AC Geometric Sums

Section 8.3 Geometric Sums

Motivating Questions

What is a finite geometric sum and how can we quickly find its value, no matter how many terms are in the sum?
How can a finite geometric sum be extended to an infinite geometric series? In what circumstances can we quickly find the value of an infinite geometric series?
How are finite and infinite geometric series connected to Taylor polynomials?

In our work in Section 8.2, we learned how to find a degree

n

polynomial approximation centered at a value

a

for a given function

f

with at least

n

derivatives. By working with several different functions and

n

-values, we’ve seen that increasing the degree of the polynomial improves the approximation, and also often helps us to see a pattern in the coefficients of the Taylor polynomials.

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For example, the degree

5

Taylor approximation of

f (x) = \ln (1 + x)

a = 0

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T_{5} (x) = x - \frac{1}{2} x^{2} + \frac{1}{3} x^{3} - \frac{1}{4} x^{4} + \frac{1}{5} x^{5},

and we see a pattern in the coefficients that allows us to easily generate

T_{6} (x),

T_{10} (x),

or indeed

T_{n} (x)

for any

n .

Note that if we want to use

T_{10} (x)

to estimate

\ln (\frac{3}{2}) = \ln (1 + \frac{1}{2}),

we need to compute the sum of

10

terms given by

🔗

\ln (\frac{3}{2}) \approx T_{10} (\frac{1}{2}) = (\frac{1}{2}) - \frac{1}{2} {(\frac{1}{2})}^{2} + \frac{1}{3} {(\frac{1}{2})}^{3} - \frac{1}{4} {(\frac{1}{2})}^{4} + \dots - \frac{1}{10} {(\frac{1}{2})}^{10} .

This computation suggests at least two questions: is there an easy way (without a computer) to determine the exact value of the

10

-term sum

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(\frac{1}{2}) - \frac{1}{2} {(\frac{1}{2})}^{2} + \frac{1}{3} {(\frac{1}{2})}^{3} - \frac{1}{4} {(\frac{1}{2})}^{4} + \dots - \frac{1}{10} {(\frac{1}{2})}^{10},

and is there a way we can make sense of continuing the sum indefinitely,

🔗

(\frac{1}{2}) - \frac{1}{2} {(\frac{1}{2})}^{2} + \frac{1}{3} {(\frac{1}{2})}^{3} - \frac{1}{4} {(\frac{1}{2})}^{4} + \dots - \frac{1}{10} {(\frac{1}{2})}^{10} + \dots ?

🔗

In this section, we investigate a special collection of similar sums that are called geometric.

🔗

Preview Activity 8.3.1.

Let’s explore sums of the form

🔗

S_{n} = 1 + \frac{1}{2} + {(\frac{1}{2})}^{2} + {(\frac{1}{2})}^{3} + \dots + {(\frac{1}{2})}^{n - 1}

🔗

(a)

Note that

S_{1} = 1,

S_{2} = S_{1} + \frac{1}{2} = 1 + \frac{1}{2} = \frac{3}{2},

and

S_{3} = S_{2} + \frac{1}{4} = \frac{3}{2} + \frac{1}{4} = \frac{7}{4} .

Using the fact that each subsequent value of

S_{n}

can be computed by adding one additional term to the preceding sum, complete the table below with the exact (fractional) value of each sum.

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$S_{1}$	$S_{2}$	$S_{3}$	$S_{4}$	$S_{5}$	$S_{6}$	$S_{7}$
$1$	$\frac{3}{2}$	$\frac{7}{4}$

(b)

What do you expect will be the values of

S_{8}

and

S_{9} ?

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(c)

Conjecture a simple formula for

S_{n}

that is given by a single fraction where the numerator and/or denominator depend on

n .

🔗

Hint: Think of the numerator and denominator separately, and think of powers of

2 .

🔗

(d)

What is happening to the value of

S_{n}

n \to \infty ?

🔗

S_{n} \to

🔗

Hint: Take the answer to part (c) and divide the numerator by the denominator. What do you get?

🔗

(e)

Now consider the sum

🔗

S_{n} = 1 + \frac{1}{4} + {(\frac{1}{4})}^{2} + {(\frac{1}{4})}^{3} + \dots + {(\frac{1}{4})}^{n - 1} .

🔗

Do you think this sum is larger or smaller than the sum we first considered?

🔗

The sum with powers of $\frac{1}{2}$ is larger.
The sum with powers of $\frac{1}{4}$ is larger.
The two sums are equal.

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What about

🔗

S_{n} = 1 - \frac{1}{2} + {(\frac{1}{2})}^{2} - {(\frac{1}{2})}^{3} + \dots + {(- \frac{1}{2})}^{n - 1} ?

🔗

Do you think this sum is larger or smaller than the sum we first considered?

🔗

The sum with powers of $\frac{1}{2}$ is larger.
The sum with powers of $- \frac{1}{2}$ is larger.
The two sums are equal.

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Subsection 8.3.1 Finite Geometric Series

Sums such as

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1 + \frac{1}{2} + {(\frac{1}{2})}^{2} + {(\frac{1}{2})}^{3} + \dots + {(\frac{1}{2})}^{n - 1},

1 + \frac{1}{4} + {(\frac{1}{4})}^{2} + {(\frac{1}{4})}^{3} + \dots + {(\frac{1}{4})}^{n - 1},

and

🔗

2 - \frac{4}{3} + \frac{8}{9} - \frac{16}{27} + \dots + 2 \cdot {(\frac{- 2}{3})}^{n - 1}

all share a similar structure: the next term in each sum is found by multiplying the last term by the same number. In the first sum, each subsequent term is found by multiplying by

\frac{1}{2};

in the second sum, by multiplying by

\frac{1}{4};

in the third sum, the multiplier is

- \frac{2}{3} .

These sums each have the form

🔗

\begin{matrix} (8.3.1) & a + a r + a r^{2} + \dots + a r^{n - 1}, \end{matrix}

which we call a finite geometric series with ratio

r .

It turns out that the value of each sum that has this form can be computed quickly without having to add all of the individual terms.

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Activity 8.3.2.

Let

a = 1

and

r = \frac{2}{5}

be real numbers and consider the corresponding finite geometric series

🔗

\begin{matrix} (8.3.2) & S_{n} = 1 + \frac{2}{5} + {(\frac{2}{5})}^{2} + \dots + {(\frac{2}{5})}^{n - 1} . \end{matrix}

🔗

Our goal in this activity is to find a shortcut formula for

S_{n}

that can be written as a single fraction that does not involve a sum of

n

terms.

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Multiply both sides of Equation (8.3.2) by $r = \frac{2}{5} .$ Write the new equation in the form

$\begin{matrix} (8.3.3) & \frac{2}{5} \cdot S_{n} = \dots . \end{matrix}$
Now subtract Equation (8.3.3) from Equation (8.3.2), and explain why it follows that

$\begin{matrix} (8.3.4) & S_{n} - \frac{2}{5} \cdot S_{n} = 1 - {(\frac{2}{5})}^{n} . \end{matrix}$
Solve Equation (8.3.4) for $S_{n}$ to find a simple formula for $S_{n}$ that does not involve adding $n$ terms.
How would your work above change if instead of the original geometric sum $S_{n},$ we considered the situation with $a = 7,$

$S_{n} = 7 + 7 \cdot \frac{2}{5} + 7 \cdot {(\frac{2}{5})}^{2} + \dots + 7 \cdot {(\frac{2}{5})}^{n - 1} ?$

🔗

The ideas in Activity 8.3.2 can be extended to the general case for any value of

a

and any value of

r \neq 1 .

In particular, replacing

1

with

a

and

\frac{2}{5}

with

r,

our work shows that by finding

r \cdot S_{n}

and then subtracting that quantity from

S_{n},

we get

🔗

S_{n} - r S_{n} = (a + a r + \dots + a r^{n - 1}) - (a r + a r^{2} + \dots + a r^{n}),

so that

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S_{n} (1 - r) = a - a r^{n},

and therefore

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S_{n} = \frac{a - a r^{n}}{1 - r} .

We summarize this result formally as follows.

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The value of a finite geometric series.

A finite geometric series

S_{n}

is a sum of the form

🔗

\begin{matrix} (8.3.6) & S_{n} = a + a r + a r^{2} + \dots + a r^{n - 1}, \end{matrix}

where

a

and

r

are real numbers such that

r \neq 1 .

The exact value of the finite geometric series

S_{n}

can be computed directly as

🔗

\begin{matrix} (8.3.7) & S_{n} = \frac{a (1 - r^{n})}{1 - r} . \end{matrix}

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Example 8.3.1. The value of a finite geometric series.

Use the shortcut formula in Equation (8.3.7) to find the exact value of the finite geometric series

🔗

1 + \frac{1}{2} + {(\frac{1}{2})}^{2} + {(\frac{1}{2})}^{3} + \dots + {(\frac{1}{2})}^{9} .

🔗

Solution.

We can view the given sum as the finite geometric series

S_{10}

that has

a = 1

and

r = \frac{1}{2} .

By Equation (8.3.7), it follows that

S_{10} = \frac{1 - 1 \cdot {(\frac{1}{2})}^{10}}{1 - \frac{1}{2}} = \frac{1 - \frac{1}{1024}}{\frac{1}{2}} = \frac{2047}{1024} = 1.9990234375 .

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Example 8.3.2. The value of another finite geometric series.

Use the shortcut formula in Equation (8.3.7) to find the exact value of the finite geometric series

🔗

2 - \frac{8}{3} + \frac{32}{9} - \frac{128}{27} + \dots + 2 \cdot {(- \frac{4}{3})}^{7} .

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Solution.

We can view the given sum as the finite geometric series

S_{8}

that has

a = 2

and

r = - \frac{4}{3} .

By Equation (8.3.7), it follows that

S_{8} = \frac{2 - 2 \cdot {(- \frac{4}{3})}^{8}}{1 + \frac{4}{3}} = - \frac{16850}{2187} \approx - 7.704618198 .

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Example 8.3.3. The value of one more finite geometric series.

Use the shortcut formula in Equation (8.3.7) to find the exact value of the finite geometric series

🔗

\frac{1}{3} + 1 + 3 + 9 + \dots + 6561 .

🔗

Solution.

Note that

6561 = 3^{8},

so since this sum begins with

\frac{1}{3} = 3^{- 1},

this is a finite geometric series with

10

terms. Hence we can view the given sum as the finite geometric series

S_{10}

that has

a = \frac{1}{3}

and

r = 3 :

S_{10} = \frac{1}{3} \cdot 3^{0} + \frac{1}{3} \cdot 3^{1} + \frac{1}{3} \cdot 3^{2} + \dots + \frac{1}{3} \cdot 3^{9} .

By Equation (8.3.7), we find that

S_{10} = \frac{\frac{1}{3} - \frac{1}{3} \cdot 3^{10}}{1 - 3} = \frac{29524}{3} .

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Subsection 8.3.2 Infinite Geometric Series

Our initial introduction to sums with many terms came from Taylor polynomial approximations such as

🔗

e^{1} \approx T_{5} (1) = 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!} .

Because the approximation gets better as we add more terms, it’s natural to think about the possibility of the sum extending forever. We begin by asking this question for a finite geometric series such as

🔗

S_{n} = 1 + \frac{1}{2} + {(\frac{1}{2})}^{2} + {(\frac{1}{2})}^{3} + \dots + {(\frac{1}{2})}^{n - 1} :

what happens if we let the sum continue indefinitely?

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Example 8.3.4. An infinite geometric series.

Can we find the value of

🔗

1 + \frac{1}{2} + {(\frac{1}{2})}^{2} + {(\frac{1}{2})}^{3} + \dots,

where the sum never terminates?

🔗

Solution.

In Preview Activity 8.3.1, we saw that for the finite geometric series

S_{n} = 1 + \frac{1}{2} + {(\frac{1}{2})}^{2} + {(\frac{1}{2})}^{3} + \dots + {(\frac{1}{2})}^{n - 1},

S_{1} = 1,

S_{2} = \frac{3}{2} = 1.5,

S_{3} = \frac{7}{4} = 1.75,

S_{4} = \frac{15}{8} = 1.875,

S_{5} = \frac{31}{16} = 1.9375,

and indeed

S_{n} = \frac{1 - {(\frac{1}{2})}^{n}}{1 - \frac{1}{2}}

so if we multiply both the numerator and denominator by

2^{n},

we find that

\begin{matrix} (8.3.8) & S_{n} = \frac{2^{n} - 1}{2^{n - 1}} . \end{matrix}

We can view

S_{n}

as a “partial sum” of the infinite geometric series

1 + \frac{1}{2} + {(\frac{1}{2})}^{2} + {(\frac{1}{2})}^{3} + \dots

whose value we seek. Plotting these partial sums on a number line, we see evidence that the value of the

n

th partial sum is approaching

2 .

Figure 8.3.5. A plot of $S_{1},$ $S_{2},$ $S_{3},$ $S_{4},$ $S_{5},$ and the number $2$ on the interval $[1, 2] .$

Indeed, we observe that each partial sum lies halfway between the preceding partial sum and the number

2 :

S_{2} = 1.5

is halfway between

S_{1} = 1

and

2;

S_{3} = 1.75

is halfway between

S_{2} = 1.5

and

2;

and so on.

We can see this more formally in Equation (8.3.8) if we divide the two terms in the numerator of

S_{n}

by the denominator. Doing so, an equivalent formula for

S_{n}

\begin{matrix} (8.3.9) & S_{n} = 2 - \frac{1}{2^{n - 1}} . \end{matrix}

In Equation (8.3.9), if we let

n \to \infty,

we have

lim_{n \to \infty} S_{n} = lim_{n \to \infty} (2 - \frac{1}{2^{n - 1}}) = 2,

since

\frac{1}{2^{n - 1}} \to 0

n

increases without bound.

Thus, it makes sense to say that the sum of the infinite geometric series

S = 1 + \frac{1}{2} + {(\frac{1}{2})}^{2} + {(\frac{1}{2})}^{3} + \dots

is finite and that

S = 2 .

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Example 8.3.4 demonstrates the general principle that we use to determine if any infinite series has a finite value: we consider the partial sum,

S_{n},

which is the finite sum of the first

n

terms, and then investigate if the partial sums converge to a single value as

n

increases without bound. For geometric series, determining whether the partial sums converge or not is straightforward.

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The value of an infinite geometric series.

An infinite geometric series

S

is a sum of the form

🔗

\begin{matrix} (8.3.10) & S = a + a r + a r^{2} + \dots + a r^{n - 1} + \dots, \end{matrix}

where

a

and

r

are real numbers. If

| r | < 1,

then the infinite geometric series converges to the finite value

🔗

\begin{matrix} (8.3.11) & S = \frac{a}{1 - r} . \end{matrix}

| r | \geq 1,

then the infinite geometric series does not converge to a finite value.

🔗

Equation (8.3.11) holds because when

| r | < 1,

it follows that

r^{n} \to 0

n \to \infty .

Indeed, when

| r | < 1,

we see by taking the limit as

n \to \infty

in Equation (8.3.6), that

🔗

lim_{n \to \infty} S_{n} = lim_{n \to \infty} \frac{a (1 - r^{n})}{1 - r} = \frac{a}{1 - r} .

In Exercise 8.3.5.7, we will explore why the infinite geometric series does not converge when

| r | = 1

and when

| r | > 1 .

🔗

As in our work in Chapter 4 with Riemann sums, we can use sigma notation (see Subsection 4.2.1) to express a sum in convenient shorthand. For instance, it’s equivalent to write

🔗

S_{n} = a + a r + a r^{2} + \dots + a r^{n - 1} = \sum_{k = 0}^{n - 1} a r^{k} .

🔗

Furthermore, we can use sigma notation for a more concise way to express an infinite geometric series. For example, it is equivalent to write

🔗

1 + \frac{1}{2} + {(\frac{1}{2})}^{2} + {(\frac{1}{2})}^{3} + \dots = \sum_{k = 0}^{\infty} {(\frac{1}{2})}^{k} .

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Activity 8.3.3.

For each of the following infinite geometric series, determine the values of

a

and

r,

compute the partial sums

S_{5}

and

S_{10}

exactly (writing each as a fraction), and if the infinite geometric series converges, find its value.

🔗

$1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \dots$
$4 - 2 + 1 - \frac{1}{2} + \frac{1}{4} - \dots$
$2 + \frac{8}{3} + \frac{32}{9} + \frac{128}{27} + \dots$
$\sum_{k = 0}^{\infty} 5 \cdot {(\frac{3}{4})}^{k}$
$\sum_{k = 1}^{\infty} - 2 \cdot {(- \frac{2}{3})}^{k}$

🔗

Subsection 8.3.3 How geometric series naturally connect to Taylor polynomials

If we take

a = 1

in Equation (8.3.11), we see that

🔗

\begin{matrix} (8.3.12) & 1 + r + r^{2} + \dots + r^{n} + \dots = \frac{1}{1 - r}, \end{matrix}

provided that

| r | < 1 .

To study this equation further, we are going to let

r

vary and thus we introduce the function

f (x) = \frac{1}{1 - x}

and replace

r

x

in Equation (8.3.12) to have

🔗

\begin{matrix} (8.3.13) & f (x) = \frac{1}{1 - x} = 1 + x + x^{2} + \dots + x^{n} + \dots, \end{matrix}

which is valid for

| x | < 1 .

One reason this equation is interesting is that we have a function

f (x)

that can be represented in two different ways: as the rational function

\frac{1}{1 - x},

and as the infinite polynomial function

1 + x + x^{2} + \dots .

🔗

For

x

such that

| x | < 1,

we know that the infinite geometric series

1 + x + x^{2} + \dots + x^{n} + \dots

converges. Thus, from Equation (8.3.13), it follows

🔗

f (x) = \frac{1}{1 - x} \approx 1 + x + x^{2} + \dots + x^{n}

for

x

near

a = 0 .

Moreover, because the infinite series converges, the larger the value of

n,

the better the approximation will be. If we plot the function

f (x) = \frac{1}{1 - x}

along with several of the polynomials that arise for different choices of

n,

say

P_{1} (x) = 1 + x,

P_{4} (x) = 1 + x + \dots + x^{4},

and

P_{7} (x) = 1 + x + \dots + x^{7},

we can see the impact of increasing the degree

n

in Figure 8.3.6.

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Figure 8.3.6. A plot of $f (x) = \frac{1}{1 - x}$ (blue, no dashes), $P_{1} (x) = 1 + x$ (light gray, small dashes), $P_{4} (x) = 1 + x + \dots + x^{4}$ (medium gray, medium dashes), and $P_{7} (x) = 1 + x + \dots + x^{7}$ (dark gray, long dashes).
🔗

In particular, we observe that as the degree of the polynomial approximation increases, the polynomial not only appears closer to

f (x) = \frac{1}{1 - x},

but does so on a wider interval of

x

-values. Since the infinite series only converges when

| x | < 1

and the function

f (x) = \frac{1}{1 - x}

is undefined when

x = 1,

we also expect the approximations to only be accurate on an interval that lies within

- 1 < x < 1 .

This reminds of our earlier work with Taylor polynomials where in images such as Figure 8.2.8, increasing the degree of the Taylor polynomial similarly improves the approximation.

🔗

In our preceding work in this subsection, we arrived at the function

f (x) = \frac{1}{1 - x}

by starting with the infinite geometric series

🔗

1 + x + x^{2} + \dots + x^{n} + \dots

and exploring its partial sums. Next, we change perspective and start with function

f (x) = \frac{1}{1 - x}

and determine the Taylor polynomial approximations to

f

that are centered at

a = 0

in order to see an interesting connection.

🔗

Activity 8.3.4.

Consider the function

f (x) = \frac{1}{1 - x} .

Find the degree

n

Taylor polynomial approximation centered at

a = 0

for

f .

🔗

To begin finding $T_{n} (x),$ do the usual work of computing the various derivatives of $f$ and their respective values at $a = 0;$ note that it’s helpful to view $f (x)$ in the form $f (x) = (1 - x)^{- 1}$ so that we can easily compute the derivatives of $f$ using the chain rule. For instance,

$f^{'} (x) = (- 1) (1 - x)^{- 2} (- 1)$

where the first “ $- 1$ ” arises from the power rule, while the second “ $- 1$ ” results from the chain rule, since $\frac{d}{d x} [1 - x] = - 1 .$ In order to see key patterns that arise, it’s helpful not to combine the products of numbers that arise in the various derivatives. Record the first five derivatives of $f (x)$ in Table 8.3.7.

Table 8.3.7. Finding the derivatives of $f (x) = \frac{1}{1 - x} .$

$f (x) =$ $\frac{1}{1 - x} = (1 - x)^{- 1}$

$f^{'} (x) =$ $(- 1) (1 - x)^{- 2} (- 1)$

$f^{″} (x) =$

$f^{‴} (x) =$

$f^{(4)} (x) =$

$f^{(5)} (x) =$
Next, evaluate the derivatives you determined in (a) at $a = 0$ and use these to find the values of the coefficients of the Taylor polynomial centered at $a = 0 .$ Record your work in Table 8.3.8.

Table 8.3.8. Finding the Taylor polynomial coefficients $c_{k} .$

$f (0) =$ $\frac{1}{1 - 0} = 1$ $c_{0} =$ $f (0) = 1$

$f^{'} (0) =$ $(- 1) (1 - 0)^{- 2} (- 1) = 1$ $c_{1} =$ $\frac{f^{'} (0)}{1!} = \frac{1}{1!} = 1$

$f^{″} (0) =$ $c_{2} =$ $\frac{f^{″} (0)}{2!} =$

$f^{‴} (0) =$ $c_{3} =$

$f^{(4)} (0) =$ $c_{4} =$

$f^{(5)} (0) =$ $c_{5} =$
What pattern do you observe in the value of $c_{k} ?$ State the degree $5$ Taylor polynomial, $T_{5} (x),$ as well as the formula you expect for the general degree $n$ Taylor polynomial, $T_{n} (x) .$
By identifying the value of $r,$ explain why the degree $n$ Taylor polynomial $T_{n} (x)$ can be thought of as a finite geometric series.
What is the Taylor series centered at $a = 0$ for $f (x) = \frac{1}{1 - x} ?$ (As we will see in the next section, the “Taylor series” of a function is the infinite series that results by simply extending the Taylor polynomials indefinitely.)

🔗

Subsection 8.3.4 Summary

A finite geometric series $S_{n}$ is a sum of the form

$S_{n} = a + a r + a r^{2} + \dots + a r^{n - 1},$

where $a$ and $r$ are real numbers such that $r \neq 1 .$ The value of the finite geometric series $S_{n}$ can be computed directly as

$S_{n} = \frac{a (1 - r^{n})}{1 - r} .$
An infinite geometric series $S$ is a sum of the form

$S = a + a r + a r^{2} + \dots + a r^{n - 1} + \dots,$

where $a$ and $r$ are real numbers. If $| r | < 1,$ then the infinite geometric series converges to the finite value

$S = \frac{a}{1 - r} .$

If $| r | \geq 1,$ then the infinite geometric series does not converge to a finite sum.
If we consider the infinite geometric series with $a = 1$ and $| r | < 1$ and replace $r$ with $x,$ we can say that

$\frac{1}{1 - x} = 1 + x + x^{2} + x^{3} + \dots + x^{n} + \dots .$

As we found in Activity 8.3.4, the Taylor series centered at $a = 0$ for $f (x) = \frac{1}{1 - x}$ is precisely the infinite geometric series

$T (x) = 1 + x + x^{2} + x^{3} + \dots + x^{n} + \dots$

and we know this series is equal to $f (x)$ for $| x | < 1 .$

🔗

Exercises 8.3.5 Exercises

1.

For each of the following, decide if the given series is a geometric series.

🔗

3 - 6 + 12 - 24 + 48 - \dots

🔗

Is this a geometric series?

🔗

If this is a geometric series, enter

🔗

the first term:

🔗

and the ratio between successive terms:

🔗

(Enter na for the first term and ratio if this is not a geometric series.)

🔗

3 + 9 x + 27 x^{2} + 81 x^{3} + 243 x^{4} + \dots

🔗

Is this a geometric series?

🔗

If this is a geometric series, enter

🔗

the first term:

🔗

and the ratio between successive terms:

🔗

(Enter na for the first term and ratio if this is not a geometric series.)

🔗

3 x^{7} + 4 x^{8} + 5 x^{9} + 6 x^{10} + \dots

🔗

Is this a geometric series?

🔗

If this is a geometric series, enter

🔗

the first term:

🔗

and the ratio between successive terms:

🔗

(Enter na for the first term and ratio if this is not a geometric series.)

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2.

For each of the finite geometric series given below, indicate the number of terms in the sum and find the sum. For the value of the sum, enter an expression that gives the exact value, rather than entering an approximation.

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4 + 4 (0.5) + 4 (0.5)^{2} + \dots + 4 (0.5)^{12}

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number of terms =

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value of sum =

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4 (0.5)^{6} + 4 (0.5)^{7} + 4 (0.5)^{8} + \dots + 4 (0.5)^{11}

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number of terms =

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value of sum =

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3.

Find the sum of each of the geometric series given below. For the value of the sum, enter an expression that gives the exact value, rather than entering an approximation.

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- 12 + 4 - \frac{4}{3} + \frac{4}{3^{2}} - \frac{4}{3^{3}} + \dots - \frac{4}{3^{15}} =

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\sum_{n = 6}^{\infty} {(\frac{1}{2})}^{n} =

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4.

Find the sum of the infinite geometric series

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1 - \frac{1}{4} + \frac{1}{16} - \frac{1}{64} + \dots

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Enter DNE if the sum does not exist.

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Sum =

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5.

Find the sum of the series

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8 + \frac{8}{6} + \frac{8}{36} + . . . + \frac{8}{6^{n - 1}} + . . . .

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Answer:

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6.

Determine whether the series converges, and if so find its sum.

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\sum_{k = 1}^{\infty} \frac{2^{k + 5}}{3^{k}} =

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(Enter DNE if the sum does not exist.)

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7.

In Equation (8.3.11), we learned that for an infinite geometric series

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S = a + a r + a r^{n} + \dots + a r^{n - 1} + \dots,

| r | < 1,

then

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

This result provides a quick way to determine the value of an infinite geometric series that converges.

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In this exercise, we explore why the infinite geometric series fails to converge when

r = 1,

r = - 1,

and

| r | > 1 .

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If we consider the finite geometric series that results when $a = 1$ and $r = 1,$ we get the sum

$S_{n} = 1 + 1 + \dots + 1,$

where there are $n$ terms in the sum.
1. Compute $S_{2},$ $S_{3},$ and $S_{4} .$ What is the general formula for $S_{n} ?$
2. Explain why the sequence $S_{n}$ does not converge to a finite value. Contrast this with, for example, our work in Example 8.3.4.
Next, consider the finite geometric series that results when $a = 1$ and $r = - 1,$ which is the sum

$S_{n} = 1 - 1 + 1 - 1 + \dots + (- 1)^{n - 1},$

where there are $n$ terms in the sum. (We view the first “ $1$ ” as resulting from “ $1 \cdot (- 1)^{0}$ ”.)
1. Compute $S_{2},$ $S_{3},$ $S_{4},$ and $S_{5} .$ What do you observe?
2. Explain why the sequence $S_{n}$ does not converge to a finite value. Contrast this with, for example, our work in Example 8.3.4.
Now consider the geometric sum with $a = 1$ and $r = 2,$ so

$S_{n} = 1 + 2 + 4 + \dots + 2^{n - 1} .$
1. Compute $S_{2},$ $S_{3},$ and $S_{4} .$ What is the general formula for $S_{n} ?$
2. What do you observe happens to these partial sums as $n$ increases without bound? What does this tell us about the infinite geometric series $1 + 2 + 4 + \dots + 2^{n - 1} + \dots ?$

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8.

It is important to understand the power of geometric growth compared to linear growth. Suppose you are hired for a job that will take you 30 days to complete and are offered two options for how you’ll be compensated.

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Option 1.: You can be paid $500 per day, or
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Option 2.: You can be paid 1 cent the first day, 2 cents the second day, 4 cents the third day, 8 cents the fourth day, and so on, doubling the amount you are paid each day.
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How much will you be paid for the job in total under Option 1?
Complete Table 8.3.13 to determine the pay you will receive under Option 2 for the first 10 days.

Table 8.3.13. Option 2 payments

Day Pay on this day Total amount paid to date

$1$ $$ 0.01$ $$ 0.01$

$2$ $$ 0.02$ $$ 0.03$

$3$

$4$

$5$

$6$

$7$

$8$

$9$

$10$
Find a formula for the amount paid on day $n,$ as well as for the total amount paid by day $n .$ Use this formula to determine which option (1 or 2) you should take.

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9.

Suppose you drop a golf ball onto a hard surface from a height

h .

The collision with the ground causes the ball to lose energy and so it will not bounce back to its original height. The ball will then fall again to the ground, bounce back up, and continue. Assume that at each bounce the ball rises back to a height

\frac{3}{4}

of the height from which it dropped. Let

h_{n}

be the height of the ball on the

n

th bounce, with

h_{0} = h .

In this exercise we will determine the distance traveled by the ball and the time it takes to travel that distance.

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Determine a formula for $h_{1}$ in terms of $h .$
Determine a formula for $h_{2}$ in terms of $h .$
Determine a formula for $h_{3}$ in terms of $h .$
Determine a formula for $h_{n}$ in terms of $h .$
Write an infinite series that represents the total distance traveled by the ball. Then determine the value of this series.
Next, let’s determine the total amount of time the ball is in the air.
1. When the ball is dropped from a height $H,$ if we assume the only force acting on it is the acceleration due to gravity, then the height of the ball at time $t$ is given by
  
  $H - \frac{1}{2} g t^{2} .$
  
  Use this formula to determine the time it takes for the ball to hit the ground after being dropped from height $H .$
2. Use your work in the preceding item, along with that in (a)-(e) above to determine the total amount of time the ball is in the air.

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You have attempted 1 of 8 activities on this page.

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$f (x) =$	$\frac{1}{1 - x} = (1 - x)^{- 1}$
$f^{'} (x) =$	$(- 1) (1 - x)^{- 2} (- 1)$
$f^{″} (x) =$
$f^{‴} (x) =$
$f^{(4)} (x) =$
$f^{(5)} (x) =$

$f (0) =$	$\frac{1}{1 - 0} = 1$	$c_{0} =$	$f (0) = 1$
$f^{'} (0) =$	$(- 1) (1 - 0)^{- 2} (- 1) = 1$	$c_{1} =$	$\frac{f^{'} (0)}{1!} = \frac{1}{1!} = 1$
$f^{″} (0) =$		$c_{2} =$	$\frac{f^{″} (0)}{2!} =$
$f^{‴} (0) =$		$c_{3} =$
$f^{(4)} (0) =$		$c_{4} =$
$f^{(5)} (0) =$		$c_{5} =$

Day	Pay on this day	Total amount paid to date
$1$	$$ 0.01$	$$ 0.01$
$2$	$$ 0.02$	$$ 0.03$
$3$
$4$
$5$
$6$
$7$
$8$
$9$
$10$