Geometric Series Formula: Sum, Convergence and Examples

A geometric series adds terms that are each a fixed multiple \( r \) (the common ratio) of the one before:

$$a + ar + ar^2 + ar^3 + \cdots = \sum_{n=0}^{\infty}ar^n$$

The geometric series formula gives the sum in one step. This guide covers both the infinite and finite versions, two derivations, how to read off \( a \) and \( r \) from any series, worked examples from easy to exam-level, and the mistakes that trip students up.

The formulas

Infinite sum (only when \( |r| < 1 \)):

$$\sum_{n=0}^{\infty}ar^n = \frac{a}{1 - r}$$

Finite sum of the first \( n \) terms (any \( r \neq 1 \)):

$$S_n = a\,\frac{1 - r^n}{1 - r}$$

If \( |r| \ge 1 \), the infinite series diverges because the terms don’t shrink to 0.

In words: the infinite sum is “first term over one minus the ratio.” Memorize it in that form, because it works no matter where the index starts.

Why it works: the picture

Take a strip of length 2. Cover half of it with a piece of length 1. Cover half of what’s left with a piece of length \( \frac12 \), then half of the remainder with \( \frac14 \), and so on. Each piece is half the previous one, and the uncovered gap halves every time. The pieces never overflow the strip, and the gap shrinks toward nothing, so the total is exactly 2.

That is \( 1 + \frac12 + \frac14 + \cdots = 2 \). The general formula says the same thing for any ratio: when each term is a fixed fraction of the last, the leftover shrinks geometrically and the sum settles at a finite value.

Derivation

Write \( S_n = a + ar + \dots + ar^{n-1} \) and multiply by \( r \): \( rS_n = ar + \dots + ar^n \). Subtract: most terms cancel, leaving \( S_n - rS_n = a - ar^n \), so \( S_n = a\frac{1 - r^n}{1 - r} \). If \( |r| < 1 \), \( r^n \to 0 \) and \( S_n \to \frac{a}{1 - r} \).

The last step is a limit at infinity: powers of a number between \( -1 \) and 1 shrink to 0. When \( |r| > 1 \) the powers blow up instead, and the partial sums have no limit.

A quicker argument. If the infinite sum \( S \) exists, notice that after the first term, the rest of the series is just \( r \) times the whole series:

$$S = a + r(a + ar + ar^2 + \cdots) = a + rS$$

Solving gives \( S(1 - r) = a \), so \( S = \frac{a}{1 - r} \). This version is elegant but assumes the sum exists, which is why the \( |r| < 1 \) condition must be checked separately.

How to find a and r

1. Write out the first two or three terms explicitly.
2. The first term you wrote is \( a \), whatever the index starts at.
3. Divide the second term by the first to get \( r \). Check with the third term divided by the second.
4. Confirm \( |r| < 1 \) before using the infinite formula.

Examples

1. \( 1 + \frac12 + \frac14 + \frac18 + \cdots = \frac{1}{1 - 1/2} = 2 \). (Zeno’s paradox resolved: infinitely many steps, finite distance.)

2. \( 3 - 1 + \frac13 - \frac19 + \cdots \): \( a = 3 \), \( r = -\frac13 \), sum \( = \frac{3}{4/3} = \frac94 \). A negative ratio makes the terms alternate, and the partial sums zigzag toward \( \frac94 \).

3. \( \sum_{n=1}^{\infty}\left(\frac12\right)^n \): the first term is \( a = \frac12 \) (not 1!), so the sum is \( \frac{1/2}{1/2} = 1 \).

4. Finite: \( 1 + 2 + 4 + \dots + 2^9 = \frac{1 - 2^{10}}{1 - 2} = 1023 \). There are 10 terms (powers 0 through 9), so \( n = 10 \).

5. Index starting at 2. \( \sum_{n=2}^{\infty}3\left(-\frac12\right)^n \). The first term is \( 3\cdot\frac14 = \frac34 \) and \( r = -\frac12 \), so

$$\text{sum} = \frac{3/4}{1 + 1/2} = \frac12$$

Repeating decimals as fractions

\( 0.272727\ldots = 0.27 + 0.0027 + \cdots \), with \( a = 0.27 \) and \( r = 0.01 \):

$$\frac{0.27}{0.99} = \frac{27}{99} = \frac{3}{11}$$

Any repeating block of \( k \) digits gives \( r = 10^{-k} \), which is why every repeating decimal is a fraction.

Word problem: the bouncing ball

A ball dropped from 10 m rebounds to 60% of its height each time. Total distance traveled: 10 m down, then each bounce goes up and down.

$$10 + 2\left(6 + 3.6 + 2.16 + \cdots\right) = 10 + 2\cdot\frac{6}{1 - 0.6} = 10 + 30 = 40\ \text{m}$$

The factor 2 counts each rebound twice, once up and once down. The initial drop is counted once, which is why it sits outside the parentheses.

Example 6 (exam-level): a series with x in it

For which \( x \) does \( \sum_{n=0}^{\infty}\left(\frac{x}{3}\right)^n \) converge, and what is its sum?

The ratio is \( r = \frac x3 \), so the series converges exactly when \( \left|\frac x3\right| < 1 \), that is \( -3 < x < 3 \). There the sum is

$$\frac{1}{1 - x/3} = \frac{3}{3 - x}$$

At \( x = 1 \), for example, the sum is \( \frac32 \). This is a power series, and the interval \( (-3, 3) \) is its interval of convergence.

Example 7 (exam-level): differentiating a geometric series

Starting from \( \sum_{n=0}^{\infty}x^n = \frac{1}{1-x} \) for \( |x| < 1 \), differentiate both sides and multiply by \( x \):

$$\sum_{n=1}^{\infty}nx^n = \frac{x}{(1-x)^2}$$

At \( x = \frac12 \), this gives \( \sum\frac{n}{2^n} = \frac{1/2}{1/4} = 2 \), a sum that is not geometric at all.

Common mistakes

• Wrong first term. When the sum starts at \( n = 1 \) or later, \( a \) is not the coefficient in front. Write out the first term, as in Examples 3 and 5.
• Using the formula when \( |r| \ge 1 \). Plugging \( r = 2 \) into \( \frac{1}{1-r} \) gives \( -1 \) for \( 1 + 2 + 4 + \cdots \), which is nonsense. That series diverges.
• Miscounting \( n \) in the finite formula. \( n \) is the number of terms, not the last exponent. Powers 0 through 9 are 10 terms.
• Losing the sign of \( r \). In \( 3 - 1 + \frac13 - \cdots \), the ratio is \( -\frac13 \), so the denominator is \( 1 + \frac13 \).
• Missing a hidden geometric series. \( \frac{5}{4^n} \) is \( 5\left(\frac14\right)^n \). Rewrite powers in the denominator as a ratio.

Where geometric series show up

Anything that repeatedly scales by the same factor produces a geometric series. A few places you’ll meet one:

• Finance. A payment of 100 dollars per year forever, discounted at 5% per year, is worth \( \sum_{n=1}^{\infty}\frac{100}{1.05^n} \) today. Here \( a = \frac{100}{1.05} \) and \( r = \frac{1}{1.05} \), and the formula simplifies to \( \frac{100}{0.05} = 2000 \) dollars.
• Probability. The chance that the first head in repeated fair coin flips comes on flip \( n \) is \( \left(\frac12\right)^n \). These probabilities add up to 1, exactly as in Example 3.
• Growth and decay. Sampling exponential growth or decay at equal time steps gives a geometric sequence, and adding up those amounts, such as drug doses that partly wear off between doses, gives a geometric series.
• Geometry. Self-similar shapes, like squares nested inside squares, have total areas and perimeters that are geometric sums.

Connection to calculus

• \( \frac{1}{1 - x} = 1 + x + x^2 + \cdots \) is the simplest Taylor series, valid for \( |x| < 1 \).
• The ratio test works by comparing a series with a geometric one.
• Integrals of decaying exponentials over \( [0, \infty) \) are the continuous cousins of geometric series; see improper integrals.

Check a series numerically

The gadget computes partial sums and tells you whether the series converges. The series convergence calculator page has the same tool at full width.

Practice problems

Try these before checking the answers.

1. \( \sum_{n=0}^{\infty}\left(\tfrac23\right)^n \)
2. \( \sum_{n=1}^{\infty}\frac{5}{4^n} \)
3. \( 0.\overline{81} \) as a fraction
4. \( 1 - \frac14 + \frac1{16} - \frac1{64} + \cdots \)
5. The sum of the first 8 terms of \( 3 + 6 + 12 + \cdots \)
6. The values of \( x \) for which \( \sum(2x)^n \) converges

Answers: (1) \( 3 \); (2) \( \frac53 \); (3) \( \frac{9}{11} \); (4) \( \frac45 \); (5) \( 765 \); (6) \( -\frac12 < x < \frac12 \).

FAQ

How do I find r?

Divide any term by the one before it: \( r = \frac{a_{n+1}}{a_n} \).

What if r = 1?

Every term is \( a \), so the partial sums are \( na \), which diverge (unless \( a = 0 \)).

What’s the difference between a geometric sequence and a geometric series?

A sequence is the list of terms \( a, ar, ar^2, \dots \). A series is their sum. The sequence can converge to 0 while the series converges to a nonzero total.

Does 1 − 1 + 1 − 1 + ⋯ have a sum?

No. Here \( r = -1 \), and the partial sums bounce between 1 and 0 forever, so the series diverges.

Can a geometric series with a negative ratio converge?

Yes, as long as \( -1 < r < 0 \). The terms alternate in sign, and the formula \( \frac{a}{1-r} \) still applies.

Further reading

• Special Series (Paul’s Online Math Notes) — geometric, telescoping and harmonic series with worked examples.
• Infinite Series (OpenStax Calculus Volume 2) — partial sums and the geometric series in a textbook setting.
• Geometric series (Wikipedia) — history from Zeno and Archimedes, plus applications in finance and beyond.