Infinite geometric series
Add infinitely many terms of a convergent geometric sequence.
Infinite geometric series is one of 1 series formula in the precalculus section of this library, and it is used at high school · ap level.
Why infinite geometric series works
The sum of the first n terms is a(1 − rⁿ)/(1 − r). When |r| < 1, raising r to higher powers drives it toward 0, so the rⁿ piece simply fades away and the partial sums close in on a/(1 − r). Adding infinitely many terms can give a finite answer because each term is a fixed fraction of the one before it.
What each symbol means
$a$ is the first term and $r$ is the common ratio.
Infinite geometric series: when it holds
Converges only when $|r|<1$.
When it stops applying
It fails for |r| ≥ 1, where the terms never fade. At r = 1 the sum grows without bound, and at r = −1 the partial sums flip between a and 0 forever without settling anywhere, so no single value can be assigned.
Infinite geometric series: a worked example
$1+\frac12+\frac14+\cdots=1/(1-1/2)=2$.
The mistake to avoid
What people do: Students apply the formula without checking that the ratio is smaller than 1 in size.
Why it goes wrong: With r = 2 and a = 1 the formula returns 1/(1 − 2) = −1 for the series 1 + 2 + 4 + 8 + ..., a negative answer for a sum of positive terms, which is nonsense.
Do this instead: Test |r| < 1 before anything else. If the terms are not shrinking, the series has no sum and the correct answer is that it diverges.
Infinite geometric series: step by step
- Name the unknown, and the unit the answer has to come out in.
- Match the symbols to your values. $a$ is the first term and $r$ is the common ratio.
- Check the conditions before substituting. Converges only when $|r|<1$.
- Substitute, keep exact values to the last line, then test the sign, size, and unit against a rough estimate — the check that catches most precalculus slips.
Where this formula fits
- Subject
- Precalculus formulas — 10 entries in this library
- Topic
- Series
- Level
- High school · AP
Formulas are easiest to keep when they sit inside a method rather than on a list. Use the links below to see where infinite geometric series comes from, to check a calculation against a tool, and to practise it until you can recall it without looking.
- Sequences and Series — the lesson behind this formula: recognize arithmetic, geometric, and recursive patterns.
- Precalculus Calculator — check your substitution and the value it produces.
- Study precalculus — the subject guide that explains the ideas these formulas compress.
- Precalculus Practice — questions that make you retrieve the formula instead of recognising it.
- All 10 precalculus formulas — the full grouped reference, or the complete formula library.
Questions about infinite geometric series
What if r is negative?
The formula still works as long as |r| < 1. With a = 1 and r = −1/2 the series 1 − 1/2 + 1/4 − ... adds to 1/(1.5) = 2/3, and the partial sums close in from both sides.
Does 0.999... really equal 1?
Yes, and this formula proves it. That decimal is the series with a = 0.9 and r = 0.1, so the sum is 0.9/0.9, which is exactly 1 with nothing left over.
What if my series starts at n = 1 instead of n = 0?
Then the first term is ar rather than a, so the sum is ar/(1 − r). The simplest fix is to identify the actual first term you are adding and use that as the numerator.
How do I turn a repeating decimal into a fraction?
Treat the repeating block as the first term and use 10 to the negative block length as the ratio. For 0.272727... take a = 0.27 and r = 0.01, giving 0.27/0.99, which is 3/11.