Determine if a geometric series converges and find its exact sum using S∞ = a₁/(1−r) when |r| < 1.
| n | Sₙ (partial sum) | S∞ − Sₙ (error) |
|---|
Partial sums Sₙ plotted against n — watch them approach the horizontal asymptote S∞ (gold dashed line).
S∞ = a₁ / (1 − r), when |r| < 1
|r| < 1 is both necessary and sufficient for the series to converge. This means −1 < r < 1.
Why r ≥ 1 diverges: When r ≥ 1, each new term is at least as large as the last — the partial sums keep growing without bound.
Why r ≤ −1 diverges: When r = −1 terms alternate +a₁, −a₁, ... and never settle. When r < −1 the magnitude of terms grows.
−1 < r < 0 (alternating): Terms shrink in absolute value and alternate sign. The series still converges — partial sums oscillate closer and closer to S∞.
1 + 1/2 + 1/4 + 1/8 + ··· = 2
Zeno's Paradox (r = 1/2, a₁ = 1): S∞ = 1/(1−1/2) = 2. Each step covers half the remaining distance, yet the total converges to exactly 2.
r = 1/3, a₁ = 1: 1 + 1/3 + 1/9 + 1/27 + ··· = 1/(1−1/3) = 3/2.
0.999… = 1: Write 0.999… as 9/10 + 9/100 + ··· Here a₁ = 9/10, r = 1/10, so S∞ = (9/10)/(1−1/10) = (9/10)/(9/10) = 1. Exactly 1!
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