Geometric Series

Intuition

The previous lesson ended on a cliffhanger. A king promised the inventor of chess one grain of wheat on the first square, two on the second, four on the third. The doubling ran across all sixty-four squares, and we found each square’s count but never added them up. This lesson adds them, and the grand total turns out to be astonishingly large. Adding the terms of a geometric sequence gives a geometric series, and two surprises are waiting. A finite geometric series, even a long one, folds into a single short formula instead of a marathon of additions. And an infinite geometric series, when its ratio is small enough, can add infinitely many positive numbers and still land on a finite total. That idea finally answers Zeno’s ancient paradox of the runner who must cross half the track, then half of what is left, and half again, forever.

Learning goals

Sum a finite series with $S_n = \tfrac{a_1(1 - r^n)}{1 - r}$
Prove it by multiplying by $r$ and subtracting
Take $S = \tfrac{a_1}{1 - r}$ when $|r| < 1$
Say why $|r| \ge 1$ leaves no finite sum
Convert a repeating decimal into an exact fraction

From a sequence to its sum

An ordered list like $3, 6, 12, 24, 48$ is a geometric sequence. Join its terms with plus signs and you get a geometric series, a single number:

3 + 6 + 12 + 24 + 48 = 93.

As with an arithmetic series, write $S_n$ for the sum of the first $n$ terms of a geometric sequence. Using the explicit term $a_k = a_1 r^{k-1}$ from the last lesson, every term is the first term times a power of the ratio:

S_n = a_1 + a_2 + \cdots + a_n = a_1 + a_1 r + a_1 r^2 + \cdots + a_1 r^{n-1}.

In the summation notation from the arithmetic-series lesson, the same sum is

S_n = \sum_{k=1}^{n} a_1 r^{k-1}.

The only thing new compared with an arithmetic series is what sits between the terms: a constant multiplier $r$ instead of a constant added difference. That single change forces a different summing trick, because the reverse-and-add pairing no longer produces equal columns when the terms are scaled rather than stepped.

A formula for the finite sum

There is a clever move that collapses the whole sum. Multiply the series by the ratio $r$ and subtract the result from the original. Because multiplying by $r$ shifts every term over by one position, almost everything cancels.

See it first with numbers. Take the six-term sum $S = 3 + 6 + 12 + 24 + 48 + 96$ , built from the ratio- $2$ sequence. Multiply every term by $2$ and line the result up under the original so matching terms sit in the same column:

\begin{aligned} S \;&=\; 3 + 6 + 12 + 24 + 48 + 96 \\[2pt] 2S \;&=\; \phantom{3 + {}} 6 + 12 + 24 + 48 + 96 + 192 \end{aligned}

Every term from $6$ through $96$ sits in both rows, so subtracting the second row from the first cancels every one of them. Only the leading $3$ from the top row and the trailing $192$ from the bottom row are left over:

S - 2S = 3 - 192 = -189, \qquad\text{so}\qquad S = 189.

That matches adding the six terms directly. The general proof below is the same cancellation, written so it works for any first term, ratio, and number of terms.

The finite sum is $S_n = \frac{a_1(1 - r^{n})}{1 - r}$ for $r \ne 1$ #

Write the sum of the first $n$ terms in full, each term the one before it times $r$ :

S_n = a_1 + a_1 r + a_1 r^2 + \cdots + a_1 r^{n-1}.

Now multiply the whole line by the common ratio $r$ . Every term picks up one more factor of $r$ , so the entire list shifts one place to the right:

r S_n = a_1 r + a_1 r^2 + a_1 r^3 + \cdots + a_1 r^{n}.

Subtract the second line from the first and watch what lines up. The $a_1 r$ in the top line cancels the $a_1 r$ in the bottom line, the $a_1 r^2$ cancels the $a_1 r^2$ , and so on all the way down. Every middle term appears in both sums, so every one of them cancels. Only two survivors are left. One is the $a_1$ at the front of the top line, which has nothing beneath it. The other is the $-a_1 r^{n}$ at the end of the bottom line, which has nothing above it:

S_n - r S_n = a_1 - a_1 r^{n}.

Factor each side. The left side has the common factor $S_n$ , and the right side has the common factor $a_1$ :

S_n(1 - r) = a_1(1 - r^{n}).

As long as $r \ne 1$ the factor $1 - r$ is not zero, so divide both sides by it to free $S_n$ :

S_n = \frac{a_1(1 - r^{n})}{1 - r}.

Check your understanding

In the proof above, multiplying $S_n$ by $r$ and subtracting leaves only two surviving terms, $a_1$ and $-a_1 r^{n}$ . Why does every other term cancel?

A Because

r

is always less than

1

, so the higher powers of

r

shrink toward

0

B Because every term from

a_1 r

through

a_1 r^{n-1}

appears in both

S_n

and

r S_n

, so subtracting removes each of them C Because the two sums have a different number of terms, so only the extra ones are left over D Because

a_1

is the only term that actually gets multiplied by

r

The one ratio this formula cannot handle is $r = 1$ , because then $1 - r = 0$ and dividing by it is illegal. That case needs no formula at all. When $r = 1$ every term equals $a_1$ , so the series is just $a_1$ added to itself $n$ times:

S_n = \underbrace{a_1 + a_1 + \cdots + a_1}_{n \text{ terms}} = n\,a_1.

When the ratio is larger than $1$ it is tidier to avoid the negative quantities in the fraction. Multiplying the top and bottom by $-1$ flips both differences and gives an equal form built from positive parts:

S_n = \frac{a_1(r^{n} - 1)}{r - 1}.

The two versions are the same number, so reach for whichever form keeps the denominator positive: the first when $r < 1$ (so $1 - r > 0$ ), the second when $r > 1$ (so $r - 1 > 0$ ).

Worked example 1 Sum the first 4 terms

Find the sum of the first $4$ terms of the geometric series $81 + 27 + 9 + 3 + \cdots$

Read off the two numbers that define it. The first term is $a_1 = 81$ , and the ratio is any term divided by the one before it, $\tfrac{27}{81} = \tfrac13$ , so $r = \tfrac13$ . You want $n = 4$ terms. Since $r < 1$ , use the form with $1 - r$ in the denominator:

S_4 = \frac{a_1(1 - r^{4})}{1 - r} = \frac{81\left(1 - \left(\tfrac13\right)^{4}\right)}{1 - \tfrac13}.

Work out the power first, $\left(\tfrac13\right)^{4} = \tfrac{1}{81}$ , then finish the arithmetic:

S_4 = \frac{81\left(1 - \tfrac{1}{81}\right)}{\tfrac23} = \frac{81 \times \tfrac{80}{81}}{\tfrac23} = \frac{80}{\tfrac23} = 120.

Adding the four terms directly, $81 + 27 + 9 + 3$ , also gives $120$ , a good check that the formula does the same job in a single step.

Worked example 2 The chessboard grand total

Total the wheat on the chessboard from the previous lesson: one grain on the first square, two on the second, four on the third. Each square’s count doubles the one before it, across all $64$ squares.

The grain counts are the geometric sequence $1, 2, 4, 8, \ldots$ , with first term $a_1 = 1$ and ratio $r = 2$ , and the last square (the $64$ th) holds $2^{63}$ grains. The grand total is the finite geometric series

S_{64} = 1 + 2 + 4 + \cdots + 2^{63}.

With $a_1 = 1$ , $r = 2$ , and $n = 64$ , the $r > 1$ form keeps everything positive:

S_{64} = \frac{a_1(r^{n} - 1)}{r - 1} = \frac{1\,(2^{64} - 1)}{2 - 1} = 2^{64} - 1.

That is $18{,}446{,}744{,}073{,}709{,}551{,}615$ grains, more than eighteen quintillion, far more wheat than has ever been harvested in all of history. Notice how the formula crushes a sum of $64$ terms into a single subtraction. The whole board comes to exactly one less than $2^{64}$ , the value of the square that would have come next.

Check your understanding

A geometric series has $a_1 = 4$ , $r = 3$ , and $n = 4$ terms. Find $S_4$ .

160

224

324

108

Adding infinitely many terms

A geometric pattern can keep going forever, so it is natural to ask what happens if you try to add all of its terms, out to infinity. This is exactly the situation behind Zeno’s ancient puzzle: a runner must cross half the track, then half of what is left, then half of that, and so on forever, an endless list of shrinking distances. For most ratios, adding forever is dull: the running total never settles on one number, so there is no finite sum to report. But when the ratio is small in size, something remarkable happens. Infinitely many positive pieces can add up to a single finite value, exactly what lets Zeno’s runner finish the race after all.

Split a square in half, then halve the leftover again and again. The pieces have area 1/2, 1/4, 1/8, 1/16, and so on, and they tile the whole square without ever spilling out, so 1/2 + 1/4 + 1/8 + ... adds up to exactly 1.

This is Zeno’s runner too: each remaining stretch of track is half of what is left, and although the list of stretches never ends, they tile the whole track exactly, so the runner does finish the race after a finite distance.

The finite formula already shows why the same thing happens for any ratio with $|r| < 1$ , not only $r = \tfrac12$ . Look again at the sum of the first $n$ terms:

S_n = \frac{a_1(1 - r^{n})}{1 - r}.

The only part that changes as you pile on more terms is the power $r^{n}$ in the numerator. Suppose $r$ lies strictly between $-1$ and $1$ , that is $|r| < 1$ . Then every multiplication by $r$ lands you on a number smaller in size than before, because multiplying by something less than $1$ in absolute value always shrinks it. Keep multiplying and $r^{n}$ marches toward zero. For $r = \tfrac12$ the powers run $\tfrac12, \tfrac14, \tfrac18, \tfrac{1}{16}, \ldots$ , each half the last and all heading to nothing.

$r^{n}$ never actually reaches $0$ after any finite number of terms, but it can be made as close to $0$ as you like by adding enough terms. So the numerator $1 - r^{n}$ gets closer and closer to $1$ , and the partial sum $S_n$ gets closer and closer to a fixed value:

S = \frac{a_1}{1 - r}.

This is the sum of the infinite geometric series. When the partial sums approach one fixed number like this, the series is said to converge to that number; when they do not, it diverges. The tail you are dropping, all the terms past the $n$ th, is itself a tiny geometric series that keeps shrinking, so the more terms you include, the closer the total sits to $\frac{a_1}{1 - r}$ . When $r$ is positive, every term shares $a_1$ ‘s sign, and the running total moves steadily toward that value without ever crossing it. When $r$ is negative, the terms alternate in sign and the running total lands just above and just below the value in turn. Those overshoots shrink to nothing, so the running total still closes in on that value.

This shortcut only makes sense when the sum $S$ really is a finite number to begin with, which is exactly the case $|r| < 1$ .

Everything here hinges on that condition. If $|r| \ge 1$ the terms do not shrink. When $r = 1$ you are adding $a_1$ to itself forever and the total moves further from zero without bound (upward if $a_1$ is positive, downward if $a_1$ is negative). When $r > 1$ the terms grow in size, so the total moves further from zero even faster. When $r = -1$ the running sum flips between $a_1$ and $0$ and never settles anywhere. When $r < -1$ the terms grow in size while flipping sign, so the running total swings between larger and larger positive and negative values and never settles either. In every one of these cases there is no finite sum, and the formula $\tfrac{a_1}{1 - r}$ does not apply. The rule to memorize is short: an infinite geometric series has a finite sum exactly when $|r| < 1$ .

Check your understanding

The series with $a_1 = 6$ and $r = -2$ has no finite sum. Which explanation is correct?

a_1 = 6

is positive, so the terms can never shrink toward zero B The ratio is negative, so the series diverges no matter its size C

|r| = 2 \ge 1

, so the terms grow in size while flipping sign, and the running total swings between larger and larger positive and negative values without ever settling D The denominator

1 - r

equals zero when

r = -2

, so the sum is undefined for that reason

Worked example 3 Two infinite sums

Find the sum of each infinite geometric series.

(a) $8 + 4 + 2 + 1 + \tfrac12 + \cdots$

The first term is $a_1 = 8$ , and each term is half the one before it, so $r = \tfrac12$ . Because $|r| < 1$ , the series has a finite sum:

S = \frac{a_1}{1 - r} = \frac{8}{1 - \tfrac12} = \frac{8}{\tfrac12} = 16.

The partial sums $8, 12, 14, 15, 15.5, \ldots$ stay just under $16$ , and they get closer and closer to $16$ the more terms you add. The infinite sum is the number they are closing in on, so the answer is exactly $16$ .

(b) $12 - 6 + 3 - \tfrac32 + \cdots$

Here $a_1 = 12$ , and each term is $-\tfrac12$ times the one before it, so $r = -\tfrac12$ . The sign does not stop it from converging, since $|r| = \tfrac12 < 1$ ; it only makes the terms alternate. Carry the sign into the formula:

S = \frac{12}{1 - \left(-\tfrac12\right)} = \frac{12}{\tfrac32} = 8.

With $a_1$ positive here, the negative ratio pulls the sum below the first term. The same formula lands the answer regardless of the signs involved, though whether a negative ratio pulls the sum up or down in general also depends on the sign of $a_1$ .

Check your understanding

Find the sum of the infinite geometric series $\frac{1}{5} + \frac{1}{25} + \frac{1}{125} + \cdots$

\frac{1}{4}

\frac{1}{5}

\frac{5}{4}

\frac{1}{20}

Repeating decimals are geometric series

A repeating decimal such as $0.7777\ldots$ hides an infinite geometric series, which is why every repeating decimal is secretly an exact fraction. Splitting the decimal into its place values lines the series up, and the infinite-sum formula turns it back into a fraction.

Worked example 4 Repeating decimals to fractions

Write each repeating decimal as an exact fraction.

(a) $0.\overline{7} = 0.7777\ldots$

Break the decimal into place values. The digit $7$ sits in the tenths place, then the hundredths, then the thousandths, so

0.\overline{7} = \frac{7}{10} + \frac{7}{100} + \frac{7}{1000} + \cdots

This is an infinite geometric series with first term $a_1 = \tfrac{7}{10}$ and ratio $r = \tfrac{1}{10}$ , since each place is a tenth of the one before it. Because $|r| < 1$ , apply the formula:

0.\overline{7} = \frac{7/10}{1 - 1/10} = \frac{7/10}{9/10} = \frac{7}{9}.

(b) $0.\overline{48} = 0.484848\ldots$

When the repeating block is two digits long, each copy is a hundredth of the last. So for a two-digit block, the first term is the block over $100$ and the ratio is $\tfrac{1}{100}$ :

0.\overline{48} = \frac{48}{100} + \frac{48}{10000} + \cdots = \frac{48/100}{1 - 1/100} = \frac{48/100}{99/100} = \frac{48}{99} = \frac{16}{33}.

The pattern is worth keeping: a one-digit repeat divides by $9$ , and a two-digit repeat by $99$ . Each result then reduces to lowest terms, here from $\tfrac{48}{99}$ down to $\tfrac{16}{33}$ .

Check your understanding

Write $0.\overline{18}$ as an exact fraction.

\frac{2}{11}

\frac{9}{50}

\frac{1}{5}

\frac{11}{2}

Worked example 5 A bouncing ball

A ball is dropped from a height of $12$ m. On each bounce it rebounds to half of its previous height. If it keeps bouncing, what is the total vertical distance it travels?

The ball first falls the full $12$ m. After that, every bounce carries it up to some peak and then back down the same distance, so each bounce past the drop contributes twice its peak height. The peak heights form a geometric sequence starting at the first rebound, $12 \cdot \tfrac12 = 6$ m, and halving each time: $6, 3, \tfrac32, \ldots$ . The total up-and-down distance from all the bounces is an infinite geometric series with $a_1 = 6$ and $r = \tfrac12$ , doubled to count both directions:

2\left(6 + 3 + \tfrac32 + \cdots\right) = 2 \cdot \frac{6}{1 - \tfrac12} = 2 \cdot 12 = 24 \text{ m}.

Add the first drop, which happened before any bounce:

12 + 24 = 36 \text{ m}.

The ball bounces infinitely many times in this idealized model, yet the rebounds shrink fast enough that the whole journey adds up to a finite $36$ m.

Common mistakes

Using the arithmetic sum formula. A geometric sum is $\frac{a_1(1 - r^{n})}{1 - r}$ , with the ratio raised to a power. Do not reach for $\frac{n(a_1 + a_n)}{2}$ : that averages evenly spaced terms, and geometric terms are not evenly spaced.
Dividing by zero when $r = 1$ . The finite formula has $1 - r$ in the denominator, so it is undefined at $r = 1$ . If every term is the same, the sum is just $n\,a_1$ . Check for $r = 1$ before you divide.
Summing to infinity when $|r| \ge 1$ . The infinite-sum formula $\frac{a_1}{1 - r}$ works only when $|r| < 1$ . If the ratio is $1$ or larger in size, the terms stop shrinking toward zero, so the running total never settles on one number (its size grows past every bound, or, for a negative ratio, it swings back and forth forever), and there is no finite answer to give.
Losing the sign of a negative ratio. A ratio like $r = -\tfrac12$ still has $|r| < 1$ , so the series converges, but the sign belongs in the formula: $\frac{a_1}{1 - (-1/2)} = \frac{a_1}{3/2}$ , not $\frac{a_1}{1/2}$ . Substitute $r$ with its sign attached.
Reading the first term of a repeating decimal wrong. For $0.\overline{7}$ the first term is $\frac{7}{10}$ and the ratio is $\frac{1}{10}$ , not $7$ and $\frac{1}{10}$ . Line the series up as $\frac{7}{10} + \frac{7}{100} + \cdots$ before you read off $a_1$ and $r$ .
Confusing the term count with the ratio. In $\frac{a_1(1 - r^{n})}{1 - r}$ the exponent $n$ counts the terms while $r$ is the multiplier. Feeding the count in where $r$ goes, or the ratio in where $n$ goes, scrambles the answer.

Key takeaways

A geometric series is the sum of the terms of a geometric sequence, $S_n = a_1 + a_1 r + \cdots + a_1 r^{n-1} = \sum_{k=1}^{n} a_1 r^{k-1}$ .
The finite sum for $r \ne 1$ is $S_n = \frac{a_1(1 - r^{n})}{1 - r}$ , equivalently $\frac{a_1(r^{n} - 1)}{r - 1}$ , proved by multiplying the series by $r$ and subtracting so the middle terms cancel. When $r = 1$ the sum is simply $S_n = n\,a_1$ .
An infinite geometric series with $|r| < 1$ converges to the finite sum $S = \frac{a_1}{1 - r}$ , because $r^{n}$ shrinks toward zero as more terms are added, so the dropped tail stops changing the total.
When $|r| \ge 1$ the terms do not shrink toward zero, so the infinite series diverges: it has no finite sum.
A repeating decimal is an infinite geometric series in disguise: $0.\overline{7} = \frac{7/10}{1 - 1/10} = \frac{7}{9}$ , which is how any repeating decimal becomes an exact fraction.

Practice

Multiple Choice Questions (MCQ)

Progressively harder sets of questions. Each opens on its own page.

Level 1 Foundational 12 questions Start → Level 2 Intermediate 12 questions Start → Level 3 Challenge 12 questions Start →

Core practice

Practice problems at the level of the course, to be worked out on paper. Hints one at a time, then the answer or the full worked solution, with your progress kept in this browser.

More practice (optional)

Extra sets, as hard as the Challenge set. Each one opens on its own page.

More resources (optional)

Other explanations of this lesson, if you want a second take.

Geometric Series (worked examples) (Purplemath)
Geometric Sequences and Sums (Math is Fun)
Series and Their Notations (free textbook) (OpenStax)
Geometric Series (Wolfram MathWorld)

Go deeper (optional)

You can skip this and keep going. Read it if you want to know more.

A second way to reach the infinite-sum formula

Here is a second way to reach $S = \dfrac{a_1}{1 - r}$ , working with the infinite sum directly instead of through the finite partial sums.

See it first with numbers, using a series that converges: $a_1 = 3$ and $r = \tfrac16$ , so $S = 3 + \tfrac12 + \tfrac{1}{12} + \cdots$ . Multiply every term by $r = \tfrac16$ and line the result up under the original so matching terms sit in the same column:

\begin{aligned} S \;&=\; 3 + \tfrac12 + \tfrac{1}{12} + \cdots \\[2pt] \tfrac16 S \;&=\; \phantom{3 + {}} \tfrac12 + \tfrac{1}{12} + \cdots \end{aligned}

Every term from $\tfrac12$ onward sits in both rows, so subtracting the second row from the first cancels all of them, leaving only the leading $3$ :

S - \tfrac16 S = 3, \qquad\text{so}\qquad \tfrac56 S = 3, \qquad\text{so}\qquad S = \tfrac{18}{5} = 3.6.

That matches the infinite-sum formula: $S = \dfrac{a_1}{1 - r} = \dfrac{3}{1 - 1/6} = \dfrac{3}{5/6} = \dfrac{18}{5}$ .

The same move works for any $|r| < 1$ . Call the full infinite sum $S$ , then multiply it by $r$ , which shifts every term one place to the right:

S = a_1 + a_1 r + a_1 r^2 + \cdots, \qquad r S = a_1 r + a_1 r^2 + a_1 r^3 + \cdots

The second sum is exactly the first with its leading $a_1$ removed, so $r S$ is $S - a_1$ . Solving that relation gives the same formula:

r S = S - a_1 \;\Longrightarrow\; S(1 - r) = a_1 \;\Longrightarrow\; S = \frac{a_1}{1 - r}.

This shortcut only makes sense when the sum $S$ really is a finite number to begin with, which is exactly the case $|r| < 1$ .

A bit of history (optional)

Most rules of area the Greeks trusted were built for straight edges: rectangles, triangles, and shapes you can cut into those. So what is the area of a region with a curved side?

Archimedes settled that question for one curve, the parabola, working in Syracuse, a Greek city, in the third century BCE. His method was to fill the region with triangles. One large triangle takes most of it. The gaps left over take smaller triangles, the gaps after those take smaller ones still, and the process never ends.

The pattern hidden in those triangles is what rescues the calculation. Each new batch of triangles contributes exactly a quarter of the batch before it. So the running total is one, plus a quarter, plus a sixteenth, and on for ever. That is a geometric series with ratio $\tfrac14$ . Archimedes proved it comes to $\tfrac43$ of that first triangle. It is an exact value, not an estimate.

He would never have written the sum the way you just saw it. Greek mathematicians distrusted the idea of literally adding up an endless list, so he reached $\tfrac43$ instead by a longer route that stayed finite the whole way, comparing areas directly rather than summing an infinite series. The number he was chasing is nevertheless yours. Put $a_1 = 1$ and $r = \tfrac14$ into $\tfrac{a_1}{1 - r}$ and out comes $\tfrac43$ . He measured a region bounded by a curve, nineteen centuries before calculus existed.