Home›Calculus 2›Partial sums of a series›Problem 7.560

Partial sums of a series

Problem 7.560 · easy

Find the first four partial sums \( \displaystyle S_1, \dots, S_4 \) of \( \displaystyle \sum_{n=1}^{\infty} \left(-1\right)^{n} \).

S_k adds the first k terms: a₁ = -1, a₂ = 1, and so on.
\[ \left. \left(-1\right)^{n} \right|_{\substack{ n=1 }} = -1 \]
S1.✓ Proved
\[ -1 + 1 = 0 \]
S2.✓ Proved
\[ -1 - 1 + 1 = -1 \]
S3.✓ Proved
\[ -1 - 1 + 1 + 1 = 0 \]
S4.✓ Proved

Answer \( S_1 = -1,\ S_2 = 0,\ S_3 = -1,\ S_4 = 0 \)

Lines: 4 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. How we verify

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
4	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
5	✓ 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 terms added in floating point

Reviewers

gpt-oss:20b: pass
qwen3.6:27b-mlx: fail (error) — The solution incorrectly calculates the partial sums by using -1 for all odd-indexed terms (e.g., S3 uses -1 - 1 + 1 instead of -1 + 1 - 1). The terms are a_n = (-1)^n, so a_1=-1, a_2=1, a_3=-1, a_4=1. The correct sums are S_1=-1, S_2=0, S_3=-1, S_4=0, but the intermediate arithmetic shown in lines 4 and 5 is factually wrong (e.g., -1-1+1 = -1 is correct arithmetic, but it represents the sum of -1, -1, 1, not the first three terms of the series).

Every verdict on record (6)

qwen3.6:27b-mlx: fail (error) 2026-10-17 — The solution incorrectly calculates the partial sums by using -1 for all odd-indexed terms (e.g., S3 uses -1 - 1 + 1 instead of -1 + 1 - 1). The terms are a_n = (-1)^n, so a_1=-1, a_2=1, a_3=-1, a_4=1. The correct sums are S_1=-1, S_2=0, S_3=-1, S_4=0, but the intermediate arithmetic shown in lines 4 and 5 is factually wrong (e.g., -1-1+1 = -1 is correct arithmetic, but it represents the sum of -1, -1, 1, not the first three terms of the series).
gpt-oss:20b: pass 2026-10-17
qwen3.6:27b-mlx: fail (error) 2026-10-15 — The solution incorrectly calculates the partial sums by using -1 for all odd-indexed terms instead of alternating signs. Specifically, S_3 should be -1 + 1 - 1 = -1 (which happens to be correct by coincidence of the sum, but the expression -1 - 1 + 1 implies a_3 = -1, which is wrong; a_3 is -1, so the terms are -1, 1, -1. Wait, a_n = (-1)^n. a_1 = -1, a_2 = 1, a_3 = -1, a_4 = 1. S_3 = -1 + 1 + (-1) = -1. The equation line 4 says '-1 - 1 + 1'. This implies the terms are -1, -1, 1. This is incorrect. The terms are -1, 1, -1. The sum is the same (-1), but the representation of the partial sum is wrong. S_4 = -1 + 1 - 1 + 1 = 0. Line 5 says '-1 - 1 + 1 + 1'. This implies terms -1, -1, 1, 1. This is incorrect. The terms are -1, 1, -1, 1. The sum is the same (0), but the expansion is wrong. The student is taught that the terms are -1, -1, 1, 1... which is false.
gpt-oss:20b: pass 2026-10-15
qwen3.6:27b-mlx: fail (error) 2026-10-14 — The calculation for S_3 is incorrect; it sums to 0, not -1. The solution incorrectly writes '-1 - 1 + 1' instead of '-1 + 1 - 1', implying a_2 was -1, which contradicts the established pattern a_2 = 1.
gpt-oss:20b: pass 2026-10-14

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/partial_sums, checked 2026-10-17 with SymPy 1.14.0.