Sum of an Infinite GP: Formula a/(1−r) with Examples

The sum of an infinite geometric progression asks what happens when the terms keep coming forever. If the common ratio satisfies |r| < 1, each term is smaller than the last and the terms fade toward zero, so the total settles at a finite limit: S = a/(1 - r). For example, 8 + 4 + 2 + 1 + 0.5 + … creeps toward 16 but never passes it.

This matters because infinite processes with shrinking steps appear throughout mathematics. Zeno's ancient paradox about never reaching a destination, repeating decimals like 0.333…, and the total effect of an economic stimulus rippling through spending rounds are all infinite geometric sums. The formula converts "forever" into one division.

Key Concepts

• Convergence condition: the formula works only when the absolute value of r is below 1. Then r^n approaches 0 as n grows, and the finite-sum formula collapses to the infinite one.
• Infinite sum: S = a/(1 - r), the limit of the partial sums.
• Divergence: if |r| is 1 or more, the terms do not fade away, so the sum never settles. A formula applied there gives nonsense.
• Partial sums: S_1, S_2, S_3, … approach S from below (for positive terms), each step halving the remaining gap, a visible picture of convergence.

Formulas

• Infinite GP sum: S = a/(1 - r), valid only when |r| < 1.
• Derivation: start from S_n = a times (1 - r^n)/(1 - r); as n grows without bound, r^n tends to 0, leaving a/(1 - r).
• Assumptions: |r| < 1 strictly; a can be any non-zero number.

Worked Example

Sum the infinite series 8 + 4 + 2 + 1 + … . Here a = 8 and r = 1/2, which satisfies |r| < 1, so S = 8/(1 - 1/2) = 8/(1/2) = 16. Watching the partial sums confirms the approach: 8, then 12, then 14, then 15, then 15.5, each step halving the distance to 16. A second use: convert 0.333… to a fraction. Write it as 3/10 + 3/100 + 3/1000 + …, a GP with a = 3/10 and r = 1/10. Then S = (3/10)/(1 - 1/10) = (3/10)/(9/10) = 3/9 = 1/3, as expected. The same idea converts any repeating decimal: 0.272727… is 27/100 + 27/10000 + …, with a = 27/100 and r = 1/100, giving (27/100)/(99/100) = 27/99 = 3/11.

Common Misconceptions

• 1 + 2 + 4 + 8 + … = 1/(1 - 2) = -1: a famous abuse. With r = 2 the condition |r| < 1 fails, the terms explode, and the formula is meaningless. An infinite sum of growing positive terms cannot be negative.
• a/(1 - r) works for finite sums: it is the limit after infinitely many terms. For exactly n terms, use S_n = a times (1 - r^n)/(1 - r); the infinite formula is only its limiting case.
• The terms must actually reach zero: they never do. The terms merely approach zero, and the infinite tail beyond any point still adds a positive amount, which is exactly why the total reaches 16 rather than stopping at 15.5.

Real-World Applications

• Zeno's paradox: covering half the remaining distance repeatedly sums to the whole: 1/2 + 1/4 + 1/8 + … = 1.
• Repeating decimals: every recurring decimal converts to a fraction through an infinite GP.
• Economics: the multiplier effect of government spending ripples as a geometric series in the fraction re-spent.
• Physics: total light reflected between two parallel mirrors, and similar infinite-bounce problems, sum geometrically.

When |r| < 1, an infinite geometric progression settles at S = a/(1 - r) because the r^n term in the finite formula vanishes in the limit. Always check the convergence condition first, keep the finite and infinite formulas distinct, and the same idea that resolves Zeno's paradox will convert repeating decimals and price every infinite shrinking process.