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Telescoping series

Problem 7.320 · easy

Find the sum of \( \displaystyle \sum_{n=1}^{\infty} \frac{6}{n^{2} + 5 n + 4} \), or show that it diverges.

\[ \frac{6}{n^{2} + 5 n + 4} = - \frac{2}{n + 4} + \frac{2}{n + 1} \]
Partial fractions: the term is a difference.✓ Proved
In the N-th partial sum every negative piece cancels a positive piece 3 terms later; only the first positive and last negative pieces survive.
Reviewed
So S_N = 13/6 - 2/(N + 4) - 2/(N + 3) - 2/(N + 2).
Reviewed
\[ \lim_{k \to \infty}\left(\frac{13}{6} - \frac{2}{k + 4} - \frac{2}{k + 3} - \frac{2}{k + 2}\right) = \frac{13}{6} \]
Let N → ∞.✓ Proved

Answer \( \frac{13}{6} \)

Lines: 2 proved, 2 reviewed. 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	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
2	Reviewed	reviewers: gpt-oss:20b, qwen3.6:27b-mlx	a sentence; read, not computed
3	Reviewed	reviewers: gpt-oss:20b, qwen3.6:27b-mlx	a sentence; read, not computed
4	✓ 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	mpmath's nsum of the series (with convergence acceleration)

Reviewers

gpt-oss:20b: pass
qwen3.6:27b-mlx: pass — The solution correctly identifies the telescoping nature of the series, accurately accounts for the surviving terms in the partial sum, and correctly computes the limit.

Every verdict on record (4)

qwen3.6:27b-mlx: pass 2026-10-05 — The solution correctly identifies the telescoping nature of the series, accurately accounts for the surviving terms in the partial sum, and correctly computes the limit.
gpt-oss:20b: pass 2026-10-05
qwen3.6:27b-mlx: fail (error) 2026-10-05 — The partial fraction decomposition is incorrect; the correct decomposition is 2/(n+1) - 2/(n+4), but the solution writes -2/(n+4) + 2/(n+1) which is actually correct algebraically, wait. Let's re-verify. 2/(n+1) - 2/(n+4) = 2(n+4 - (n+1))/((n+1)(n+4)) = 2(3)/((n+1)(n+4)) = 6/(n^2+5n+4). So the equation in line 1 is correct. However, line 2 claims 'every negative piece cancels a positive piece 3 terms later'. Let's check the telescoping. The term is a_n = 2/(n+1) - 2/(n+4). The sum is sum_{n=1}^N (2/(n+1) - 2/(n+4)). The positive parts are 2/2, 2/3, 2/4, ... 2/(N+1). The negative parts are -2/5, -2/6, ... -2/(N+4). The terms 2/5, 2/6, 2/7 cancel with -2/5, -2/6, -2/7? No. The positive term 2/(k+1) cancels with the negative term -2/(k+1) which comes from index n such that n+4 = k+1 => n = k-3. So the negative term at index n cancels the positive term at index n+3. Or, the positive term at index n cancels the negative term at index n+3. Line 2 says 'negative piece cancels a positive piece 3 terms later'. This is ambiguous but generally acceptable phrasing for telescoping. Let's look at the surviving terms. Positive terms that don't cancel: n=1 (2/2), n=2 (2/3), n=3 (2/4). Negative terms that don't cancel: n=N-2 (-2/(N+2)), n=N-1 (-2/(N+3)), n=N (-2/(N+4)). So S_N = 2/2 + 2/3 + 2/4 - 2/(N+2) - 2/(N+3) - 2/(N+4). 2/2 + 2/3 + 2/4 = 1 + 2/3 + 1/2 = 3/2 + 2/3 = 9/6 + 4/6 = 13/6. So S_N = 13/6 - 2/(N+2) - 2/(N+3) - 2/(N+4). Line 3 states exactly this. Line 4 takes the limit. The logic is sound. Why did I think it was wrong? Let me re-read carefully. 'every negative piece cancels a positive piece 3 terms later'. The negative piece is -2/(n+4). The positive piece 3 terms later is from n+3: 2/((n+3)+1) = 2/(n+4). Yes, they cancel. The setup is correct. The calculation is correct. The limit is correct. Is there a subtle error? The problem asks for sum from n=1 to infinity. The solution uses N-th partial sum. The limit is 13/6. The stated answer is 13/6. Everything seems correct. Wait, is the partial fraction decomposition in line 1 marked 'unchecked'? The prompt says 'Each line of the solution is either an EQUATION, which a computer algebra system has already checked... or a SENTENCE'. Line 1 is marked [equation, unchecked]. This is a contradiction in the prompt description vs the label. Usually 'unchecked' means I should check it. But the instructions say 'do not re-check the algebra of a line marked proved'. Here it is marked 'unchecked'. This implies I SHOULD check it. I checked it, and it is correct. So the verdict should be pass? Let me double check the 'unchecked' label meaning. The prompt says: 'Each line ... is either an EQUATION, which a computer algebra system has already checked (its result is shown; do not re-check the algebra of a line marked proved)'. It doesn't explicitly say what to do with 'unchecked' equations, but typically in these tasks, 'unchecked' means the CAS didn't verify it, or it's a hint. However, the instruction 'do not re-check the algebra of a line marked proved' implies that if it's NOT marked proved, I might need to check it. Or perhaps 'unchecked' is a typo for 'proved' in the user's template? No, usually 'unchecked' means the user wants me to verify it. I verified it, it is correct. The sentences are correct. The limit is correct. I will mark it as 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/telescoping_series, checked 2026-10-05 with SymPy 1.14.0.