Algebra
Intermediate

Infinite Geometric Series

Sum of an infinite geometric series that converges.

Formula

S=a11−r,∣r∣<1S = \frac{a_1}{1 - r}, \quad |r| < 1

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Variables

a_1First term
rCommon ratio (

Example

1 + ½ + ¼ + ... = 2

How it works

An infinite geometric series adds terms that keep getting multiplied by the same factor r each time: a1, a1r, a1r^2, and so on. When the absolute value of r is less than 1, the terms shrink toward 0, so the total sum settles to a fixed value rather than growing without bound. The formula comes from comparing the series to a shifted copy of itself and using the fact that the leftover tail vanishes as the number of terms becomes very large.

When to use it

Use this when a quantity is repeated over and over by a constant ratio and the repeated pieces get smaller, such as in some finance, probability, decay, or limit problems. It only applies when the common ratio satisfies |r| < 1, so the terms actually converge.

Worked example, step by step

Find the sum of 12 + 6 + 3 + 1.5 + ... Step 1: Identify the first term and ratio: a1 = 12, r = 6/12 = 1/2 Step 2: Check convergence: |r| = 1/2 < 1, so the formula applies Step 3: Substitute into S = a1 / (1 - r) S = 12 / (1 - 1/2) Step 4: Simplify the denominator S = 12 / (1/2) Step 5: Divide S = 24

Common mistakes to avoid

  • Using the formula when |r| is 1 or greater, which makes the infinite sum diverge
  • Forgetting that a1 is the first term, not the ratio or the second term
  • Substituting the wrong sign for r, especially when the terms alternate

Did You Know?

An infinite sum can add up to a finite number — Zeno's paradox of never reaching the wall is resolved by 1/2+1/4+1/8+··· = 1.

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Reviewed & maintained by the FormulaWon editorial teamLast reviewed September 2026