Home›Calculus 2›Error bounds for series›Problem 7.301

Error bounds for series

Problem 7.301 · easy

How many terms of \( \displaystyle \sum_{n=1}^{\infty} \frac{\left(-1\right)^{n + 1}}{n!} \) 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}.
\[ \left. \frac{1}{n!} \right|_{\substack{ n=5 }} = \frac{1}{120} \]
b_5 = 1/120 < 1/100.✓ Proved
\[ \left. \frac{1}{n!} \right|_{\substack{ n=4 }} = \frac{1}{24} \]
but b_4 = 1/24 ≥ 1/100, so N = 4 is the smallest that works.✓ Proved

Answer \( N = 4 \)

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: pass
qwen3.6:27b-mlx: fail (error) — The solution claims N=4 is the smallest N that works, but for N=3, the error bound is b_4 = 1/24 which is > 1/100, so N=3 fails. However, for N=4, the error bound is b_5 = 1/120 which is < 1/100, so N=4 works. Wait, let's re-read carefully. The question asks for N such that |R_N| <= 1/100. The estimate is |R_N| <= b_{N+1}. So we need b_{N+1} < 1/100. b_1=1, b_2=1/2, b_3=1/6, b_4=1/24, b_5=1/120. b_5 < 1/100 is true. b_4 < 1/100 is false. So we need N+1 >= 5, i.e., N >= 4. So N=4 is indeed the smallest integer. The logic in step 3 says 'b_4 >= 1/100, so N=4 is the smallest that works'. This phrasing is slightly confusing. It implies that because b_4 fails, N=4 works? No, b_4 corresponds to the error bound for N=3 (since |R_3| <= b_4). Since b_4 > 1/100, N=3 does not guarantee the error. Since b_5 < 1/100, N=4 does guarantee the error. The sentence 'b_4 = 1/24 >= 1/100, so N = 4 is the smallest that works' is logically disjointed. It states a fact about b_4 and concludes about N=4 without explicitly linking b_4 to N=3's failure. A student might think b_4 is the bound for N=4. The bound for N=4 is b_5. The sentence is misleading because it doesn't clearly state that N=3 fails because its bound b_4 is too large. It just says b_4 is large, therefore N=4 is the answer. It skips the explicit check that N=3 fails. However, is it an error? It's a bit of a logical leap. Let's look closer. Step 2 shows b_5 < 1/100. This implies N=4 works. Step 3 shows b_4 >= 1/100. This implies N=3 does NOT work (since |R_3| <= b_4 is not sufficient to guarantee < 1/100, wait. The estimate is an upper bound. If the upper bound is > 1/100, we cannot guarantee the error is < 1/100 using this test. So N=3 is not guaranteed. Thus N=4 is the smallest. The sentence 'so N=4 is the smallest that works' is the correct conclusion from the two facts. Is it misleading? It might be clearer to say 'Since b_4 > 1/100, N=3 does not guarantee the error. Since b_5 < 1/100, N=4 does. Thus N=4 is the smallest.' The current text is concise but potentially confusing regarding which N corresponds to which b. Specifically, it doesn't explicitly say 'N=3 fails'. It just says 'b_4 ... so N=4 ...'. A student might not make the connection that b_4 is the bound for N=3. This is a 'misleading' or 'style' issue? Or 'error'? If a student reads 'b_4 >= 1/100, so N=4 is the smallest', they might think the condition is b_N >= 1/100? No. They might think the condition for N is b_N. If they think the condition is b_N < 1/100, then b_4 < 1/100 is false, so N=4 doesn't work? No, the prompt says use |R_N| <= b_{N+1}. The solution uses this. The sentence is just poorly phrased. It's not strictly wrong, but it's not clear. However, looking at the strict instructions: 'nothing would teach a student something false'. Does it teach something false? It might teach that you check b_N against the tolerance for N. If a student thinks 'I need b_N < tol', then for N=4, b_4=1/24 > 1/100, so they'd think N=4 fails. But the solution says N=4 works. This is contradictory if the student has the wrong model. But the solution itself is correct. The sentence is just a summary. 'b_4 >= 1/100' is a fact. 'N=4 is the smallest that works' is a fact. The 'so' implies causality. The causality is: N=3 fails because b_4 >= 1/100. N=4 works because b_5 < 1/100. The sentence only mentions b_4. It omits the link to N=3. This is a 'style' or 'misleading' issue. I will call it 'misleading' because it obscures the index shift.

Every verdict on record (4)

qwen3.6:27b-mlx: fail (error) 2026-10-05 — The solution claims N=4 is the smallest N that works, but for N=3, the error bound is b_4 = 1/24 which is > 1/100, so N=3 fails. However, for N=4, the error bound is b_5 = 1/120 which is < 1/100, so N=4 works. Wait, let's re-read carefully. The question asks for N such that |R_N| <= 1/100. The estimate is |R_N| <= b_{N+1}. So we need b_{N+1} < 1/100. b_1=1, b_2=1/2, b_3=1/6, b_4=1/24, b_5=1/120. b_5 < 1/100 is true. b_4 < 1/100 is false. So we need N+1 >= 5, i.e., N >= 4. So N=4 is indeed the smallest integer. The logic in step 3 says 'b_4 >= 1/100, so N=4 is the smallest that works'. This phrasing is slightly confusing. It implies that because b_4 fails, N=4 works? No, b_4 corresponds to the error bound for N=3 (since |R_3| <= b_4). Since b_4 > 1/100, N=3 does not guarantee the error. Since b_5 < 1/100, N=4 does guarantee the error. The sentence 'b_4 = 1/24 >= 1/100, so N = 4 is the smallest that works' is logically disjointed. It states a fact about b_4 and concludes about N=4 without explicitly linking b_4 to N=3's failure. A student might think b_4 is the bound for N=4. The bound for N=4 is b_5. The sentence is misleading because it doesn't clearly state that N=3 fails because its bound b_4 is too large. It just says b_4 is large, therefore N=4 is the answer. It skips the explicit check that N=3 fails. However, is it an error? It's a bit of a logical leap. Let's look closer. Step 2 shows b_5 < 1/100. This implies N=4 works. Step 3 shows b_4 >= 1/100. This implies N=3 does NOT work (since |R_3| <= b_4 is not sufficient to guarantee < 1/100, wait. The estimate is an upper bound. If the upper bound is > 1/100, we cannot guarantee the error is < 1/100 using this test. So N=3 is not guaranteed. Thus N=4 is the smallest. The sentence 'so N=4 is the smallest that works' is the correct conclusion from the two facts. Is it misleading? It might be clearer to say 'Since b_4 > 1/100, N=3 does not guarantee the error. Since b_5 < 1/100, N=4 does. Thus N=4 is the smallest.' The current text is concise but potentially confusing regarding which N corresponds to which b. Specifically, it doesn't explicitly say 'N=3 fails'. It just says 'b_4 ... so N=4 ...'. A student might not make the connection that b_4 is the bound for N=3. This is a 'misleading' or 'style' issue? Or 'error'? If a student reads 'b_4 >= 1/100, so N=4 is the smallest', they might think the condition is b_N >= 1/100? No. They might think the condition for N is b_N. If they think the condition is b_N < 1/100, then b_4 < 1/100 is false, so N=4 doesn't work? No, the prompt says use |R_N| <= b_{N+1}. The solution uses this. The sentence is just poorly phrased. It's not strictly wrong, but it's not clear. However, looking at the strict instructions: 'nothing would teach a student something false'. Does it teach something false? It might teach that you check b_N against the tolerance for N. If a student thinks 'I need b_N < tol', then for N=4, b_4=1/24 > 1/100, so they'd think N=4 fails. But the solution says N=4 works. This is contradictory if the student has the wrong model. But the solution itself is correct. The sentence is just a summary. 'b_4 >= 1/100' is a fact. 'N=4 is the smallest that works' is a fact. The 'so' implies causality. The causality is: N=3 fails because b_4 >= 1/100. N=4 works because b_5 < 1/100. The sentence only mentions b_4. It omits the link to N=3. This is a 'style' or 'misleading' issue. I will call it 'misleading' because it obscures the index shift.
gpt-oss:20b: pass 2026-10-05
qwen3.6:27b-mlx: fail (error) 2026-10-05 — The solution claims N=4 is the smallest N that works, but b_4 = 1/24 > 1/100, so N=4 does not guarantee the error is below 1/100. The smallest N satisfying b_{N+1} < 1/100 is N=3 (since b_4 = 1/24 is not < 1/100, wait, b_{N+1} for N=3 is b_4=1/24 which is > 1/100. For N=4, b_5=1/120 < 1/100. So N=4 IS the answer. However, the reasoning in line 3 says 'b_4 >= 1/100, so N=4 is the smallest'. This logic is flawed because b_4 corresponds to the error bound for N=3, not N=4. The condition for N=4 is b_5 < 1/100. The check for N=3 would be b_4 < 1/100, which fails. So N=4 is indeed the first one. But the sentence 'b_4 >= 1/100, so N=4 is the smallest' is confusingly phrased. It implies checking b_4 determines N=4. Actually, checking b_5 determines N=4 works. Checking b_4 determines N=3 fails. The sentence conflates the index of the term with the index of the partial sum. It says 'b_4 ... so N=4'. It should say 'b_4 ... so N=3 fails, thus we check N=4'. As written, it suggests b_4 is the bound for N=4, which is false. b_{N+1} for N=4 is b_5. So the sentence is misleading/wrong in its implication.
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.