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9.4 Series and Their Notations
Chapter 9 Sequences, Probability, and Counting
Theory
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.
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
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
Series (cont.)
⚫ Example: Evaluate
=

5
2
1
k
k
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
=
= + + + +
= + + + +
=

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.
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
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
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
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
+
= + =
Arithmetic Series (cont.)
⚫ Example: Find S33 for the series with a1 = 9 and d = –2.
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
−
= = −
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 = −
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
= −
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
= −
=
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 =
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
− = +
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
= −
= −
Using Summation Notation
⚫ Example: Evaluate ( )
10
1
4 8
i
i
=
+

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 = + =
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
=
+ = = + = =

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
=
+

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
= + −
− =
− =
=
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
− = + − −
− = − −
− =
=
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
−
= = −
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
Geometric Series (cont.)
⚫ Example: Find S34 for the geometric series with a1 = 7
and r = 1.03.
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
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
−
= =
−
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?
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
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
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
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
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
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
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
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.
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
−
=
−
=
Classwork
⚫ College Algebra 2e
⚫ 9.4: 6-24 (even); 9.3: 18-30 (even); 9.2: 44-54 (even)
⚫ Quiz 9.3

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9.4 Series and Their Notations

  • 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