Infinite Geometric Series
Simple Explanation
When a geometric series continues forever, and its common ratio satisfies |r| < 1 (so each term shrinks toward zero), the running total approaches a fixed, finite limit: S∞ = a₁/(1−r), even though infinitely many terms are being added.
Why Do We Need It?
This is a genuinely surprising result — adding infinitely many positive numbers can still produce a finite total, as long as the terms shrink fast enough — with real applications in repeating decimals and physics.
Formula
Sum of an Infinite Geometric Series
S∞ = a₁ / (1 − r) (only valid when |r| < 1)
Gives the finite total that a geometric series with shrinking terms approaches as more and more terms are added, forever.
- S∞
- — the sum of the infinite series
- a₁
- — the first term
- r
- — the common ratio, with |r| < 1 required for the sum to exist
When to use it: Whenever a geometric series continues forever and its common ratio satisfies |r| < 1, so its terms shrink toward zero.
Worked Example
Find the sum of an infinite geometric series
Find the sum of 8 + 4 + 2 + 1 + ... continuing forever.
Why Does This Work?
From the finite sum formula, Sₙ = a₁(1−rⁿ)/(1−r) — as n grows larger and larger, if |r| < 1 then rⁿ shrinks closer and closer to 0 (since repeatedly multiplying a number smaller than 1 by itself makes it smaller each time), so (1−rⁿ) approaches 1, and the whole expression approaches a₁(1)/(1−r) = a₁/(1−r) — the finite limiting value the partial sums are heading toward.
Real-Life Example
Converting a repeating decimal to a fraction
0.333... can be written as the infinite series 3/10 + 3/100 + 3/1000 + ..., a geometric series with a₁=3/10, r=1/10.
S∞ = (3/10)/(1−1/10) = (3/10)/(9/10) = 1/3 — confirming that 0.333... exactly equals 1/3, using the infinite geometric series formula.
Practice
Find the sum of 12 + 6 + 3 + 1.5 + ... continuing forever.
HardWhy does 5 + 10 + 20 + 40 + ... NOT have a finite infinite sum?
HardCommon mistake
Applying the infinite sum formula when |r| ≥ 1 — the formula is only valid (and the sum only exists) when |r| < 1, since otherwise the terms don't shrink and the total grows without bound.
Quick Review
- S∞ = a₁/(1−r), only valid when |r| < 1.
- As n→∞, rⁿ→0 when |r|<1, which is why the partial sums approach a fixed finite value.
- If |r| ≥ 1, no finite sum exists — the series diverges.