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Series rules

Geometric Series Formula: Sum a/(1 − r), Convergence and Examples

The geometric series formula: a + ar + ar² + ... = a/(1 − r) when |r| < 1. Partial sums, the convergence condition and worked examples.

Series rules Updated 3 min read

At a glance

\sum_{n=0}^{\infty} a\,r^n = \frac{a}{1 - r}, \quad \lvert r\rvert < 1
First term
a, the term you get when the exponent is 0
Common ratio
r, the factor that turns each term into the next
Converges when
\lvert r\rvert < 1
Diverges when
\lvert r\rvert \ge 1 (and a \ne 0)
Partial sum
S_N = a\,\dfrac{1 - r^N}{1 - r} for the first N terms, r \ne 1
On this page
  1. The geometric series formula
  2. Why the formula works
  3. How to use it
  4. Worked examples
  5. Finite vs infinite sums
  6. Common mistakes
  7. Where it leads

A geometric series adds terms that each come from the one before by multiplying by the same number r. It is one of the few infinite series whose exact sum you can write down, which makes it the benchmark that every convergence test is compared against.

The geometric series formula

With first term a and common ratio r:

a + ar + ar^2 + ar^3 + \cdots = \sum_{n=0}^{\infty} a\,r^n = \frac{a}{1-r}, \qquad \lvert r\rvert < 1

If \lvert r\rvert \ge 1 and a \ne 0, the series diverges.

The sum of the first N terms is finite for every r \ne 1:

S_N = a + ar + \cdots + ar^{N-1} = a\,\frac{1 - r^N}{1 - r}

Why the formula works

Multiply S_N by r and subtract. Almost every term cancels:

S_N - rS_N = a - ar^N \quad\Longrightarrow\quad S_N = a\,\frac{1 - r^N}{1 - r}

When \lvert r\rvert < 1, r^N \to 0 as N \to \infty, so S_N \to \frac{a}{1-r}. When \lvert r\rvert \ge 1, r^N does not go to zero and the partial sums have no finite limit.

Partial sums of 1 + 1/2 + 1/4 + ... approaching 2
Each partial sum S_N = 2 - 2^{1-N} closes half of the remaining gap to the limit \frac{1}{1 - 1/2} = 2.

How to use it

  1. Find a: the first term of the series as written. Plug in the starting index.
  2. Find r: divide any term by the previous one.
  3. Check \lvert r\rvert < 1. If not, stop: the series diverges.
  4. Compute \frac{a}{1-r}.

Worked examples

Example 1: a classic. 1 + \frac{1}{2} + \frac{1}{4} + \cdots has a = 1, r = \frac{1}{2}:

\sum_{n=0}^{\infty} \left(\frac{1}{2}\right)^n = \frac{1}{1 - \frac{1}{2}} = 2

Example 2: the index does not start at 0. In \sum_{n=1}^{\infty} 3\left(\frac{2}{3}\right)^n the first term is 3\cdot\frac{2}{3} = 2, not 3:

\sum_{n=1}^{\infty} 3\left(\frac{2}{3}\right)^n = \frac{2}{1 - \frac{2}{3}} = 6

Example 3: a negative ratio. \sum_{n=0}^{\infty} \left(-\frac{1}{3}\right)^n has a = 1, r = -\frac{1}{3}:

\frac{1}{1 - \left(-\frac{1}{3}\right)} = \frac{3}{4}

Example 4: a repeating decimal. 0.272727\ldots = \frac{27}{100} + \frac{27}{10000} + \cdots has a = \frac{27}{100} and r = \frac{1}{100}:

\frac{27/100}{99/100} = \frac{27}{99} = \frac{3}{11}

Example 5: divergence. \sum_{n=0}^{\infty} \left(\frac{5}{4}\right)^n has r = \frac{5}{4} > 1, so it diverges. The terms grow instead of shrinking.

Finite vs infinite sums

The finite formula works for every ratio except r = 1, so even a series with r = 3 has a perfectly good sum of the first ten terms: \sum_{n=0}^{9} 3^n = \frac{1 - 3^{10}}{1 - 3} = 29524. Only the infinite sum needs \lvert r\rvert < 1. When r = 1 every term equals a, so the first N terms add to Na, and the infinite series diverges.

Common mistakes

  1. Using the coefficient instead of the first term. When the sum starts at n = 1 or later, compute the first term directly.
  2. Ignoring the condition. \frac{a}{1-r} with r = 2 gives -a, a meaningless number. Always check \lvert r\rvert < 1 first.
  3. Mistaking other series for geometric. \sum \frac{1}{n^2} and \sum \frac{1}{2n} have no constant ratio, so the formula does not apply.
  4. Dropping a sign in r. For an alternating geometric series, r is negative, and 1 - r becomes 1 + \lvert r\rvert.

Where it leads

Geometric series are the model for the ratio test, which asks whether a series eventually behaves like a geometric one with \lvert r\rvert < 1. They also give the first power series, \frac{1}{1-x} = \sum x^n for \lvert x\rvert < 1, the starting point for power series. For more examples, read the full guide to geometric series.

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Frequently asked questions

What is the formula for the sum of an infinite geometric series?

If the first term is a and the common ratio r satisfies \lvert r\rvert < 1, the sum is \frac{a}{1-r}. If \lvert r\rvert \ge 1 the series diverges.

How do you find r in a geometric series?

Divide any term by the term before it. For 6 + 2 + \frac{2}{3} + \cdots, the ratio is \frac{2}{6} = \frac{1}{3}.

Why does a geometric series diverge when |r| is at least 1?

If \lvert r\rvert \ge 1 the terms ar^n do not shrink to zero, so by the nth-term test the series cannot converge.

How do you write 0.999... as a geometric series?

0.999\ldots = \frac{9}{10} + \frac{9}{100} + \cdots, a geometric series with a = \frac{9}{10} and r = \frac{1}{10}. Its sum is \frac{9/10}{9/10} = 1.

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