Mastering the Sum of Geometric Series Formula in Algebra

In the case where the terms of a sequence increase by a constant ratio, the sum of the sequence can be modeled by geometric series. This lesson will introduce how to find the sum of geometric series for both finite and infinite cases.

Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.

Geometric Sequence
Explicit Rule of Geometric Sequences
Series
Sigma Notation

Challenge

Finding the Sum

Consider the following sum of a sequence.

\frac{1}{2} + \frac{1}{2 0} + \frac{1}{2 0 0} + \frac{1}{2 0 0 0} + \dots

a Is the sum finite or infinite?

b If the sum is finite, what is its value? Round the answer to two decimals.

Discussion

Geometric Series

A geometric series is the sum of the terms of a geometric sequence.

Geometric Sequence 1, 2, 4, 8, \dots Geometric Series 1 + 2 + 4 + 8 + \dots

The explicit rule of the geometric sequence above is

a_{n} = 2^{n - 1} .

This rule can be used to write the series using sigma notation.

Sum of Infinite Terms i = 1 \sum \infty 2^{i - 1} Sum of n Terms i = 1 \sum n 2^{i - 1}

The sum of an infinite or a finite geometric series can be found by using its corresponding formula.

Pop Quiz

Identifying Geometric Series

The following applet shows the first five terms of a sum. Identify whether the given sum is a geometric series or not.

Discussion

Finite Geometric Series

Let $a_{1}$ and $r$ be the first term and the common ratio, respectively, of a geometric sequence with $n$ terms, where $r \neq = 1 .$ The sum of the related finite geometric series can be found by using the following formula.

$S_{n} = \frac{a _{1} ( 1 - r ^{n} )}{1 - r},$ $r \neq = 1$

Proof

Finite Sum

To derive the formula for a geometric series, a geometric sequence of

n

terms with the first term

a_{1}

and the common ratio

r

will be considered.

a_{1}, a_{1} r, a_{1} r^{2}, \dots, a_{1} r^{n - 1}

All the terms will be added to express the series and

a_{1}

will be factored out.

S_{n} = a_{1} + a_{1} r + a_{1} r^{2} + \dots + a_{1} r^{n - 1}

FactorOut

Factor out $a_{1}$

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

Then, both sides of the equation will be multiplied by

(1 - r)

and the resulting equation simplified.

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

MultEqn

$LHS \cdot (1 - r) = RHS \cdot (1 - r)$

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

Distr

Distribute $(1 - r)$

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

Multiply

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

Notice that the like terms are ordered in minus-plus pairs. This means that after simplifying, they will cancel out and only the first and last terms will remain.

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

AddSubTerms

Add and subtract terms

S_{n} (1 - r) = a_{1} (1 - r^{n})

DivEqn

$LHS / (1 - r) = RHS / (1 - r)$

S_{n} = \frac{a _{1} ( 1 - r ^{n} )}{1 - r}

The formula for the sum of a finite geometric series has been derived.

$S_{n} = \frac{a _{1} ( 1 - r ^{n} )}{1 - r}$

Example

The Number of Cases of Sickness at School

Tearrik goes to North High School. One day he suddenly started feeling sick at school. Later it was determined that according to school records, Tearrik's three best friends started feeling sick the next day, nine more students stayed home sick on the third day, and so on.

Within a few days, it was discovered that a new kind of viral disease was spreading rapidly through the school, with each infected student transmitting the virus to three other students each day.

a Write an explicit rule to represent the number of students who become infected on a given day.

b Calculate the total number of infected students that there would be in

7

days if the school does not suspend classes.

Hint

a The number of newly-infected students each day makes a geometric sequence.

b Use the formula for the sum of a geometric series.

Solution

a It is given that the number of newly-infected students is triple the number of students that were infected the day before.

This means that a common ratio exists between the number of newly-infected students each day. Therefore, their sum represents a geometric series. With this in mind, an explicit rule can be written to find the number of newly-infected students on the

n^{th}

day.

a_{n} = a_{1} \cdot r^{n - 1}

Since only Tearrik fell ill on the first day, the first term of the sequence is

a_{1} = 1

and the common ratio

r = 3 .

Substitute these values into the formula to find the explicit rule for this sequence.

a_{n} = a_{1} \cdot r^{n - 1}

SubstituteII

$a_{1} = 1$ , $r = 3$

a_{n} = 1 \cdot 3^{n - 1}

IdPropMult

Identity Property of Multiplication

a_{n} = 3^{n - 1}

The number of newly-infected students on the

n^{th}

day can be represented by the formula

a_{n} = 3^{n - 1} .

b To calculate the total number of infected students on any given day, the formula for the sum of a finite geometric series will be used.

S_{n} = \frac{a _{1} ( 1 - r ^{n} )}{1 - r}

Since the total number of infected students on the

7^{th}

day is required,

n = 7

will be substituted into the formula. Remember that the first term

a_{1}

1

and the common ratio

r

3 .

S_{n} = \frac{a _{1} ( 1 - r ^{n} )}{1 - r}

SubstituteValues

Substitute values

S_{7} = \frac{1 ( 1 - 3 ^{7} )}{1 - 3}

▼

Evaluate right-hand side

IdPropMult

Identity Property of Multiplication

S_{7} = \frac{1 - 3 ^{7}}{1 - 3}

ExpandFrac

$\frac{a}{b} = \frac{a \cdot ( - 1 )}{b \cdot ( - 1 )}$

S_{7} = \frac{3 ^{7} - 1}{3 - 1}

CalcPow

Calculate power

S_{7} = \frac{2 1 8 7 - 1}{3 - 1}

SubTerms

Subtract terms

S_{7} = \frac{2 1 8 6}{2}

CalcQuot

Calculate quotient

S_{7} = 1093

If the school does not suspend classes, there will be

1093

infected students in

7

days.

Example

Geometric Series in Sigma Notation

As the number of infected students increased, the school decided to start a quarantine period. During the quarantine period at North High School, lessons started being taught online. In one of his math lessons, Tearrik's math teacher introduced a geometric series an example written using sigma notation.

$Two example of series shown in the computer screen. Series 1: \sum \limits_{i=1}^{12} 6 \t 3^{i} Series 2: \sum \limits_{n=1}^{7} \t 3^{i}$

Calculate the given sum.

Hint

Use the formula for the sum of a finite geometric series.

Solution

The given sum is a finite geometric series. Therefore, the formula for the sum of a finite geometric series can be used to find this sum.

S_{n} = \frac{a _{1} ( 1 - r ^{n} )}{1 - r}, r \neq = 1

In this formula,

a_{1}

is the first term,

r

is the common ratio, and

n

is the number of terms. First, substitute

i = 1

into the expression given on the right side of the sigma notation to find

a_{1} .

6 \cdot 3^{i}

Substitute

$i = 1$

6 \cdot 3^{1}

ExponentOne

$a^{1} = a$

6 \cdot 3

Multiply

18

The first term is

18 .

Notice that the summation index

i

is placed as a single exponent of

3,

meaning that the next term can be found by multiplying the previous one by

3 .

This makes the common ratio

r = 3 .

Another way to find

r

is to divide the second term by the first. Start by substituting

i = 2

into the given expression.

6 \cdot 3^{i}

Substitute

$i = 2$

6 \cdot 3^{2}

CalcPow

Calculate power

6 \cdot 9

Multiply

54

Now calculate the ratio of

a_{2}

a_{1} .

\frac{a _{2}}{a _{1}} = \frac{5 4}{1 8} \Rightarrow r = 3

The common ratio was found to be

3 .

Since the lower limit of the sigma notation is

1

and the upper limit of the sigma notation is

12,

there are

12

terms in the series in total, so

n = 12 .

The sum can be calculated with these values. To do so, substitute all the values into the formula for

S_{n}

and simplify.

S_{n} = \frac{a _{1} ( 1 - r ^{n} )}{1 - r}

SubstituteValues

Substitute values

S_{12} = \frac{1 8 ( 1 - 3 ^{12} )}{1 - 3}

▼

Evaluate right-hand side

UseCalc

Use a calculator

S_{12} = \frac{1 8 ( 1 - 5 3 1 4 4 1 )}{1 - 3}

SubTerms

Subtract terms

S_{12} = \frac{1 8 ( - 5 3 1 4 4 0 )}{- 2}

Multiply

S_{12} = \frac{- 9 5 6 5 9 2 0}{- 2}

DivNegNeg

$\frac{- a}{- b} = \frac{a}{b}$

S_{12} = \frac{9 5 6 5 9 2 0}{2}

CalcQuot

Calculate quotient

S_{12} = 4782960

Pop Quiz

Calculating Finite Geometric Series

Calculate the sum of all the terms of the given finite geometric series written in summation notation. Remember that the formula for the sum of a geometric series can be used rather than adding the terms one by one.

Discussion

Infinite Geometric Series

Let $a_{1}$ and $r$ be the first term and the common ratio, respectively, of a geometric sequence with $n$ terms, where $r \neq = 1 .$ For an infinite series, if the common ratio $r$ is greater than $- 1$ and less than $1$ — in other words, if $∣ r ∣ < 1$ — then the sum can be found by using the following formula.

$S_{\infty} = \frac{a _{1}}{1 - r},$ $- 1 < r < 1$

This means that the sum converges on a number. If the common ratio $r$ is less than or equal to $- 1$ or greater than or equal to $1$ — if $∣ r ∣ \geq 1$ — then the sum diverges. In such cases, there is no sum for the infinite geometric series.

Proof

Infinite Sum

To derive the formula for the sum of an infinite geometric series with

- 1 < r < 1,

the standard formula for the finite geometric series will be considered.

S_{n} = \frac{a _{1} ( 1 - r ^{n} )}{1 - r}

Since

r

is a number between

- 1

and

1,

the value of

r^{n}

becomes very small as the value of

n

increases. In other words, it gets closer to

0

n

approaches infinity.

r^{n} 1 n \to \infty 0

Therefore,

r^{n} = 0

can be substituted into the standard formula and the resulting equation simplified.

S_{n} = \frac{a _{1} ( 1 - r ^{n} )}{1 - r}

SubstituteII

$r^{n} = 0$ , $n = \infty$

S_{\infty} = \frac{a _{1} ( 1 - 0 )}{1 - r}

SubTerm

Subtract term

S_{\infty} = \frac{a _{1} ( 1 )}{1 - r}

IdPropMult

Identity Property of Multiplication

S_{\infty} = \frac{a _{1}}{1 - r}

The formula for the sum of an infinite geometric series with

- 1 < r < 1

has been derived.

$S_{\infty} = \frac{a _{1}}{1 - r}$

Example

Solving Examples of Geometric Series

In the next math lesson, Tearrik's math teacher continued with the topic geometric series and this time she introduced the first few terms of two different geometric series as examples.

Two example of series shown in the computer screen. Series 1: 1+2/3+ 4/9+ 8/27 + 16/81 + ··· Series 2: -10/4+50/16-250/64+1250/256-···

a Determine whether the geometric series in Example I converges or diverges. If it converges, find its sum.

b Determine whether the geometric series in Example II converges or diverges. If it converges, find its sum.

Answer

a Does the series converge or diverge? Converge

Sum: $3$

b Does the series converge or diverge? Diverge

Sum: No sum.

Hint

a Consider the common ratio of the series. When does a series converge or diverge?

b What is the common ratio of the given series?

Solution

a To determine whether the sum of the series converges to a number or if the series does not have a sum, start by looking at the common ratio between the terms of the given series in Example I.

As shown, the common ratio is

\frac{2}{3} .

Since the absolute value of

\frac{2}{3}

less than

1,

it means that the sum of the series does converge to a number.

∣ r ∣ = ∣ ∣ ∣ ∣ ∣ \frac{2}{3} ∣ ∣ ∣ ∣ ∣ \Rightarrow ∣ r ∣ = \frac{2}{3} < 1

Therefore, it is possible to find its sum by using the formula for the sum of an infinite series.

S_{\infty} = \frac{a _{1}}{1 - r}

To find the sum, substitute

a_{1} = 1

and

r = \frac{2}{3}

into the formula and evaluate.

S_{\infty} = \frac{a _{1}}{1 - r}

SubstituteII

$a_{1} = 1$ , $r = \frac{2}{3}$

S_{\infty} = \frac{1}{1 - \frac{2}{3}}

▼

Evaluate right-hand side

OneToFrac

Rewrite $1$ as $\frac{3}{3}$

S_{\infty} = \frac{1}{\frac{3}{3} - \frac{2}{3}}

SubFrac

Subtract fractions

S_{\infty} = \frac{1}{\frac{1}{3}}

DivOneByFracD

$\frac{1}{a / b} = \frac{b}{a}$

S_{\infty} = \frac{3}{1}

SimpQuot

Simplify quotient

S_{\infty} = 3

The sum of the infinite geometric series given in Example I is

3 .

b Once again, start by identifying the common ratio of the given series. One way to find the common ratio is to divide the second term by the first term.

\frac{a _{2}}{a _{1}}

SubstituteII

$a_{1} = - \frac{1 0}{4}$ , $a_{2} = \frac{5 0}{1 6}$

\frac{- \frac{5 0}{1 6} -}{- \frac{1 0}{4}}

▼

Simplify

DivFracByFracD

$\frac{a / b}{c / d} = \frac{a}{b} \cdot \frac{d}{c}$

\frac{5 0}{1 6} \cdot (- \frac{4}{1 0})

MoveNegFracToNum

Put minus sign in numerator

\frac{5 0}{1 6} \cdot (\frac{- 4}{1 0})

MultFrac

Multiply fractions

\frac{- 2 0 0}{1 6 0}

ReduceFrac

$\frac{a}{b} = \frac{a / 4 0}{b / 4 0}$

\frac{- 5}{4}

MoveNegNumToFrac

Put minus sign in front of fraction

- \frac{5}{4}

The common ratio of the given series in Example II is

- \frac{5}{4} .

Now the absolute value of the common ratio will be found.

∣ r ∣ = ∣ ∣ ∣ ∣ ∣ - \frac{5}{4} ∣ ∣ ∣ ∣ ∣ \Rightarrow ∣ r ∣ = \frac{5}{4} > 1

Since the absolute value of the common ratio is greater than

1,

the series diverges. Therefore, it is not possible to find a sum for this series.

Pop Quiz

Does It Converge or Diverge?

Determine whether the given infinite geometric series converge or diverge. Remember that if the common ratio $∣ r ∣ < 1,$ then the infinite series converges to a number, and that if $∣ r ∣ \geq 1,$ then the series diverges.

Discussion

Partial Sum of an Infinite Geometric Series

If the common ratio of an infinite geometric series is less than or equal to $- 1$ or greater than or equal to $1,$ the sum of the series does not exist. However, it is possible to find a partial sum or the sum of the first several terms in the series. This partial series can be thought of as a finite series. As such, its sum can be found using the formula for a finite geometric series.

$S_{n} = \frac{a _{1} ( 1 - r ^{n} )}{1 - r},$ $r \neq = 1$

Example

Calculating the Distance Traveled by the Ball

After recovering from his illness, Tearrik returns to school and continues to play basketball with his best friend Tadeo. Suppose that after the ball hits the rim of the basket, the ball falls $3$ meters and rebounds to $85 %$ of the height of the previous bounce.

the path of the ball after hitting the rim

a Find the total vertical distance traveled by the ball before it comes to rest.

b What would the sum of the vertical distance traveled by the ball be if Tadeo caught the ball at the top of the fourth bounce? Round the answer to two decimal places.

Hint

a The vertical distance traveled by the ball for one bounce will be

2

times

0.85

of the previous bounce since the ball rises and then falls the same distance.

b Use the formula for the partial sum of an infinite series.

Solution

a It is given that after each bounce, the ball rises

85 %

of the height of the previous bounce. Since the initial height of the ball is

3

meters, the first bounce will be

85 % = 0.85

times the initial height.

Initial Height : First Bounce : 3 m 3 (0.85) m

For the second bounce, the height of the ball will be

0.85

times the first bounce,

3 (0.85)

meters.

Second Bounce : 3 (0.85) (0.85) \Rightarrow 3 (0.85)^{2} m

The heights of the other bounces can be written by considering this pattern.

Because the ball rises and then falls the same distance after each bounce, the vertical distance traveled is

2

times

0.85

of the previous bounce. Since the initial height of the ball is

3

meters, the total distance traveled by the ball can be written as follows.

3 + 2 \cdot 3 (0.85) + 2 \cdot 3 (0.85)^{2} + \dots ⇓ 3 + 6 (0.85) + 6 (0.85)^{2} + \dots

Since a common ratio

r = 0.85

exists between the distances that the ball travels, the distances traveled starting with the first bounce represent a geometric series. This sum is considered infinite because it is assumed that the ball could continue to bounce in increasingly small increments forever. Now, express the sum using summation notation.

3 + n = 1 \sum \infty 6 (0.85)^{n}

Note that even though the sum of the series is infinite, it can still be found by using the formula for the sum of an infinite series because the absolute value of the common ratio

∣ r ∣ = 0.85

is less than

1 .

S_{\infty} = \frac{a _{1}}{1 - r}

The first term of the infinite geometric series is

6 (0.85)

and the common ratio is

0.85 .

Now, substitute these values into the formula!

n = 1 \sum \infty 6 (0.85)^{n} = \frac{6 ( 0 . 8 5 )}{1 - 0 . 8 5}

Then, evaluate the right hand side.

n = 1 \sum \infty 6 (0.85)^{n} = \frac{6 ( 0 . 8 5 )}{1 - 0 . 8 5}

▼

Evaluate right-hand side

Multiply

n = 1 \sum \infty 6 (0.85)^{n} = \frac{5 . 1}{1 - 0 . 8 5}

SubTerm

Subtract term

n = 1 \sum \infty 6 (0.85)^{n} = \frac{5 . 1}{0 . 1 5}

CalcQuot

Calculate quotient

n = 1 \sum \infty 6 (0.85)^{n} = 34

The sum of the series is

34 .

Now the initial

3 -

meter height of the ball will be added.

34 + 3 = 37

This means that the ball will have traveled

37

meters when it comes to rest.

b This time the ball is caught at the top of the fourth bounce. To find the vertical distance traveled by the ball, start by considering the vertical distance traveled for each bounce until the end of the third bounce. Then, add the fourth vertical distance only once since the ball will rise and but be caught before it falls again.

3 + 2 \cdot 3 (0.85) + 2 \cdot 3 (0.85)^{2} + 2 \cdot 3 (0.85)^{3} ⇓ 3 + 6 (0.85) + 6 (0.85)^{2} + 6 (0.85)^{3}

Notice that this sum represents a partial sum of the infinite series considered in Part A. To calculate this partial sum, the formula for the sum of a finite geometric series will be used, as the sum is calculated up until

n = 3 .

S_{n} = \frac{a _{1} ( 1 - r ^{n} )}{1 - r}

With this in mind, substitute

n = 3,

a_{1} = 6 (0.85),

and

r = 0.85

into the formula and evaluate the result.

S_{n} = \frac{a _{1} ( 1 - r ^{n} )}{1 - r}

SubstituteValues

Substitute values

S_{3} = \frac{6 ( 0 . 8 5 ) ( 1 - ( 0 . 8 5 ) ^{3} )}{1 - 0 . 8 5}

▼

Evaluate right-hand side

Multiply

S_{3} = \frac{( 5 . 1 ) ( 1 - ( 0 . 8 5 ) ^{3} )}{1 - 0 . 8 5}

CalcPow

Calculate power

S_{3} = \frac{( 5 . 1 ) ( 1 - 0 . 6 1 4 1 2 5 )}{1 - 0 . 8 5}

SubTerms

Subtract terms

S_{3} = \frac{( 5 . 1 ) ( 0 . 3 8 5 8 7 5 )}{0 . 1 5}

Multiply

S_{3} = \frac{1 . 9 6 7 9 6 2 \dots}{0 . 1 5}

UseCalc

Use a calculator

S_{3} = 13.11975

Now, add the fourth vertical distance

3 (0.85)^{4}

to this sum.

13.11975 + 3 (0.85)^{4}

CalcPow

Calculate power

13.11975 + 3 \cdot 0.522006 \dots

Multiply

13.11975 + 1.566018 \dots

AddTerms

Add terms

14.685768 \dots

RoundDec

Round to $2$ decimal place(s)

\approx 14.69

Finally, the initial

3 -

meter height will be added.

14.69 + 3 = 17.69

The ball will travel vertically about

17.69

meters in total until Tadeo catches it at the top of the fourth bounce.

Example

Who Will Save More Money?

Now that they are completely recovered, Tearrik and Tadeo decide to save money so they can go to the NBA finals next year, $15$ months from now. They start chatting about their own ways to save money.

Conversation between Tadeo and Tearrik about the way of saving money

Who will save more money in

15

months?

Hint

Recall that a series is geometric if it has a common ratio.

Solution

To determine who will save more money in $15$ months, the total amounts saved will be calculated one at a time.

Tadeo's Savings

Tadeo is planning to save

100

dollars every month. This means that his savings will increase linearly by a constant rate of

100

dollars. Therefore, multiply

100

dollars by

15

months to find the total amount of money he will have saved in

15

months.

100 \cdot 15 = 1500

After

15

months, Tadeo will have saved

1500

dollars.

Tearrik's Savings

Tearrik will start by saving only $$ 0.50$ the first month. Then, he will increase the amount of money he saves every month by doubling the previous amount.

Since there is a common ratio between the amount of money saved each month, this sum represents a geometric series. Therefore, the formula for the sum of a finite geometric series will be used to calculate the total amount of money Tearrik will have saved in

15

months.

S_{n} = \frac{a _{1} ( 1 - r ^{n} )}{1 - r}

Now, substitute

n = 15,

a_{1} = 0.50,

and

r = 2

into this formula and simplify.

S_{n} = \frac{a _{1} ( 1 - r ^{n} )}{1 - r}

SubstituteValues

Substitute values

S_{15} = \frac{0 . 5 0 ( 1 - 2 ^{15} )}{1 - 2}

▼

Evaluate right-hand side

UseCalc

Use a calculator

S_{15} = \frac{0 . 5 0 ( 1 - 3 2 7 6 8 )}{1 - 2}

SubTerms

Subtract terms

S_{15} = \frac{0 . 5 0 ( - 3 2 7 6 7 )}{- 1}

Multiply

S_{15} = \frac{- 1 6 3 8 3 . 5}{- 1}

DivNegNeg

$\frac{- a}{- b} = \frac{a}{b}$

S_{15} = \frac{1 6 3 8 3 . 5}{1}

DivByOne

$\frac{a}{1} = a$

S_{15} = 16383.5

After

15

months, Tearrik will have saved

16383.5

dollars.

Conclusion

Notice that Tearrik's savings will increase rapidly although he starts with an extremely small amount. On the other hand, Tadeo will save the same amount of money each month and his savings will therefore increase at a constant rate. Finally, after $15$ months, Tearrik will have saved much more money than Tadeo.

Closure

Finding the Sum

At the beginning of the lesson, the following sum of a sequence was presented.

\frac{1}{2} + \frac{1}{2 0} + \frac{1}{2 0 0} + \frac{1}{2 0 0 0} + \dots

a Is the sum finite or infinite?

b If the sum is finite, what is its value? Round the answer to two decimals.

Hint

a Check the common ratio between the terms of the sum.

b Recall the formula for the sum of an infinite series.

Solution

a As can be seen, the given sequence has infinitely many terms. That said, it can still have a finite sum. To check whether the sum converges to a number or not, the common ratio between the terms will be identified. To do so, the ratio of the

second

term

to the

first

term

will be calculated.

\frac{a _{2}}{a _{1}}

SubstituteII

$a_{1} = \frac{1}{2}$ , $a_{2} = \frac{1}{2 0}$

\frac{\frac{1}{2 0}}{\frac{1}{2}}

DivFracByFracD

$\frac{a / b}{c / d} = \frac{a}{b} \cdot \frac{d}{c}$

\frac{1}{2 0} \cdot \frac{2}{1}

MultFrac

Multiply fractions

\frac{2}{2 0}

ReduceFrac

$\frac{a}{b} = \frac{a / 2}{b / 2}$

\frac{1}{1 0}

Notice that this ratio exists between each pair of consecutive terms.

Therefore, the sum represents a geometric series. Notice that the absolute value of the common ratio

∣ r ∣ = ∣ ∣ ∣ \frac{1}{1 0} ∣ ∣ ∣

is less than

1 .

∣ r ∣ = ∣ ∣ ∣ ∣ ∣ \frac{1}{1 0} ∣ ∣ ∣ ∣ ∣ \Rightarrow ∣ r ∣ = \frac{1}{1 0} < 1

This means that the sum is finite and can be calculated by using the formula for the sum of an infinite series.

b As it is mentioned in Part A, the sum is finite. Having a geometric series with a common ratio less than

1,

the sum of this series will be calculated by using the formula for the sum of an infinite series.

S_{\infty} = \frac{a _{1}}{1 - r}

Substitute

a_{1} = \frac{1}{2}

and

r = \frac{1}{1 0}

into the formula.

S_{\infty} = \frac{a _{1}}{1 - r}

SubstituteII

$a_{1} = \frac{1}{2}$ , $r = \frac{1}{1 0}$

S_{\infty} = \frac{\frac{1}{2}}{1 - \frac{1}{1 0}}

▼

Evaluate right-hand side

OneToFrac

Rewrite $1$ as $\frac{1 0}{1 0}$

S_{\infty} = \frac{\frac{1}{2}}{\frac{1 0}{1 0} - \frac{1}{1 0}}

SubFrac

Subtract fractions

S_{\infty} = \frac{\frac{1}{2}}{\frac{9}{1 0}}

DivFracByFracD

$\frac{a / b}{c / d} = \frac{a}{b} \cdot \frac{d}{c}$

S_{\infty} = \frac{1}{2} \cdot \frac{1 0}{9}

MultFrac

Multiply fractions

S_{\infty} = \frac{1 0}{1 8}

ReduceFrac

$\frac{a}{b} = \frac{a / 2}{b / 2}$

S_{\infty} = \frac{5}{9}

UseCalc

Use a calculator

S_{\infty} = 0.555555 \dots

RoundDec

Round to $2$ decimal place(s)

S_{\infty} \approx 0.56

The sum of the geometric series converges to the decimal number

0.56 .

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Catch-Up and Review

Proof

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Proof

Answer

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Tadeo's Savings

Tearrik's Savings

Conclusion

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Solution