Error bounds for series
Problem 7.303 · 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}.
- \[ \left. \frac{1}{n!} \right|_{\substack{ n=8 }} = \frac{1}{40320} \]b_8 = 1/40320 < 1/10000.✓ Proved
- \[ \left. \frac{1}{n!} \right|_{\substack{ n=7 }} = \frac{1}{5040} \]but b_7 = 1/5040 ≥ 1/10000, so N = 7 is the smallest that works.✓ Proved
Answer \( N = 7 \)
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 conclusion that N=7 works is incorrect: b_7=1/5040≈0.000198>1/10000, so the error after 7 terms is not guaranteed below 1/10000. The correct smallest N is 8.qwen3.6:27b-mlx: fail (error) — The solution claims N=7 is the smallest integer satisfying the condition, but b_7 = 1/5040 > 1/10000, so N=7 does not guarantee the error is below 1/10000. The correct answer is N=6, since b_7 is the error bound for the sum of the first 6 terms, and 1/5040 > 1/10000 is false? Wait. 1/5040 is approx 0.000198, which is > 0.0001. So N=6 fails. N=7 gives error bound b_8 = 1/40320 < 0.0001. So N=7 is correct. However, the text says 'b_7 ... so N=7 is the smallest that works'. This is logically garbled. It implies checking b_7 determines N=7. Actually, checking b_{N+1} < epsilon determines N. For N=6, b_7 > epsilon (fail). For N=7, b_8 < epsilon (pass). The sentence 'b_7 ... so N=7 is the smallest that works' is misleading/wrong reasoning. It should say 'Since b_7 > 1/10000, N=6 is not sufficient. Since b_8 < 1/10000, N=7 is sufficient.' The current phrasing suggests b_7 is the bound for N=7, which is false (b_8 is the bound for N=7). Or it suggests b_7 failing means N=7 works, which is a non-sequitur without checking N=7's bound. Let's look closer. 'b_7 ... so N=7 is the smallest that works'. This is a logical error. The failure of b_7 (as a bound for N=6) doesn't prove N=7 works; one must check b_8. The solution does check b_8 in line 2, but the conclusion in line 3 attributes the sufficiency of N=7 to the value of b_7. This is confusing and technically incorrect reasoning.
Every verdict on record (4)
qwen3.6:27b-mlx: fail (error) 2026-10-05 — The solution claims N=7 is the smallest integer satisfying the condition, but b_7 = 1/5040 > 1/10000, so N=7 does not guarantee the error is below 1/10000. The correct answer is N=6, since b_7 is the error bound for the sum of the first 6 terms, and 1/5040 > 1/10000 is false? Wait. 1/5040 is approx 0.000198, which is > 0.0001. So N=6 fails. N=7 gives error bound b_8 = 1/40320 < 0.0001. So N=7 is correct. However, the text says 'b_7 ... so N=7 is the smallest that works'. This is logically garbled. It implies checking b_7 determines N=7. Actually, checking b_{N+1} < epsilon determines N. For N=6, b_7 > epsilon (fail). For N=7, b_8 < epsilon (pass). The sentence 'b_7 ... so N=7 is the smallest that works' is misleading/wrong reasoning. It should say 'Since b_7 > 1/10000, N=6 is not sufficient. Since b_8 < 1/10000, N=7 is sufficient.' The current phrasing suggests b_7 is the bound for N=7, which is false (b_8 is the bound for N=7). Or it suggests b_7 failing means N=7 works, which is a non-sequitur without checking N=7's bound. Let's look closer. 'b_7 ... so N=7 is the smallest that works'. This is a logical error. The failure of b_7 (as a bound for N=6) doesn't prove N=7 works; one must check b_8. The solution does check b_8 in line 2, but the conclusion in line 3 attributes the sufficiency of N=7 to the value of b_7. This is confusing and technically incorrect reasoning.gpt-oss:20b: fail (error) 2026-10-05 — The conclusion that N=7 works is incorrect: b_7=1/5040≈0.000198>1/10000, so the error after 7 terms is not guaranteed below 1/10000. The correct smallest N is 8.qwen3.6:27b-mlx: fail (error) 2026-10-05 — The solution claims N=7 is the smallest N that works, but b_7 = 1/5040 > 1/10000, so the error bound for N=6 (which is b_7) is not sufficient. However, the error bound for N=7 is b_8 = 1/40320 < 1/10000, so N=7 is indeed the correct answer. Wait, let me re-read carefully. The alternating series estimate says |R_N| <= b_{N+1}. We need b_{N+1} < 1/10000. For N=6, b_7 = 1/5040 > 1/10000. For N=7, b_8 = 1/40320 < 1/10000. So N=7 is the smallest integer such that the error bound is less than 1/10000. The logic in step 3 says 'b_7 >= 1/10000, so N=7 is the smallest that works'. This phrasing is slightly confusing because b_7 corresponds to the error bound for N=6. It should say 'Since b_7 > 1/10000, N=6 is not sufficient. Since b_8 < 1/10000, N=7 is sufficient.' The conclusion is correct, but the reasoning in sentence 3 is imprecise/misleading about which N corresponds to which term. Actually, looking closer, step 3 says 'b_7 ... >= 1/10000, so N=7 is the smallest that works'. This implies that because b_7 is too big, we need to go to the next N. But b_7 is the bound for N=6. So N=6 fails. N=7 uses b_8. The sentence skips the explicit check of b_8 for N=7 in the text, relying on step 2. Step 2 checks b_8. Step 3 checks b_7. The logic is: Check N=7 -> bound is b_8 (step 2). Check N=6 -> bound is b_7 (step 3). Since b_7 is too big, N=6 doesn't work. Since b_8 is small enough, N=7 works. The sentence 'so N=7 is the smallest that works' follows from the fact that N=6 failed and N=7 passed. This is logically sound, though the phrasing 'b_7 ... so N=7' is a bit of a jump. Is it an error? No. Is it misleading? Maybe slightly. But wait, is N=7 the answer? Yes. Is the reasoning correct? Yes. Let's look for a real error. Ah, the problem asks for N terms. The error is bounded by b_{N+1}. We need b_{N+1} < 1/10000. 1/7! = 1/5040 > 0.0001. 1/8! = 1/40320 < 0.0001. So we need N+1 >= 8, so N >= 7. The smallest N is 7. The solution is correct. Why would it fail? Let me re-read the prompt constraints. 'Check that... each conclusion follows from the lines before it'. Step 3 says 'b_7 >= 1/10000, so N=7 is the smallest that works'. This conclusion requires knowing that N=7 *does* work. Step 2 establishes that b_8 < 1/10000, which means N=7 works. Step 3 establishes that b_7 >= 1/10000, which means N=6 does *not* work. Combining these, N=7 is the smallest. The sentence in step 3 only explicitly mentions the failure of the previous term, but implicitly relies on step 2 for the success of the current term. This is standard mathematical shorthand. It is not an error. It is not misleading. It is style? Maybe. But usually 'pass' is for sound solutions. This is sound. Let me double check the 'unchecked' equations. 1/40320 is approx 2.48e-5. 1/10000 is 1e-4. 2.48e-5 < 1e-4. Correct. 1/5040 is approx 1.98e-4. 1.98e-4 > 1e-4. Correct. The logic holds. Verdict: 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.