Home›Calculus 2›Summing a series by differentiating or integrating›Problem 7.685

Summing a series by differentiating or integrating

Problem 7.685 · hard

Find the sum of \( \displaystyle \sum_{n=1}^{\infty} \frac{\left(- \frac{1}{5}\right)^{n}}{n} \).

Integrate Σ xⁿ⁻¹ = 1/(1 − x) from 0 to x term by term: Σ xⁿ/n = −ln(1 − x) for |x| < 1.
Reviewed
\[ \frac{d}{d x} \left(- \ln{\left(1 - x \right)}\right) = \frac{1}{1 - x} \]
−ln(1 − x) is the antiderivative of 1/(1 − x) that is 0 at x = 0.✓ Proved
\[ \left. - \ln{\left(1 - x \right)} \right|_{\substack{ x=- \frac{1}{5} }} = - \ln{\left(6 \right)} + \ln{\left(5 \right)} \]
Put x = -1/5.✓ Proved
\[ \sum_{n=1}^{\infty} \frac{\left(- \frac{1}{5}\right)^{n}}{n} = - \ln{\left(6 \right)} + \ln{\left(5 \right)} \]
Summed directly, the series agrees.✓ Proved

Answer \( - \ln{\left(6 \right)} + \ln{\left(5 \right)} \)

✓ Nihil obstat Lines: 3 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. How we verify

The full receipt

Line	Status	Checked by	Detail
1	Reviewed	reviewers: gpt-oss:20b, qwen3.6:27b-mlx	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
answer, a second way	✓ Checked independently	sympy 1.14.0 + mpmath 1.3.0	the first 399 terms added in floating point

Reviewers

gpt-oss:20b: pass
qwen3.6:27b-mlx: pass — The solution correctly identifies the series as the Taylor expansion of -ln(1-x) and verifies the convergence condition |x| < 1 for x = -1/5. The algebraic simplification is correct.

Every verdict on record (6)

qwen3.6:27b-mlx: pass 2026-10-18 — The solution correctly identifies the series as the Taylor expansion of -ln(1-x) and verifies the convergence condition |x| < 1 for x = -1/5. The algebraic simplification is correct.
gpt-oss:20b: pass 2026-10-18
qwen3.6:27b-mlx: pass 2026-10-16 — The solution correctly identifies the series as the Maclaurin expansion for -ln(1-x) and correctly substitutes x = -1/5. The logic is sound and the result is correct.
gpt-oss:20b: pass 2026-10-16
qwen3.6:27b-mlx: pass 2026-10-15 — The solution correctly identifies the series as the Taylor expansion of -ln(1-x) and correctly substitutes x = -1/5. The algebraic simplification to ln(5/6) is correct.
gpt-oss:20b: pass 2026-10-15

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_sum_by_calculus, checked 2026-10-18 with SymPy 1.14.0.