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Sequences and seriesAlgebraCollegeabout 3:06

Evaluate the sum of (-1)^n/n from n=1 to infinity: animated lesson

A narrated, step-by-step sequences and series video lesson with captions. Final answer: The series converges conditionally (alternating series test); sum ≈ -0.6931471806 (matches −ln(2) to 10+ digits).

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Video with a synthetic English voice and captions. Pick another language for subtitles in that language, or open the interactive version.

Video narrated by a synthetic voice (Piper TTS, en_US-ljspeech-high voice (LJ Speech dataset, public domain)); the steps come from our maths engine and the answer was checked independently. Download MP4 (1.1 MB) · Captions (WebVTT)

Steps on screen

1. The problem
Worked example
$\sum_{n=1}^{\infty} \frac{{\left(-1\right)}^{n}}{n}$
2. Steps, part 1
1. $\sum_{n=1}^{\infty} \frac{{\left(-1\right)}^{n}}{n}$
  Write the series
2. $\lim_{n\to\infty} \frac{1}{n} = 0$
  nth-term test
3. Alternating series test
4. $\lim_{n\to\infty} \frac{a_n}{1/n} = 1$
  Limit comparison with a p-series
3. Steps, part 2
1. Absolute or conditional?
2. Estimate the sum numerically
4. The answer
$\sum_{n=1}^{\infty} \frac{{\left(-1\right)}^{n}}{n} \ \text{converges conditionally} \approx -0.6931471806$
alternating series test ⇒ converges conditionally, S ≈ -0.6931471806 (matches −ln(2) to 10+ digits)
5. Check
- Partial sums (supporting evidence, not a proof): S₁₀ = -0.64563492, S₁₀₀ = -0.68817218, S₁₀₀₀ = -0.69264743, S₁₀₀₀₀ = -0.69309718

Transcript

The problem. Here is the problem: Σ (negative 1) ^ n / n, n equals 1 to ∞. Decide whether Σ (negative 1) to the power n / n (n from 1 to ∞) converges or diverges.

Steps, part 1. Write the series. Σ a(n) for n equals 1, 2, 3, … with a(n) equals (negative 1) to the power n / n. First terms: negative 1, 0.5, negative 0.333333, 0.25, …. nth-term test. lim |a(n)| equals 0, so the divergence test is inconclusive; the terms do shrink, so a finer test is needed. Alternating series test. The signs alternate and the magnitudes b(n) equals 1 / n decrease (checked on 310 indices up to 10 to the power 6) and tend to 0, so the series converges. Limit comparison with a p-series. Absolute values Σ 1 / n: the terms shrink roughly like 1/n (estimated from a(2N)/a(N)), so compare with Σ 1/n: lim a(n) times n equals 1. The limit is a positive finite number, so the series behaves exactly like the p-series Σ 1/n, which diverges because p less than or equal to 1.

Steps, part 2. Absolute or conditional?. Σ |a(n)| diverges (limit comparison with Σ 1/n), so the series converges only conditionally. Estimate the sum numerically. Partial sums S N were computed for several N and extrapolated to N → ∞ (Richardson extrapolation; alternating terms are averaged first). Two independent extrapolations agree to better than 10 to the power minus 10, giving S approximately negative 0.6931471806. Any closed form shown is a numerical identification, not a proof.

The answer. So the answer is The series converges conditionally (alternating series test); sum approximately negative 0.6931471806 (matches negative ln(2) to 10+ digits). A series converges when its partial sums approach a finite number. The terms must tend to 0 for that to happen, but that alone is not enough (the harmonic series diverges), so a comparison, ratio, root, alternating or integral test is used to decide.

Check. Partial sums (supporting evidence, not a proof): S₁₀ equals negative 0.64563492, S₁₀₀ equals negative 0.68817218, S₁₀₀₀ equals negative 0.69264743, S₁₀₀₀₀ equals negative 0.69309718.

About sequences and series

Evaluate finite sums and arithmetic and geometric series.

Evaluate the sum of (-1)^n/n from n=1 to infinity (Lesson)