Sequences and Series
I.
General Terms
A sequence of numbers, is simply a list of numbers, generated by some rule.
Examples
A. 3,6,9,12,……
B. 2,8,18,32,…...
Each term in a sequence can be referred to by it's place in the sequence, i.e. first
term, third term, nthterm. In the examples above
A. The 4th term is 12
B. The 2nd term is 8
The rule used to generate a sequence, is often described by referring to the nth term.
In the examples above the nth terms are as follows. More on how to find the nth
term will be discussed later.
A. The nth term is 3n so the 8th term is 24
B. The nth term is 2n2 so the 4th term is 32
A sequence can develop in 4 ways
Divergent
The terms keep growing
Convergent
The terms converge on a
single value, in this case 0.
The sequence repeats itself
after a set number of terms.
The sequence oscillates
between 2 values.
Periodic
Oscillating
If you add the terms of a sequence together, you get a series
A. 3 + 6 + 9 +12 + 15
B. 2 + 8 + 18 + 32
Arithmetic Sequences and Series
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I.
Arithmetic Sequence
Sequences of numbers that follow a pattern of adding a fixed number from one
term to the next are called arithmetic sequences.
Definition
A sequence with general term
an+1 = an + d
is called an arithmetic sequence.
an = nth term and d = common
difference
Examples
Find the general (nth) term for the following arithmetic sequences:
A.
B.
C.
2,6,10,14,18,22, ...
-5,-3,-1,1,3,...
1,4,7,10,13,16,...
Solution
All of these have one thing in common. To get to the next term we add a fixed
number (d). Let a1 = 1st term and an = nth term
A.
a1 = 2
a2 = 2 + 4 = 6
a3 = 2 + 2(4) = 10
a4 = 2 + 3(4) = 14
and so on .
an = a1 + (n – 1)4
If we let d = 4 this becomes an = a1 + (n – 1)d
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Definition
The nth or general term of an arithmetic sequence is given by
an = a1 + (n - 1)d
So in our example a1 = 2 and d = 4 so an = 2 + (n – 1)4 = 2 + 4n – 4 = 4n - 2
B.
a1 = -5 and d = 2
an = -5 + (n – 1)(2) = -5 + 2n – 2 = 2n - 7
C.
a1 = 1 and d = 3
an = 1 + (n – 1)(3) = 1 + 3n – 3 = 3n - 2
Exercises: Find an
1.
II.
-1,10,21,32,43,54,...
2.
3,0,-3,-6,-9,-12,...
The Arithmetic Series
The following theorem provides us with an easy way to calculate the arithmetic
series.
Theorem
If
an = a1 + (n - 1)d
is an arithmetic sequence then the sum of the sequence is
Sn 
n
n
(a1  an )  [2a  (n  1)d ]
2
2
This can be proven but let’s just convince ourselves that it works.
It is easy to determine the sum of the following arithmetic sequence:
3 + 5 + 7 = 15 This is an arithmetic series with common difference 2
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Now let’s use the formula
We have a1 = 3, an = 7, d = 2
Sn 
3
(3  7)  15
2
Examples:
Find the following sums
A.
B.
3 + 7 + 11 + 15 + ... + 35
-2 + 1 + 4 + 7 + …+ 25
Solutions:
A. We have
a1 = 3, an = 35, d = 4
To find n we note that
35 = 3 + (n - 1)4
so that
32 = (n - 1)4 and n = 9
Now we are ready to use the formula
9
(3  35)  171
2
B. We have a1 = -2, an = 25 and n = 10
Sn 
So that S10 
10
(2  25)  5(23)  115
2
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Application
Suppose that you play black jack at Harrah's on June 1 and lose $1,000. Tomorrow
you bet and lose $15 less. Each day you lose $15 less that your previous loss.
What will your total losses be for the 30 days of June?
Solution
This is an arithmetic series with
a1 = 1000
and
d = -15
We can calculate
a30 = 1000 - 15(30 - 1) = 565
Now we use the formula
S30 = 30/2 (1000 + 565) = 23,475
You will lose a total of $23,475 during June.
Exercises: Find the sums
3.
5 + 10 + 15 +.... + 500
4.
3 + 6 + 9 + .... + 99
5.
-5 + -15 + -25 + -35 + ... + -95
6.
What is the sum of the numbers 1 to 100?
7.
Find the sum of the first 27 terms of the series that starts 7 + 3 – 1 - 5
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Geometric Sequences and Series
I.
Geometric Sequence
Sequences of numbers that follow a pattern of multiplying a fixed number from one
term to the next are called geometric sequences.
Definition
A sequence with general term
an+1 = an r
is called an geometric sequence.
an = nth term and r = common ratio
Examples
Find the general (nth) term for the following geometric sequences:
A. 2,6,18,54, ...
B. 27,9,3,1,…..
C. 16,-8,4,-2,1,...
Solution
All of these have one thing in common. To get to the next term we multiply a fixed
number (r). Let a1 = 1st term and an = nth term
A. a1 = 2
a2 = 2(3) = 6
a3 = 2(3)2 = 18
a4 = 2(3)3 = 54
and so on .
an = a1 (3)n-1
If we let r = 3 this becomes an = a1 rn-1
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Definition
The nth or general term of an geometric sequence is given by
an = a rn-1 where a is the 1st term
C.
a = 27 and r = 1/3
an = 27(1/3)n-1 = 33(31)(1/3)n = 81(1/3)n
D.
a = 16 and d = -1/2
an = 16(-1/2)n-1 = 24(-2)1(-1/2)n = (-2)5(-1/2)n
Exercises: Find an
8. 1, 2, 4, 8, 16, 32,,...
9. 24, 12, 6, 3, 3/2, 3/4, …
10. 1/2, -1, 2, -4, 8, ...
II.
The Finite Geometric Series
The following theorem provides us with an easy way to calculate the arithmetic
series.
Theorem
If
an = a rn-1
is a geometric sequence then the sum of the sequence is
n
1 r n
S n   an  a
i 1
 1 r



This can be proven but let’s just convince ourselves that it works.
It is easy to determine the sum of the following geometric sequence:
3 + 6 + 12 = 21 This is a geometric series with common ratio 2
Page 7
Now let’s use the formula
We have a = 3, an = 12, r = 2
 1  23 
7
  3
S n  3
 3(7)  21
1
 1 2 
Examples:
Find the following sums
A. 1st 5 terms of -6 + 18 – 54 + ….
B. 5 + 10 + 20 + 40 + ... + 2560
Solutions:
A.
We have a = -6, r = -3 and n = 5
 1  (3)5 
  6 244   366
So that  2(3)   6
 1  (3) 
 4 
r 1


5
B.
n
We have a = 5, r = 2, an = 2560
To find n we use an = a rn-1
2560 = 5(2)n-1
512 = (2)n-1
512 = 2n2-1
512 = 2n(1/2)
1024 = 2n
210 = 2n
n = 10
Now we are ready to use the formula
 1  210   1  1024 
  5
S10  5
  5(1023)  5115
 1 2   1 
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Exercises: Find the sums
11.
12.
13.
14.
First 20 terms of 2 + 4 + 6 + ….
3 - 6 + 12 - .... - 96
1 1 1
First 15 terms of  1     ......
2 4 8
1 + 1/3 + 1/9 + .... + 3 (1/3)12
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III.
The Infinite Geometric Series
Theorem
If
an = a rn-1
is a geometric sequence and r  1 then the sum of the infinite
sequence is

S n   an 
i 1
a
1 r
Examples:
Find the following sums
A.  2  4  8  16  .....
B. 24 + 12 + 6 + 3 + 3/2 + ¾ + …….
Solutions:
A. We have a = -2, r =-2
This infinite series diverges because r = -2 and  2 is not less than 1. There is no
sum.
B. We have a = 24, r = 1/2
So that S n 
24
 48
1  12
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Application
How much is going to taxes? Suppose that we track a tax refund of $100. Each time
money is spent 8% goes towards taxes and the rest gets spent again. How much of the
original $100 will go back to taxes?
Solution
a=8
r = 0.92
(The next amount to be taxed is 92% less than the current amount)
a
S
=
1-r
8
=
8
=
1 - 0.92
= $100
0.08
Hence all of the refund will eventually find its way back to the government coffers.
Exercises: Find the sums if possible
15.
3 + 1 + 1/3 + 1/9 + ……..
16.
28 – 14 + 7 - 7/2 …….
Page 11