WEBVTT

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Here is the problem: Σ 1 over 3 ^ n, n equals 1 to ∞.

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Decide whether Σ 1 over 3 to the power n (n from 1 to ∞) converges or diverges.

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Write the series. Σ a(n) for n equals 1, 2, 3, … with a(n) equals 1 over 3 to the power n. First terms: 0.333333, 0.111111, 0.037037, 0.0123457, ….

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Recognise a geometric series. Each term is the previous one times r equals 1 over 3 (checked for the first 40 terms), with first term a equals 1 over 3. Since |r| less than 1 the series converges, and its sum is a/(1 minus r) equals 1 over 2 approximately 0.5.

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So the answer is The series converges (geometric series (|r| less than 1)); sum equals 1 over 2.

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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.

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Partial sums (supporting evidence, not a proof): S₁₀ equals 0.49999153, S₁₀₀ equals 0.5, S₁₀₀₀ equals 0.5, S₁₀₀₀₀ equals 0.5.