Can the sum of an infinite series ever be negative?

Yes. If a_1 < 0 or if r is negative (causing alternating signs), the sum can be negative. For example, the series -2 + 1 - \frac{1}{2} + \frac{1}{4} - \cdots has a_1 = -2 and r = -\frac{1}{2}, giving S = \frac{-2}{1-(-1/2)} = \frac{-2}{3/2} = -\frac{4}{3}, which is negative.

What does it mean for a series to converge?

A series converges when the sum of infinitely many terms equals a finite number. For geometric series, convergence happens exactly when |r| < 1. If |r| \geq 1, the series diverges—the sum either grows without bound or oscillates without settling to a single value.

Is there a difference between an infinite geometric series and an infinite geometric sequence?

Yes. A geometric sequence is just a list of numbers (like 1, 2, 4, 8, 16, \ldots). A geometric series is the sum of those numbers (like 1 + 2 + 4 + 8 + 16 + \cdots). This formula finds the sum, so you use it for a series, not a sequence.

Why does the formula have (1-r) in the denominator instead of just r?

The formula comes from subtracting S = a_1 + a_1 r + a_1 r^2 + \cdots from rS = a_1 r + a_1 r^2 + a_1 r^3 + \cdots to get S - rS = a_1. This simplifies to S(1-r) = a_1, so S = \frac{a_1}{1-r}. The (1-r) appears naturally from this algebra, not by arbitrary choice.

Infinite Geometric Series

Q: Can the sum of an infinite series ever be negative?

Yes. If a_1 < 0 or if r is negative (causing alternating signs), the sum can be negative. For example, the series -2 + 1 - \frac{1}{2} + \frac{1}{4} - \cdots has a_1 = -2 and r = -\frac{1}{2}, giving S = \frac{-2}{1-(-1/2)} = \frac{-2}{3/2} = -\frac{4}{3}, which is negative.

Q: What does it mean for a series to converge?

A series converges when the sum of infinitely many terms equals a finite number. For geometric series, convergence happens exactly when |r| < 1. If |r| \geq 1, the series diverges—the sum either grows without bound or oscillates without settling to a single value.

Q: Is there a difference between an infinite geometric series and an infinite geometric sequence?

Yes. A geometric sequence is just a list of numbers (like 1, 2, 4, 8, 16, \ldots). A geometric series is the sum of those numbers (like 1 + 2 + 4 + 8 + 16 + \cdots). This formula finds the sum, so you use it for a series, not a sequence.

Q: Why does the formula have (1-r) in the denominator instead of just r?

The formula comes from subtracting S = a_1 + a_1 r + a_1 r^2 + \cdots from rS = a_1 r + a_1 r^2 + a_1 r^3 + \cdots to get S - rS = a_1. This simplifies to S(1-r) = a_1, so S = \frac{a_1}{1-r}. The (1-r) appears naturally from this algebra, not by arbitrary choice.

Infinite Geometric Series gives you the sum of all terms in a geometric sequence when the common ratio lies strictly between negative one and one.

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

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What each symbol means

What Infinite Geometric Series takes

Symbol	Meaning
$S$	The sum of all infinitely many terms in the series; it represents the finite value you're calculating.
$a_1$	The first term of the geometric sequence; changing it scales the entire sum proportionally.
$r$	The common ratio between consecutive terms; the formula requires $\|r\| < 1$ or the infinite series will diverge to infinity.

When to use it

When you need to sum infinitely many terms of a geometric sequence and the ratio between consecutive terms satisfies $|r| < 1$ .

Level

Usually taught in: Algebra II

Worked examples

1. Sum a series with ratio 1/2

Problem

Find the sum of the infinite geometric series

1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots

$a_1 = 1, \quad r = \frac{1}{2}$
The first term is 1, and each term is half the previous one, so $r = 1/2$ . Verify $|r| = 1/2 < 1$ , so the series converges.
$S = \frac{a_1}{1-r}$
Apply the infinite geometric series formula.
$S = \frac{1}{1 - \frac{1}{2}} = \frac{1}{\frac{1}{2}} = 2$
Substitute and simplify: the denominator becomes $1/2$ , and dividing by $1/2$ is the same as multiplying by 2.

Answer: $S = 2$

This formula works because $|r| < 1$ ensures that each successive term shrinks toward zero, so the infinite sum settles on a finite value.

2. Sum a series with negative ratio

Problem

Find the sum of the series

8 - 4 + 2 - 1 + \frac{1}{2} - \cdots

$a_1 = 8, \quad r = \frac{-4}{8} = -\frac{1}{2}$
The first term is 8. The second term is $-4$ , so the common ratio is $r = (-4) \div 8 = -1/2$ . Check: $|r| = |-1/2| = 1/2 < 1$ ✓
$S = \frac{a_1}{1-r} = \frac{8}{1-(-\frac{1}{2})}$
Apply the formula, being careful with the negative sign: $1 - r = 1 - (-1/2)$ .
$S = \frac{8}{1+\frac{1}{2}} = \frac{8}{\frac{3}{2}} = 8 \cdot \frac{2}{3} = \frac{16}{3}$
Simplify the denominator: $1 - (-1/2) = 1 + 1/2 = 3/2$ . Then divide by multiplying by the reciprocal.

Answer: $S = \frac{16}{3}$

The negative ratio causes the series to alternate in sign, but as long as $|r| < 1$ , the sum is still finite. The denominator $1 - r$ becomes larger because we are subtracting a negative number.

3. Total area of nested garden plots

Problem

You design a series of concentric rectangular garden plots. The innermost plot has area 100 square meters. Each larger plot adds a new rectangular border, with area half that of the previous contribution: the first border adds 50 square meters, the next adds 25 square meters, and this pattern continues infinitely. What is the total area of all plots combined?

$a_1 = 100, \quad r = \frac{1}{2}$
The innermost plot contributes 100 square meters. Each successive layer adds half the area of the previous contribution, so $r = 1/2$ . Since $|r| = 1/2 < 1$ , the infinite sum converges.
$S = \frac{a_1}{1-r} = \frac{100}{1-\frac{1}{2}}$
The total area is the sum of infinitely many layers, each decreasing in size. Use the infinite geometric series formula.
$S = \frac{100}{\frac{1}{2}} = 100 \cdot 2 = 200$
Simplify: the denominator is $1 - 1/2 = 1/2$ , and dividing by $1/2$ means multiplying by 2.

Answer: $S = 200 \text{ square meters}$

Even though we add infinitely many plots, the total area stays finite because each new plot contributes less and less. The formula gives us the answer directly: 200 square meters, which is exactly twice the innermost plot's area.

Common mistakes

Where Infinite Geometric Series usually goes wrong

The mistake	Why it is wrong	The fix
Using $S = \frac{a_1}{1-r}$ for a geometric series with $r = 1.5$ as if the formula applies.	The formula only works when $\|r\| < 1$ ; if $\|r\| \geq 1$ , the series diverges and has no finite sum.	Before using the formula, always verify that $\|r\| < 1$ . If $\|r\| \geq 1$ , the series diverges to infinity or oscillates without converging.
Finding the common ratio as $r = 1/4$ when the series is $12 + (-4) + \frac{4}{3} + \cdots$ .	It is easy to miscompute the ratio when dividing a negative by a positive: $(-4) \div 12 = -1/3$ , not $1/4$ , and forgetting the negative sign leads to the wrong answer.	Carefully compute $r = \frac{a_2}{a_1}$ with correct signs: $r = \frac{-4}{12} = -\frac{1}{3}$ , then verify $\|r\| = 1/3 < 1$ for convergence.
Using $S = \frac{a_1}{1-r}$ for a finite geometric series with exactly 5 terms.	The infinite series formula only applies when terms continue forever; if there are finitely many terms, you need a different formula that includes the number of terms.	For a finite series with $n$ terms, use $S = a_1 \cdot \frac{1-r^n}{1-r}$ . Use $S = \frac{a_1}{1-r}$ only when the series is truly infinite and $\|r\| < 1$ .

Tips and when to use something else

Always check that $|r| < 1$ before applying the formula; this ensures the series converges to a finite sum instead of diverging.
The first term $a_1$ can be negative, and $r$ can be negative—the formula still works as long as $|r| < 1$ . A negative $r$ causes alternating signs.
If the series is finite (stops after $n$ terms), do not use this formula; instead use the finite geometric series formula $S = a_1 \cdot \frac{1-r^n}{1-r}$ .
A repeating decimal like $0.\overline{3} = 0.333\ldots$ is actually an infinite geometric series: $\frac{3}{10} + \frac{3}{100} + \frac{3}{1000} + \cdots$ with $a_1 = 0.3$ and $r = 0.1$ .

Related formulas

Geometric SequenceThe formula a_n = a_1 r^{n-1} finds any term in a geometric sequence, where each term is a constant multiple of the previous one.Geometric SeriesSums the first n terms of a geometric sequence—those where each term is the previous one times a constant—using a single formula instead of adding.Sigma NotationSigma notation compresses long sums into a compact formula, letting you write and calculate the sum of many terms efficiently and accurately.Telescoping SeriesA telescoping series simplifies sums where consecutive terms partially cancel out, leaving only the first and last terms to compute.Ratio TestDetermine whether an infinite series converges by checking if the ratio of consecutive terms approaches a value less than one.

Frequently asked questions

Can the sum of an infinite series ever be negative?: Yes. If $a_1 < 0$ or if $r$ is negative (causing alternating signs), the sum can be negative. For example, the series $-2 + 1 - \frac{1}{2} + \frac{1}{4} - \cdots$ has $a_1 = -2$ and $r = -\frac{1}{2}$ , giving $S = \frac{-2}{1-(-1/2)} = \frac{-2}{3/2} = -\frac{4}{3}$ , which is negative.
What does it mean for a series to converge?: A series converges when the sum of infinitely many terms equals a finite number. For geometric series, convergence happens exactly when $|r| < 1$ . If $|r| \geq 1$ , the series diverges—the sum either grows without bound or oscillates without settling to a single value.
Is there a difference between an infinite geometric series and an infinite geometric sequence?: Yes. A geometric sequence is just a list of numbers (like $1, 2, 4, 8, 16, \ldots$ ). A geometric series is the sum of those numbers (like $1 + 2 + 4 + 8 + 16 + \cdots$ ). This formula finds the sum, so you use it for a series, not a sequence.
Why does the formula have $(1-r)$ in the denominator instead of just $r$ ?: The formula comes from subtracting $S = a_1 + a_1 r + a_1 r^2 + \cdots$ from $rS = a_1 r + a_1 r^2 + a_1 r^3 + \cdots$ to get $S - rS = a_1$ . This simplifies to $S(1-r) = a_1$ , so $S = \frac{a_1}{1-r}$ . The $(1-r)$ appears naturally from this algebra, not by arbitrary choice.

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Reviewed 2026-09-18