Infinite Geometric Series

The most common mistake students make about an infinite geometric series is thinking the formula S = a₁/(1 − r) always gives the answer. It does not. The formula only works when the series converges, and convergence requires |r| < 1. Apply it to a divergent series and you get a nonsense number, not the sum. Decide whether an infinite sum exists and, if it does, how to find it.

An infinite geometric series is the sum of the terms of an infinite geometric sequence. Each term after the first is the previous term multiplied by the common ratio r. The series is written as a₁ + a₁r + a₁r² + a₁r³ + … . Whether that sum is finite depends entirely on r.

What A Partial Sum Tells You

A partial sum Sₙ is the sum of the first n terms of a geometric series. Its formula, from OpenStax Algebra and Trigonometry 2e section 13.4, is Sₙ = a₁(1 − rⁿ)/(1 − r) for r ≠ 1. As n increases, Sₙ approaches a limit if |r| < 1. That limit is the infinite sum. If no limit exists, the series diverges and has no finite sum.

For example, take a₁ = 100 and r = 0.5. The partial sums are: S₁ = 100, S₂ = 150, S₃ = 175, S₄ = 187.5, S₅ = 193.75. They approach 200 but never exceed it. The infinite sum is 200. For a₁ = 100 and r = 2, the partial sums are: S₁ = 100, S₂ = 300, S₃ = 700, S₄ = 1500. They grow without bound. No finite sum exists.

The convergence condition is the single criterion from Stewart Calculus section 11.2: if |r| < 1, the series converges; if |r| ≥ 1, the series diverges. Check this before you compute anything.

The requirement |r| < 1 comes from the behaviour of rⁿ as n → ∞. When |r| < 1, rⁿ shrinks to zero. In the partial sum formula Sₙ = a₁(1 − rⁿ)/(1 − r), the term rⁿ vanishes, leaving S = a₁/(1 − r). When |r| > 1, rⁿ grows without bound, so Sₙ does too. When r = 1, the formula is undefined, you must use Sₙ = n·a₁, which also grows without bound. When r = −1, the partial sums oscillate between a₁ and 0, never settling on a single number. Stewart Calculus section 11.2 classifies all three divergent cases: r = 1, r = −1, and |r| > 1 all diverge.

One edge case students miss: r = 0. A geometric sequence with r = 0 is a₁, 0, 0, 0, … . OpenStax Algebra and Trigonometry 2e section 13.3 excludes r = 0 by requiring a nonzero common ratio, but the formula aₙ = a₁·0ⁿ⁻¹ still works for n = 1, and the series a₁ + 0 + 0 + … converges to a₁. If you encounter r = 0, the infinite sum is a₁, but treat it as a degenerate case, not a standard geometric series.

The infinite sum formula S = a₁/(1 − r) comes directly from the partial sum formula. Start with Sₙ = a₁(1 − rⁿ)/(1 − r). Take the limit as n → ∞. For |r| < 1, rⁿ → 0, so S = a₁/(1 − r). This derivation appears in OpenStax Algebra and Trigonometry 2e section 13.4 and Stewart Calculus section 11.2.

Stewart gives the example a = 5, r = 1/2. The infinite sum is 5/(1 − 0.5) = 10. Check with partial sums: S₁ = 5, S₂ = 7.5, S₃ = 8.75, S₄ = 9.375, S₅ = 9.6875. They approach 10.

Worked Example: Convergent

Find the sum of the infinite geometric series 12, 4, 4/3, 4/9, … . First identify a₁ = 12. Find r by dividing the second term by the first: 4 ÷ 12 = 1/3. Since |1/3| < 1, the series converges. Apply S = 12/(1 − 1/3) = 12/(2/3) = 18. The infinite sum is 18.

Every repeating decimal is a convergent geometric series. Write the decimal as a sum of fractions, identify a₁ and r, then apply S = a₁/(1 − r).

Take 0.7777… . Write it as 0.7 + 0.07 + 0.007 + … . Here a₁ = 0.7 and r = 0.1. Since |0.1| < 1, the series converges. The sum is 0.7/(1 − 0.1) = 0.7/0.9 = 7/9. So 0.7777… = 7/9.

Worked Example: Repeating Decimal To Fraction

Convert 0.454545… to a fraction. Write 0.45 + 0.0045 + 0.000045 + … . a₁ = 0.45, r = 0.01. The sum is 0.45/(1 − 0.01) = 0.45/0.99 = 45/99 = 5/11. Check: 5 ÷ 11 = 0.454545… . The method works for any repeating decimal.

Zeno's paradox of Achilles and the tortoise is the classic example of a convergent geometric series. Achilles runs a distance d to catch the tortoise, which moves ahead by d/10, then d/100, then d/1000, and so on. The distances form a geometric series with a₁ = d and r = 1/10. The total distance is d/(1 − 1/10) = 10d/9, a finite number. The paradox is resolved: the infinite number of steps sums to a finite distance.

Other real-world examples of sum of infinite geometric series include bouncing balls: a ball dropped from h metres rebounds to rh metres, then r²h metres, etc. The total vertical distance is h + 2rh + 2r²h + … for r < 1. That sum is h + 2rh/(1 − r).

In finance, the present value of a perpetuity, a payment P every period, discounted at rate i, is the infinite sum P/(1 + i) + P/(1 + i)² + … = P/i, provided the interest rate does not change.

When |r| ≥ 1, the infinite geometric series has no finite sum. Stewart Calculus section 11.2 gives four examples: r = 1/2 converges to 10 for a = 5; r = 1 diverges; r = −1 diverges by oscillation; r = 2 diverges by growth.

Worked Example: Divergent

Decide whether the series 8, 12, 18, 27, … has a finite sum. Find r = 12 ÷ 8 = 1.5. Since |1.5| > 1, the series diverges. Do not apply S = a₁/(1 − r).There is no infinite sum.

Another failure: a series with r = −1.2 and a₁ = 10.The partial sums also oscillate without settling: S₁ = 10, S₂ = −2, S₃ = 12.4, S₄ = −4.88. The series diverges.

The single thing that most often goes wrong: a student checks r < 1 instead of |r| < 1. A common ratio of −2 satisfies r < 1 but |r| = 2 ≥ 1, so the series diverges. Always check the absolute value.

The infinite sum formula S = a₁/(1 − r) is exact only for |r| < 1. It is not an approximation. But applying it without first checking convergence is the single most common error on this topic. Always check |r| < 1 before computing. If you do not, you risk reporting a finite sum for a divergent series, a mistake that a calculator will not catch for you.