7.2 Series Learning Objectives

CA7-20 Sequences and Series

7.2 Series

Learning Objectives

Use summation notation.

Find the sum of a finite arithmetic sequence.

Solve applications of arithmetic series.

Find the value of an infinite geometric series with a finite sum.

Find the sum of a finite geometric sequence.

Solve applications of geometric series.

As we get older, we benefit or suffer from our accumulated behavior; we may enjoy retirement if we have

saved yearly or we may suffer from diseases related to being overweight if we have consumed too many

calories daily. If a life is a sequence of events, then the future is influenced by the accumulation of those

events. In mathematics, the sum of numbers in a sequence is called a series. Adding finitely many of

those numbers results in a finite series, while adding infinitely many of those numbers is an infinite

series. We express a series, or the sum of a sequence, using the following notation.

Summation Notation

Definition 7.3. Summation Notation: Consider the sequence na .

The sum of the terms na , from n j to n p , can be written as 1

p

n j j p

n j

a a a a

.

The sum 1 2j j ja a a is written as 1 2n j j j

n j

a a a a

.

The symbol Σ is the Greek capital letter sigma and is read as ‘sigma’ or ‘summation’. The letter n is

the index. The value j is called the lower limit and p is called the upper limit.

Note that in the upper limit of the summation p

n

n j

a

we did not write n p , since n being the index is

clear from the lower limit of the summation. It is, hower, acceptable to write the upper limit at n p if

you prefer. Also note that we will refer to the value of a series, if it exists, as a sum. The word sum may

refer to the value of a finite or infinite summation.

7.2 Series CA7-21

Example 7.2.1. Find the following sums.

1. 6

2

3 5n

n

2. 4

1

52

100nn

3.

1

4

0

12

1

k

k

k

xk

Solution.

1. To evaluate the series 6

2

3 5n

n

, we must find the sum of the sequence generated by the formula

3 5na n , where the first term has 2n and the last term has 6n . We can do this by finding

each term and adding them together.

6

2

3 5 3 2 5 3 3 5 3 4 5 3 5 5 3 6 5

1 4 7 10 13

35

n

n

52 3 4 6a a a a a

2. For 4

1

52

100nn

, we add the sequence generated by 52

100n n

a , starting with 1n and going to 4n .

4

1 2 3 41

52 52 52 52 52 52 52 52 52

100 100 100 100 100 100 10000 1000000 100000000nn

1 2 3 4a a a a

After obtaining a common denominator and simplifying, we have 52525252

0.52525252100000000

.

3. The series

1

4

0

12

1

k

k

k

xk

is asking us to add the sequence

11

21

k

k

ka xk

, where we start

with 0k and end with 4k , resulting in the following.

0 1 1 1 2 1 3 1 4 1

0 1 2 3 41 1 1 1 12 2 2 2 2

0 1 1 1 2 1 3 1 4 1x x x x x

0 1 2 3 4a a a a a

After simplifying, we have

1

42 3 4

0

1 1 1 1 12 1 2 2 2 2

1 2 3 4 5

k

k

k

x x x x xk

In each of the problems in the previous example, we were given the formula for na , along with index

values for the first and last term, and we evaluated the first and last terms along with the terms in between

the two. Now we will go in the opposite direction. In other words, we start with the terms and generate

the summation notation. We make use of our knowledge of arithmetic and geometric sequences, looking


for a common difference or common ratio as we go from term to term to determine the type of sequence.

Once we know what type of sequence we are dealing with, we can refer to the formulas from Section 7.1.

Example 7.2.2. Write the following sums using summation notation. Assume the terms in each sum

result from an arithmetic or a geometric sequence.

1. 5 2 1 22 2. 5 5 5

52 4 8

3. 5 5 5 5

52 4 8 1024

Solution.

1. For 5 2 1 22 , we first determine a general formula for its terms. Since the difference

between the first and second term is 3, and the difference between the second and third term is 3,

we will assume the pattern is arithmetic. We still need to find a formula for the terms and the

position number for the last term, 22. For the formula, we use the formula for an arithmetic

sequence, 1 1na a n d , with 1 5a and 3d , to get

5 1 3

3 8

n

n

a n

a n

Now, to determine the value of n for the term 22, we set 22na in the formula we just found.

3 8

22 3 8

10

na n

n

n

In summation notation, 10

1

5 2 1 22 3 8n

n

.

2. Again, we need to determine if the pattern of 5 5 5

52 4 8

is arithmetic or geometric. We

quickly recognize that it is not arithmetic since there is no common difference. There is, however,

a common ratio of 1

2 so it is geometric. The lack of a last term, indicated by ‘ ’, indicates we

are adding infinitely many terms. We simply need to find the formula for na . Using the formula

for a geometric sequence, 1

1

n

na a r , along with 1 5a and 1

2r , we find

11

52

n

na

.

Finally, using summation notation,

1

1

5 5 5 15 5

2 4 8 2

n

n

7.2 Series CA7-23

3. The terms in the series 5 5 5 5

52 4 8 1024

are the same as in part 2 of this example.

However, this time there is a last term, 5

1024. From part 2, we have

1?

1

15

2

n

n

, so we need only

to determine the position of 5

1024. Checking different n values, we see that

15 1

51024 2

n

holds for 11n .

111

1

5 5 5 5 15 5

2 4 8 1024 2

n

n

Although the summation notation is new, it is just a way of writing addition for many terms that are each

generated in the same way. Because it is just addition, the usual properties of addition hold. For

example, the order of addition does not matter. Below, we list properties of summation, using

mathematical notation, and noting that p may represent an integer or infinity. In the case where p ,

these properties hold if each infinite sum is defined. We will discuss certain infinite sums later. They will

be discussed extensively in Calculus.

Properties of Summation

p p p

n n n n

n j n j n j

a b a b

1

p ph

n n n

n j n j n h

a a a

, for any integer j h p

p p

n n

n j n j

ca c a

, for any constant c

p p h

n n h

n j n j h

a a

, for any integer h (if p , replace p h with )

We now turn our attention to finding sums, limiting our sums to arithmetic and geometric series. We

begin with arithmetic series.

Arithmetic Series

There is a story often told in math classes about Carl Friedrich Gauss (1777 – 1855), one of the world’s

most famous mathematicians, when he was about 9 years old. The legend goes like this: Gauss’s teacher


often gave Gauss and his classmates tedious arithmetic tasks. On one such occasion, the teacher asked

students to add all of the integers from 1 to 100:

1+2+3+4+ +97+98+99+100

All of the students, except Gauss, added the numbers one at a time. When the teacher asked him why he

wasn’t working on the problem, Gauss replied that he already knew the answer, 5050. The teacher was

astonished and asked how he knew the answer without doing the computation. Gauss explained that if

you add the first term and the last term you get 101; then add the second term and the second to the last

term, you again get 101; and then the third and third to last term, again 101. As a matter of fact, you

always get 101 if you continue in this manner. There are a total of 50 sums of 101 (a sum is made up of a

pair of terms), so the sum is 50 101 5050 .

Figure 7.2. 1

In other words, we are taking the first term plus the last term and then multiplying that sum by half the

number of terms in the series. If we have n terms, this is equivalent to 12

n

na a . Before moving on,

let’s quickly check that this works for an odd number of terms, say 3 4 5 6 7 .

Figure 7.2. 2

With these five terms, we find 5

3 7 252

, since the middle term is half the sum of the first and last

terms, so the formula 12

n

na a appears to work for an odd number of terms as well, bringing us to the

following theorem.

4+97=101

3+98=101

2+99=101

1+100=101

1 + 2 + 3 + 4 + + 97 + 98 + 99 + 100

3+7=10

4+6=10

5=1/2 of 10

3 + 4 + 5 + 6 + 7

7.2 Series CA7-25

Theorem 7.2. Arithmetic Sum:

Consider the arithmetic sequence ka with first term 1a and common difference d . Then

1 1ka a k d , 1, 2, 3,k , and the sum of the first n terms, 1 2

1

n

n k n

k

S a a a a

, is

1 12 12 2

n n

n nS a a a n d , 2n

As verification of this theorem, consider the finite arithmetic sequence 1 1ka a k d , 1, 2, ,k n

and its sum 1 1 1 1 1

1

1 2 1n

n

k

S a k d a a d a n d a n d

. To determine the

value of the sum, we write the sum twice, the second time in reverse order, and add the two.

1 1 1 1

1 1 1 1

1 1 1 1

2 1

1 2

2 2 1 2 1 2 1 2 1

n

n

n

S a a d a n d a n d

S a n d a n d a d a

S a n d a n d a n d a n d

The right side is the sum of n copies of 12 1a n d , so

1

1

2 2 1

2 12

n

n

S n a n d

nS a n d

Using the fact that the last, or nth term in the sequence is 1 1na a n d , we can also write this sum as

12

n n

nS a a .

Example 7.2.3. Find the value of the arithmetic3 sum 12

1

4 5k

k

.

Solution. We could write out each of the terms and then add them all up, but this would be tedious, so

we choose to use the formula. We can see that there are 12 terms, so 12n . We proceed with finding

the value of the first and the last term by setting 1k and 12k , respectively.

1

12

1: 4 1 5 1

12 : 4 12 5 43

k a

k a

The sum is 12

121 43 252

2S .

3 To verify that this sum is arithmetic, we find that we have a common difference, i.e. 1 4k ka a .


Example 7.2.4. The first term of an arithmetic sequence is 15 and its 10th term is −12. Find the sum

of its first 10 terms. Find the sum of its first 20 terms.

Solution. We are given 1 15a and

10 12a . The sum of the first 10 terms is

10

1015 12

2

15

S

To find the sum of the first 20 terms, we need to find the value of 20a . We start by finding the value of

d . Using 1 1na a n d with 1 15a , 10 12a and 10n , we have

12 15 10 1

27 9

3

d

d

d

Then 20 15 3 20 1 42a . With this and the sum formula, we find

20

2015 42

2

270

S

Example 7.2.5. The first term of an arithmetic sequence is −12 and its common difference is 5. How

many terms of this sequence must we add to get a sum of 345?

Solution. We are given 1 12a and 5d , and we are looking for n so that 345nS where

12

n n

nS a a . By substituting 1 1na a n d , we have

1 1

1

12

2 12

n

nS a a n d

na n d

Now we set 345nS , 1 12a and 5d to solve for n .

2

2 12 1 5 3452

24 5 5 690

5 29 690 0

nn

n n

n n

7.2 Series CA7-27

229 29 4 5 690

2 5

29 14641

10

29 121

10

n

n

n

We get 9.2n or 15n . Since n must be a whole number, we find 15n and conclude that adding

15 terms in the sequence will result in 345.

Applications of Arithmetic Series

Example 7.2.6. On the Sunday after a minor surgery, Margaret is able to walk a half-mile. Each

Sunday, she walks an additional quarter mile. After 8 weeks, what will be the total number of miles she

has walked?

Solution. This problem can be modeled as an arithmetic series with 1

1

2a and

1

4d . We are

looking for the total number of miles walked after 8 weeks, so we know that 8n and that we are

looking for 8S . Using the formula for nS , with 8n , we have everything we need to find the sum of

miles walked except the value of 8a . We find 8a using the formula 1 1na a n d .

8

1 1 98 1

2 4 4a

We can now use the formula for an arithmetic sum to determine the number of miles Margaret walked.

8

8 1 911

2 2 4S

Margaret walked a total of 11 miles.

Before moving on to geometric series, we ponder the results of summing infinitely many arithmetic

terms. Suppose Gauss had been asked to add up all of the natural numbers.

1 2 3 4 5

This would mean adding progressively larger and larger numbers or, in other words, bigger and bigger

terms. The sum just keeps getting larger. This sum is said to ‘go to infinity’ or diverge. This always

happens for arithmetic sequences where 1 0a , even if 0d . For example, if 1 1a and 0d we have

1 1 1 . We leave it to the reader to think about whether summing infinitely many arithmetic terms

can ever result in a finite number.


Geometric Series

We begin by adding a finite number of terms of the form 1

1

k

ka a r , where 1, 2, ,k n . The sum of

these terms is 1 2 2 1

1 1 1 1 1 1

1

nk n n

n

k

S a r a a r a r a r a r

. To find the value of nS , we use a bit of

arithmetic; we multiply the terms of the series by r , subtract the result from nS to give us n nS rS , then

proceed to solve for nS .

2 2 1

1 1 1 1 1

2 2 1

1 1 1 1 1

1 1

n n

n

n n n

n

n

n n

S a a r a r a r a r

rS a r a r a r a r a r

S rS a a r

Thus,

1 1

1 1

11 1

1

1

1 1

n

n n

n

n

nn

n

S rS a a r

S r a a r

a ra a rS

r r

Theorem 7.3. Geometric Sum (finite number of terms):

Consider the geometric sequence ka with first term 1a and common ratio r , 1r . Then

1

1

k

ka a r , 1, 2, 3,k , and the sum of the first n terms, 1 2

1

n

n k n

k

S a a a a

, is

1 1

1

n

n

a rS

r

In the case when 1r , 1ka a and 1 1 1 1 1

times

n

n

S a a a a na .

Example 7.2.7. Find the value of the sum 5 5 5 5

52 4 8 1024

.

Solution. This is a finite geometric sum with 1 5a and 1

2r , since

5 15

2 2

and

5 5 1

4 2 2

. We need only to know the number of terms, n , to find the sum with the formula in

Theorem 7.3. To determine the value of n , we use the formula 1

1

n

na a r with 1 5a and 5

1024na .

7.2 Series CA7-29

1

1

1010 11 1

sin

5 15

1024 2

1 1

1024 2

1 1

2 2ce

2 1024

n

n

n

We have 1 10 11n n . We may now use the formula 1 1

1

n

n

a rS

r

with 11n ,

1 5a and

1

2r to find the sum.

11

11

1 15 1 5 12 10245 2 34152048

31 2048 3 10241

22

S

An interesting question, to be explored further in Calculus, is ‘when is it possible to add an infinite

number of numbers and get a finite sum?’ We talked about adding an infinite number of terms in an

arithmetic sequence. Now we examine the sum of infinitely many geometric terms. Suppose you add all

of the terms starting with 1

2 where each subsequent term is half of the previous term:

1 1 1 1

2 4 8 16 .

As an illustration, in the following diagram the indicated square has a side length of 1, and we sum areas

within the square as we move from left to right.

Figure 7.2. 3

1

2

Figure 7.2. 4

1 1

2 4

Figure 7.2. 5

1 1 1

2 4 8

Figure 7.2. 6

1 1 1 1

2 4 8 16

Notice that we keep adding smaller and smaller pieces and that the ‘pieces’ seem to be filling in a

‘missing piece’. The sum appears to be getting closer and closer to 1, and indeed the sum is 1.

1

2

1

2

1

4

1

2

1

4 1

8

1

2

1

4 1

8

1

16


While the geometric series 1 1 1 1

12 4 8 16 , this is obviously not the case for all geometric series.

For example, the geometric series 1 2 4 8 diverges, much like the arithmetic series we discussed

earlier. As it turns out, the series 1

1

1

k

k

a r

has a finite value when 1 1r ; we say the series

converges. For all other values of r , the infinite geometric series will not have a finite value. As to what

happens and why, we leave that discussion to your Calculus course.

Theorem 7.4. Geometric Series (infinite number of terms):

Consider the geometric sequence na with first term 1a and common ratio r , 1 1r . Then

1

1

k

ka a r , 1, 2, 3,k , and the sum of this infinite geometric series, 1 2 3

1

k

k

S a a a a

, is

1

1

aS

r

The sum 1

1

1

k

k

a r

is not defined for 1r , assuming 1 0a .

Example 7.2.8. Find the value of the series, if it exists.

1. 5 5 5

52 4 8

2.

1

1

2 3

3 2

k

k

3.

1

1

3 2

2 3

k

k

Solution.

1. The series 5 5 5

52 4 8

is geometric with 1 5a and 1

2r . Because 1r , the series will

converge to a value. We use the formula in Theorem 7.4 to find that value.

5 5 5 5 5 10

51 32 4 8 3

12 2

2. The series

1

1

2 3

3 2

k

k

is written in summation notation. Since 3

2 is the number being raised to a

power, this means we are multiplying each term by 3

2 to get to the next term. This tells us that the

ratio is 3

2, which is larger than the absolute value of 1, so the series will diverge.

7.2 Series CA7-31

3. Again, the series is written in summation notation. However, for

1

1

3 2

2 3

k

k

, we have 2

3r ,

which has an absolute value less than 1, so the series will converge to a value. We find 1 1

1

3 2 3

2 3 2a

and use the formula from Theorem 7.4 to find the sum.

1

1

33 2 3 3 92

22 3 2 1 21

3

k

k

Applications of Geometric Series

A rational number is a real-valued number that is the ratio of two integers or, in decimal notation, has a

decimal expansion that ends or repeats. The following application of geometric series allows us to write a

repeating decimal as a ratio of two integers.

Example 7.2.9. Write the rational number 4.52 4.525252 as a ratio of two integers.

Solution. We begin by noting that the decimal portion of this number can be written as a geometric

series, to which we may apply the geometric series formula.

1

geometric series

4.52 4.525252

4 0.525252

4 0.52 0.0052 0.000052

52

10041

1100

52 10.52 and 0.01

100 100a r

We find 52 448

4.52 499 99

.

An interesting fact that can be shown with the method used in Example 7.2.9 is that 0.9 0.999 1 .

Try it on your own!

An important applicaton of the geometric sum formula is the investment plan called an annuity.

Annuities differ from the kind of investments we studied in Section 4.5 in that payments are deposited

into the account on an on-going basis. In the following example, we look at an ordinary annuity, which


requires equal payments made at the end of consecutive periods, each having the same length. To find the

value of an annuity, we must sum all of the payments and the interest earned.

Example 7.2.10. A deposit of $50 is made at the end of each month in a savings account that offers

6% annual interest, compounded monthly. Find the value of the savings account after 30 years.

Solution.

We begin by assessing the values, at the end of 30 years, of the individual monthly deposits of $50. In

Section 4.5, we used the compound interest formula

1

n tr

A Pn

for principal P , annual interest rate

r , and n compoundings per year over a period of t years. In this example, we have 0.06r , 12n ,

and 50P , so 12

120.0650 1 50 1.005

12

tt

A

. Since 12t is the number of months that compounding

takes place, and payments are made at the end of each month, our initial deposit of $50 will gain interest

over 12 30 1 359 months; the second deposit of $50 will gain interest over 358 months, the third over

357 months, etc.

Deposit # Months Compounding Resulting Value

1 359 359

50 1.005

2 358 358

50 1.005

3 357 357

50 1.005

360 0 0

50 1.005

The sum of these values that result from monthly deposits is

359 358 357 0

0 1 2 359

50 1.005 50 1.005 50 1.005 50 1.005

50 1.005 50 1.005 50 1.005 50 1.005

This is the sum of a geometric sequence with 1 50a , 1.005r and 360nt .

360

360

1 1 using form

50 1 1.005

1 1.005

50225

ula 1

.75

n t

nt

a r

rS S

Thus, the account contains $50,225.75 after 30 years.

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7.2 Series Learning Objectives