* 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
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
+
= + =
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
= −
= −
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
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
−
=
−
=