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.

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}.

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 keeps the arithmetic positive: the first when $r < 1$ , the second when $r > 1$ .

Worked example 1 Sum the first 6 terms

Find the sum of the first $6$ terms of the geometric series $3 + 6 + 12 + 24 + \cdots$

Read off the two numbers that define it. The first term is $a_1 = 3$ , and the ratio is any term divided by the one before it, $\tfrac{6}{3} = 2$ , so $r = 2$ . You want $n = 6$ terms. Since $r > 1$ , use the positive form:

S_6 = \frac{a_1(r^{6} - 1)}{r - 1} = \frac{3(2^{6} - 1)}{2 - 1}.

Work out the power first, $2^{6} = 64$ , then finish the arithmetic:

S_6 = \frac{3(64 - 1)}{1} = 3 \times 63 = 189.

Adding the six terms directly, $3 + 6 + 12 + 24 + 48 + 96$ , also gives $189$ , 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.

Adding infinitely many terms

A geometric sequence never runs out of terms, so you can ask what happens if you try to add all of them, out to infinity. For most ratios the answer is dull: the running total grows past every bound and there is no finite number 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.

The finite formula already shows why. 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.

Once $r^{n}$ has shrunk to practically nothing, the numerator $1 - r^{n}$ is practically just $1$ , and the running total settles right next to a fixed value:

S = \frac{a_1(1 - 0)}{1 - r} = \frac{a_1}{1 - r}.

This is the sum of the infinite geometric series. 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 is positive and the running total climbs toward that value from below without passing 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.

A quick self-check reaches the same answer by pure algebra. 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 $|r| < 1$ .

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 runs off without bound. When $r > 1$ the terms grow, so the total grows 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 lurches further from zero at every step 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 $|r| < 1$ .

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.

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 pieces $8, 4, 2, 1, \ldots$ shrink fast enough that all of them together fall just short of $16$ and settle exactly there.

(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.

A negative ratio pulls the sum below the first term rather than above it, but the same formula still lands the answer.

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 $\tfrac{16}{33}$ .

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 do not shrink, the total grows without bound, 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 $|r| < 1$ has the finite sum $S = \frac{a_1}{1 - r}$ , because the tail $r^{n}$ shrinks toward zero as more terms are added and stops changing the total.
When $|r| \ge 1$ the terms do not shrink, so the infinite series 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.
The chessboard total is the finite series $1 + 2 + 4 + \cdots + 2^{63} = 2^{64} - 1$ , and idealized bouncing balls, Zeno’s race, and Archimedes’ parabola are the same formula at work.

Geometric Series

Learning goals

From a sequence to its sum

A formula for the finite sum

Adding infinitely many terms

Repeating decimals are geometric series

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)