Evaluate the sum of (2/3)^n from n=0 to infinity: animated lesson
A narrated, step-by-step sequences and series video lesson with captions. Final answer: The series converges (geometric series (|r| < 1)); sum = 3.
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 (0.5 MB) · Captions (WebVTT)
Steps on screen
- 1. The problemWorked example
- 2. Step by step
- Write the series
- Recognise a geometric series
- 3. The answer
geometric series (|r| < 1) ⇒ converges, S = 3
- 4. Check
- Partial sums (supporting evidence, not a proof): S₁₀ = 2.9479754, S₁₀₀ = 3, S₁₀₀₀ = 3, S₁₀₀₀₀ = 3
Transcript
The problem. Here is the problem: Σ (2 over 3) ^ n, n equals 0 to ∞. Decide whether Σ (2 over 3) to the power n (n from 0 to ∞) converges or diverges.
Step by step. Write the series. Σ a(n) for n equals 0, 1, 2, … with a(n) equals (2 over 3) to the power n. First terms: 1, 0.666667, 0.444444, 0.296296, …. Recognise a geometric series. Each term is the previous one times r equals 2 over 3 (checked for the first 40 terms), with first term a equals 1. Since |r| less than 1 the series converges, and its sum is a/(1 minus r) equals 3.
The answer. So the answer is The series converges (geometric series (|r| less than 1)); sum equals 3. 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 2.9479754, S₁₀₀ equals 3, S₁₀₀₀ equals 3, S₁₀₀₀₀ equals 3.
About sequences and series
Evaluate finite sums and arithmetic and geometric series.
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