* Use summation notation. * Use the formula for the sum of the first n terms of an arithmetic series. * Use the formula for the sum of the first n terms of a geometric series. * Use the formula for the sum of an infinite geometric series. * Solve annuity problems.

1. 9.4 Series and Their Notations
Chapter 9 Sequences, Probability, and Counting
Theory

2. Concepts and Objectives
⚫ The objectives for this section are
⚫ Use summation notation.
⚫ Use the formula for the sum of the first n terms of an
arithmetic series.
⚫ Use the formula for the sum of the first n terms of a
geometric series.
⚫ Use the formula for the sum of an infinite geometric
series.
⚫ Solve annuity problems.

3. Series
⚫ A series is the indicated sum of the terms of a sequence.
⚫ The sum of part of a series is called a partial sum.
⚫ The sum of the first n terms of a series is called the nth
partial sum of that series. It is usually represented by Sn.
⚫ Example: For the sequence 3, 5, 7, 9, …, find S4.
= + + + =
4 3 5 7 9 24
S

4. Series (cont.)
⚫ Series are usually written using summation notation.
We use the Greek letter  (sigma) to represent this.
A finite series is an expression of the form
An infinite series is an expression of the form
The letter k is called the index of summation.
=
= + + + = 
1 2
1
...
n
n n k
k
S a a a a


=
= + + + + = 
1 2
1
... ...
n k
k
S a a a a

5. Series (cont.)
⚫ Example: Evaluate
=

5
2
1
k
k

6. Series (cont.)
⚫ Example: Evaluate
This notation tells us to find the sum of k2 from k = 1 to
k = 5. We find the terms of the series by substituting
k = 1, 2, 3, 4, 5 into the function k2 and then add the
terms to find the sum.
=

5
2
1
k
k
5
2 2 2 2 2 2
1
1 2 3 4 5
1 4 9 16 25
55
k
k
=
= + + + +
= + + + +
=


7. Arithmetic Series
⚫ An arithmetic series is a series (sum) which results from
adding the terms of an arithmetic sequence.
⚫ While it is possible to find a partial sum by writing down
all of the terms and adding them up, it would easier to
find a formula for Sn.

8. Arithmetic Series (cont.)
⚫ Consider the arithmetic sequence 7, 13, 19, 25, 31, …
What if we wanted to find the sum of the first 100
terms?
We can find the 100th term using the formula from
before:
Since d = 6, we can work backwards from 601 to find the
last few terms as well, so our sum looks like:
( )( )
100 7 100 1 6 601
a = + − =
= + + + + + + + +
100 7 13 19 25 ... 583 589 595 601
S

9. Arithmetic Series (cont.)
Now, what do you notice if we add the terms together
starting from the outside and working inside?
= + + + + + + + +
100 7 13 19 25 ... 583 589 595 601
S

10. Arithmetic Series (cont.)
Now, what do you notice if we add the terms together
starting from the outside and working inside?
We have 50 sums (100  2), so our sum is
50(608)=30,400. This leads us to our formula.
= + + + + + + + +
100 7 13 19 25 ... 583 589 595 601
S
608
608
608
608

11. Arithmetic Series (cont.)
⚫ The formula for the nth partial sum of an arithmetic
series is
( )
( )
1
1 or
2 2
n
n n n
n a a
n
S a a S
+
= + =

12. Arithmetic Series (cont.)
⚫ Example: Find S33 for the series with a1 = 9 and d = –2.

13. Arithmetic Series (cont.)
⚫ Example: Find S33 for the series with a1 = 9 and d = –2.
First, we need to find a33:
Now, we can use the formula:
( )( )
33 9 33 1 2
55
a = + − −
= −
( )
33
33 9 55
759
2
S
−
= = −

14. Partial Sums
⚫ Example: The sum of the first 17 terms of an arithmetic
sequence (S17) is 187. If , find a1 and d.
17 13
a = −

15. Partial Sums
⚫ Example: The sum of the first 17 terms of an arithmetic
sequence (S17) is 187. If , find a1 and d.
17 13
a = −
( )
17 1 17
17
2
S a a
= +
( )
1
17
187 13
2
a
= −
( )
1
374 17 13
a
= −

16. Partial Sums
⚫ Example: The sum of the first 17 terms of an arithmetic
sequence (S17) is 187. If , find a1 and d.
17 13
a = −
( )
17 1 17
17
2
S a a
= +
( )
1
17
187 13
2
a
= −
( )
1
374 17 13
a
= −
1
1
374 17 221
17 595
a
a
= −
=

17. Partial Sums
⚫ Example: The sum of the first 17 terms of an arithmetic
sequence (S17) is 187. If , find a1 and d.
17 13
a = −
( )
17 1 17
17
2
S a a
= +
( )
1
17
187 13
2
a
= −
( )
1
374 17 13
a
= −
1
1
374 17 221
17 595
a
a
= −
=
1 35
a =

18. Partial Sums
⚫ Example: The sum of the first 17 terms of an arithmetic
sequence (S17) is 187. If , find a1 and d.
17 13
a = −
( )
17 1 17
17
2
S a a
= +
( )
1
17
187 13
2
a
= −
( )
1
374 17 13
a
= −
1
1
374 17 221
17 595
a
a
= −
=
1 35
a =
( )
17 1 17 1
a a d
= + −
13 35 16d
− = +

19. Partial Sums
⚫ Example: The sum of the first 17 terms of an arithmetic
sequence (S17) is 187. If , find a1 and d.
17 13
a = −
( )
17 1 17
17
2
S a a
= +
( )
1
17
187 13
2
a
= −
( )
1
374 17 13
a
= −
1
1
374 17 221
17 595
a
a
= −
=
1 35
a =
( )
17 1 17 1
a a d
= + −
13 35 16d
− = +
16 48
3
d
d
= −
= −

20. Using Summation Notation
⚫ Example: Evaluate ( )
10
1
4 8
i
i
=
+


21. Using Summation Notation
⚫ Example: Evaluate
This sum contains the first 10 terms of the arithmetic
sequence having
First term
and
Last term
( )
10
1
4 8
i
i
=
+

( )
1 12
4 1 8
a = + =
( )
10 8
4 10 8 4
a = + =

22. Using Summation Notation
⚫ Example: Evaluate
This sum contains the first 10 terms of the arithmetic
sequence having
First term
and
Last term
Thus,
( )
10
1
4 8
i
i
=
+

( )
1 12
4 1 8
a = + =
( )
10 8
4 10 8 4
a = + =
( ) ( ) ( )
10
0
1
1 4
4 8 5 60 3 0
1
1
0
2
2 8
0
i
i S
=
+ = = + = =


23. Using Summation Notation (cont.)
⚫ To convert an arithmetic series into summation
notation:
⚫ Identify the common difference, d.
⚫ Subtract d from the first term; call this a0.
⚫ The formula that goes in the summation notation will
be a0 + dk.
⚫ Example: Write using summation
notation.
⚫ d = 8; a0 = 10 – 8 = 2:
10 18 26 162
+ + + +
( )
1
2 8
n
k
k
=
+


24. Using Summation Notation (cont.)
⚫ Example: Write using summation
notation.
⚫ d = 8; a1 = 10; a0 = 10 – 8 = 2
⚫ To find n:
⚫ Therefore,
10 18 26 162
+ + + +
( )
20
1
2 8
k
k
=
+

( )( )
( )
162 10 1 8
8 1 152
1 19
20
n
n
n
n
= + −
− =
− =
=

25. Arithmetic Series (cont.)
⚫ Example: Find the sum of the series.
In this case, we know a1 and an, but we don’t know n.
We can see from the sequence that d = –5.
20 15 10 50
+ + + + −
( )( )
( )
50 20 1 5
70 5 1
1 14
15
n
n
n
n
− = + − −
− = − −
− =
=

26. Arithmetic Series (cont.)
⚫ Example: Find the sum of the series.
Now, we can substitute n into the formula:
20 15 10 50
+ + + + −
( )
15
15 20 50
225
2
S
−
= = −

27. Geometric Series
⚫ The nth partial sum of a geometric series is given by the
formula
⚫ For some reason, I’m always tempted to try to factor the
fraction further. It doesn’t factor.
 
−
=  
−
 
1
1
1
n
n
r
S a
r

28. Geometric Series (cont.)
⚫ Example: Find S34 for the geometric series with a1 = 7
and r = 1.03.

29. Geometric Series (cont.)
⚫ Example: Find S34 for the geometric series with a1 = 7
and r = 1.03.
Using the formula, we have:
 
−
=  
−
 
34
34
1 1.03
7
1 1.03
S
 404.111

30. Geometric Series (cont.)
⚫ Example: Find the sum of the geometric series .
To use the formula, we have to interpret the summation
notation. a1 is the value when k = 1, so
If we compare the formula in the summation notation
with the geometric sequence formula ( ), we
can see that r = 2. So,
6
1
3 2k
k=


1
1 3 2 6
a =  =
1
1
n
n
a a r −
=
( )
6
6
6 1 2
378
1 2
S
−
= =
−

31. Geometric Series
⚫ Example: 50238.14 is the approximate value of a partial
sum in the geometric series with a1 = 150 and r = 1.04.
Which term is it?

32. Geometric Series
⚫ Example: 50238.14 is the approximate value of a partial
sum in the geometric series with a1 = 150 and r = 1.04.
Which term is it?
 
−
=  
−
 
1
1
1
n
n
r
S a
r
 
−
=  
−
 
1 1.04
50238.14 150
1 1.04
n

33. Geometric Series
⚫ Example: 50238.14 is the approximate value of a partial
sum in the geometric series with a1 = 150 and r = 1.04.
Which term is it?
 
−
=  
−
 
1
1
1
n
n
r
S a
r
 
−
=  
−
 
1 1.04
50238.14 150
1 1.04
n
( )( )
−
= −
50238.14 0.04
1 1.04
150
n
=
1.04 14.39683...
n
1–1.04 = –0.04

34. Geometric Series
(cont.) Taking the log of each side gets n out of the
exponent.
So n = 68. (n always has to be a natural number.)
=
log1.04 log14.39683...
n
=
log1.04 log14.39683...
n
= =
log14.39683...
68.000001...
log1.04
n

35. Convergent Geometric Series
⚫ It should be obvious that the partial sums of a geometric
sequence such as the last example will continue to
increase as n increases.
⚫ Now, let’s look at a different sequence:
The first six partial sums would look like:
1 1 1 1
2, 1, , , , , ...
2 4 8 16
=
1 2
S =
2 3
S =
3
7
2
S =
4
15
4
S =
5
31
8
S =
6
63
16
S

36. Convergent Geometric Series
⚫ If we were to graph this sequence of
partial sums, we can see that it
approaches the line y = 4.
⚫ Using some algebra, we can transform
the series:
−
 
 
−
 
 
 
 
 
= = − 
   
−
 
 
2
1
1
1
2
2 4
1 2
1
2
n
n
n
S

37. Convergent Geometric Series
⚫ With this rewritten formula, we can see that as n
increases, (½)n-2 gets closer and closer to 0. (Check out
the value of ½ raised to larger and larger powers.)
⚫ Therefore, we say that the limit of Sn as n increases
without bound (or approaches infinity) is 4 or
⚫ In order for the common ratio term to go to 0 as n
increases, the denominator of the partial sums formula
must be a proper fraction. That is, . This is called a
convergent geometric series. A series that does not
converge diverges.
→
=
lim 4
n
n
S
1
r

38. Convergent Geometric Series
⚫ The formula for the sum of a convergent geometric
series is
Example: In our previous sequence, a1 = 2 and r = ½:
= 
−
1
, where 1
1
a
S r
r
= = =
−
2 2
4
1 1
1
2 2
S

39. Solving Annuity Problems
⚫ An annuity is an investment in which the purchaser
makes a sequence of periodic, equal payments. The
value of the annuity will be a geometric series.
⚫ Given an initial deposit and an interest rate, to find the
value of an annuity:
⚫ Determine a1, the value of the initial deposit.
⚫ Determine n, the number of deposits.
⚫ Determine r – 1 + annual rate  number of time
compounded
⚫ Substitute into Sn formula and simplify.

40. Solving Annuity Problems (cont.)
⚫ Example: A deposit of $100 is placed into a college fund
at the beginning of every month for 10 years. The fund
earns 9% annual interest, compounded monthly. How
much is in the account after 10 years?
a1 = 100; n = 10  12 = 120;
0.09
1 1.0075
12
r = + =
( )
120
120
100 1 1.0075
1 1.0075
19,351.43
S
−
=
−
=

41. Classwork
⚫ College Algebra 2e
⚫ 9.4: 6-24 (even); 9.3: 18-30 (even); 9.2: 44-54 (even)
⚫ Quiz 9.3