallest.' It is not an error. It is just concise. Wait, is it misleading? A student might learn 'check b_N'. No, the context is clear. Let's look for a real error. 1/10000 = 0.0001. 1/10648 ≈ 0.0000939 < 0.0001. 1/9261 ≈ 0.0001079 > 0.0001. The math is correct. The logic is correct. Is there a style issue? 'Subs' is not standard math notation, but the prompt says equations are checked by CAS. The sentences are what I review. Sentence 1 is correct. Sentence 3 is a bit abrupt but logically sound. I will pass it.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.