Algebra
Advanced

Infinite Geometric Series

Sum of an infinite geometric series that converges.

Formula

S=a1−r,  ∣r∣<1S = \frac{a}{1-r}, \; |r|<1

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Variables

aFirst term
rCommon ratio

Example

1+½+¼+... = 2

How it works

This formula gives the total of a geometric pattern that keeps shrinking by the same factor each time, so the added terms become too small to change the sum. The reason it works is that if you multiply the whole infinite sum by the common ratio r and subtract, almost all terms cancel, leaving only the first term a; this cancellation is valid only when the terms actually approach 0, which happens when |r|<1. In that case, the infinite sum settles to a finite limit of a/(1-r).

When to use it

Use this whenever a quantity repeats with constant proportional decay and you need the long-run total, such as discounted cash flows, repeated reflections, exponential-decay approximations, or accumulating probabilities in a geometric process. It is especially useful when the pattern is geometric and extends indefinitely but still converges.

Worked example, step by step

Find the sum of 12 + 6 + 3 + 1.5 + ... Here, a = 12 and r = 6/12 = 1/2 Check convergence: |r| = 1/2 < 1, so the series converges Use S = a / (1 - r) S = 12 / (1 - 1/2) S = 12 / (1/2) S = 24 So the infinite geometric series sums to 24

Common mistakes to avoid

  • Using the formula when |r| is 1 or greater, which makes the series diverge instead of converging
  • Forgetting that a is the first term of the series, not the ratio between terms
  • Substituting r with the second term instead of the common ratio between consecutive terms

Did You Know?

An infinite series of shrinking terms can add up to a finite total — the resolution of Zeno's ancient paradoxes.

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Reviewed & maintained by the FormulaWon editorial teamLast reviewed September 2026