Sum of an Infinite GP - Formula & Examples

TL;DR

An infinite geometric progression has a finite sum, S∞=\frac{a}{1 - r}, but only when the common ratio satisfies |r| < 1. Here a is the first term and r is the common ratio; if |r| ≥ 1 the series diverges and has no finite sum. This article explains why shrinking terms add up to a limit, derives the formula, and works examples, including how a repeating decimal like 0.999… turns out to equal exactly 1.

What Is The Sum Of An Infinite GP?

The sum of an infinite GP is the finite value that the running total of an unending geometric progression approaches, provided the terms shrink fast enough. For the progression a, ar, ar², ar³,… continuing forever, the sum is

S∞=\frac{a}{1 - r}, \quad |r| < 1

The condition |r| < 1 is not decoration. It is the entire difference between a series that settles on a number and one that runs off to infinity. A geometric progression is a sequence where each term is the previous one times a fixed ratio r; you can review the basics in the geometric sequence explainer. When |r| < 1, each term is a fraction of the last, so the terms melt toward zero.

Why Does The Series Converge Only When |r| < 1?

This is the question that actually matters, and it has a clean answer. Start from the finite sum of a GP:

Sn=\frac{a(1 - r^n)}{1 - r}

Now ask what happens to rⁿ as n grows without bound.

• If |r| < 1, raising a fraction to higher and higher powers drives it toward zero. For example (\frac{1}{2})^{10} = \frac{1}{1024}, and it only gets tinier. So rⁿ → 0.

• If |r| ≥ 1, the terms do not shrink. They stay the same size (r=1) or grow (|r|>1), so the running total never settles.

Substitute rⁿ → 0 into the finite formula:

S∞=\frac{a(1 - 0)}{1 - r} = \frac{a}{1 - r}

That is the whole derivation. The infinite formula is the finite formula with its one shrinking piece sent to zero. The condition |r| < 1 is just the requirement that the piece can shrink to zero.

Variable Glossary:

What Happens When |r| ≥ 1?

The series diverges - it has no finite sum. If r=2, the progression 3, 6, 12, 24,… keeps growing and the total races to infinity; if r=1, every term equals a, and adding a forever also gives infinity (unless a=0); if r=−1, the partial sums bounce between two values and never settle. In all of these, the formula \frac{a}{1 - r} gives a nonsense answer or no answer, so you must not apply it. Checking |r| < 1 before reaching for the formula is the single non-negotiable step.

Examples Of The Sum Of An Infinite GP

Example 1

Find the sum of 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots.

First term a=1, ratio r=\frac{1}{2}. Since |r| < 1, a sum exists:

S∞=\frac{1}{1 - \frac{1}{2}} = \frac{1}{\frac{1}{2}} = 2.

Final answer: S∞ = 2.

Example 2

Find the sum of \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \cdots.

Here a=\frac{1}{3} and r=\frac{1}{3}:

S∞=\frac{\frac{1}{3}}{1 - \frac{1}{3}} = \frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2}.

Final answer: S∞ = \frac{1}{2}.

Example 3

Find the sum of 3 + 6 + 12 + 24 + ⋯.

This series adds only positive, growing numbers, so a total of −3 is impossible - a sum smaller than the first term cannot be right. Here r=2, and |r|=2≥1, so the series diverges and there is no finite sum. Final answer: the series diverges; no sum exists.

Example 4

A bouncing ball. A ball is dropped from 10 metres. Each bounce reaches \frac{3}{5} of the previous height. The total vertical distance travelled before it comes to rest is:

The total distance is the first drop plus twice the sum of the bounce heights:

total distance = 10 + 2 × \frac{6}{1 - \frac{3}{5}} = 10 + 2 × 15 = 40 m.

Final answer: 40 metres.

Example 5

Show that 0.999…=1.

0.999… can be written as a series:

0.999… = \frac{9}{10} + \frac{9}{100} + \frac{9}{1000} + \cdots.

This is a GP with a=\frac{9}{10}, r=\frac{1}{10}:

S∞=\frac{\frac{9}{10}}{1 - \frac{1}{10}} = 1.

Final answer: 0.999… = 1 exactly.

Key Takeaways

• The sum of an infinite GP is S∞=\frac{a}{1-r}, valid only when |r| < 1.
• If |r| ≥ 1, the series diverges - check this first.