Concept explainerCalculus IIFoundational8 min read

Geometric series: convergence, sum, and structure

A constant ratio makes an infinite sum exactly solvable when repeated scaling shrinks toward zero.

\sum_{n=0}^{\infty}ar^n=\frac{a}{1-r}\quad(|r|<1)

Concept explainer

What the idea means, why its conditions matter, and where it connects.

Reviewed July 11, 2026

ResultStart here

\boxed{\sum_{n=0}^{\infty}ar^n=\frac{a}{1-r}\ \text{for }|r|<1}

If |r| is at least 1, the terms do not shrink in the required way and the geometric series diverges.

The partial-sum formula

For a finite geometric sum, multiplying by r and subtracting makes almost every term cancel. The remaining expression gives S_N = a(1 − r^(N+1))/(1 − r).

The infinite sum is the limit of these partial sums, not a separate arithmetic operation.

Why the ratio condition is exact

When |r| < 1, the remaining power r^(N+1) approaches zero. When |r| is 1 or larger, the terms fail to approach zero or grow in magnitude, so convergence is impossible.

Recognize shifted indexing

A series may begin at n = 1 or use a shifted exponent. Identify the first actual term a and the common ratio r from consecutive terms instead of forcing a memorized index pattern.

Worked exampleSum a repeating decimal

Write 0.272727… as a fraction.

0.272727\ldots=\frac{27}{100}+\frac{27}{100^2}+\cdots

Each repeating block moves two decimal places.

a=\frac{27}{100},\qquad r=\frac1{100}

Identify the first term and ratio.

\frac{a}{1-r}=\frac{27/100}{99/100}

Apply the geometric sum.

Result

\boxed{\frac3{11}}

Watch for

Common mistakes

Using a/(1−r) when |r| ≥ 1.
Confusing the coefficient in a formula with the first actual term.
Ignoring an index shift that changes a.

Keep

Three takeaways

A constant ratio is the defining signal.
Infinite sums are limits of partial sums.
Find a and r from the written terms.