Summing a series by differentiating or integrating
Problem 7.686 · hard
Find the sum of \( \displaystyle \sum_{n=1}^{\infty} \frac{2^{- n}}{n} \).
- 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
- \[ \left. - \ln{\left(1 - x \right)} \right|_{\substack{ x=1 \cdot \frac{1}{2} }} = \ln{\left(2 \right)} \]Put x = 1/2.✓ Proved
- \[ \sum_{n=1}^{\infty} \frac{2^{- n}}{n} = \ln{\left(2 \right)} \]Summed directly, the series agrees.✓ Proved
Answer \( \ln{\left(2 \right)} \)
✓ 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: passqwen3.6:27b-mlx: pass — The solution correctly identifies the series as the Taylor expansion of -ln(1-x) evaluated at x=1/2. The steps are logically sound and the result 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) evaluated at x=1/2. The steps are logically sound and the result is correct.gpt-oss:20b: pass 2026-10-18qwen3.6:27b-mlx: pass 2026-10-16 — The solution correctly identifies the series as the Taylor expansion of -ln(1-x) evaluated at x=1/2. The steps are logically sound and the final result is correct.gpt-oss:20b: pass 2026-10-16qwen3.6:27b-mlx: pass 2026-10-15 — The solution correctly identifies the power series for -ln(1-x) and evaluates it at x=1/2. The convergence at the boundary point x=1/2 is valid by Abel's Theorem, though not explicitly cited, the result 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.