Section 9.1 – Sequences
Sequence
A sequence {𝑎𝑛 } is a list of numbers written
in an explicit order.
𝑎𝑛 = {𝑎1 , 𝑎2 , 𝑎3 , … , 𝑎𝑛 , … }
First
Term
Second
Term
nth
Term
Generally, we will concentrate on infinite
sequences, that is, sequences with domains that
are infinite subsets of the positive integers.
Recursive Formula
A formula that requires the previous
term(s) in order to find the value of the
next term.
Example: Find a Recursive Formula for the
sequence below.
2, 4, 8, 16, …
𝑎1 = 2
𝑎𝑛 = 2 ∙ 𝑎𝑛−1
Explicit Formula
A formula that requires the number of the
term in order to find the value of the next
term.
Example: Find an Explicit Formula for the
sequence below.
2, 4, 8, 16, …
The Explicit
Formula is also
known as the
General or nth
Term equation.
an   2 
n
Arithmetic Sequences
A sequence which has a constant difference
between terms. The rule is linear.
Example: 1, 4, 7, 10, 13,…
n
a(n)
1
1
2
4
3
7
4
10
5
13
+3
+3
+3
+3
(generator is +3)
Discrete
Explicit Formula
an  3n  2
Recursive Formula
a1  1
an  an 1  3
Sequences typically start with n=1
White Board Challenge
Write an equation for the nth term of the
sequence:
Then find the n=0 term.
n=0
n=1
n=2
n=3
n=4
40, 36, 32, 28, 24, …
–4
First find the generator
a(0) is not in
the sequence!
Do not include
it in tables or
graphs!
an  4n  40
Geometric Sequences
A sequence which has a constant ratio
between terms. The rule is exponential.
Example: 4, 8, 16, 32, 64, …
n
t(n)
1
4
2
8
3
16
4
32
5
64
x2
x2
x2
x2
(generator is x2)
Discrete
0
Explicit Formula
n
n
a  2  2
1
2
3
4
5
6
Recursive Formula
a1  4
an  2  an 1
Sequences typically start with n=1
White Board Challenge
Write an equation for the nth term of the
sequence:
Then find the n=0 term.
n=0
3
,
5
n=1
n=2
n=3
n=4
3, 15, 75, 375, …
a(0) is not in
the sequence!
Do not include
it in tables or
graphs!
x5
First find the generator
3 n
an   5 
5
New Sequences
The previous sequences were the only ones
taught in Algebra 2. But, it is possible for a
sequence to be neither arithmetic nor geometric.
Example: Find a formula for the general term of
the sequence below
n=1 n=2 n=3
n=4
n=5
3 4 5
6
7
,− ,
,
−
,
,…
5 25 125 625 3125
 1
n1
n2
n
5
White Board Challenge
Example: Find a formula for the general term of
the sequence below
n=1 n=2
n=3
n=4
n=5
1 1 1 1
1, , , , , …
3 5 7 9
1
an 
2n  1
Monotonic Sequence
A sequence is monotonic if it is either
increasing (if 𝑎𝑛 < 𝑎𝑛+1 for all 𝑛 ≥ 1) or
decreasing (if 𝑎𝑛 > 𝑎𝑛+1 for all 𝑛 ≥ 1).
Example 1: Find the first 4 terms of 𝑎𝑛 =
to see how the sequence is monotonic.
1
11
,
2
2 1
1
2
2
3
3
31
,
3
4
, , ,
4
5
,
4
4 1
𝑛
𝑛+1
Example 2
Prove the sequence 𝑎𝑛 =
3
𝑛+5
is decreasing.
If the sequence is decreasing, 𝒂𝒏 > 𝒂𝒏+𝟏 for all 𝒏.
3
IF: an 
n5
3
3

THEN: an 1 
 n  1  5 n  6
Since the denominator
is smaller:
3
3

n5 n6
Therefore, 𝒂𝒏 is
decreasing.
OR
an  an 1
Bounded Sequence
A sequence {𝑎𝑛 } is bounded above if there is
a number 𝑀 such that
𝑎𝑛 ≤ 𝑀
for all 𝑛 ≥ 1
A sequence {𝑎𝑛 } is bounded below if there is
a number 𝑚 such that
𝑚 ≤ 𝑎𝑛
for all 𝑛 ≥ 1
If it is bounded above and below, then {𝑎𝑛 } is a
bounded sequence.
Example
Determine if the sequences below bounded below,
bounded above, or bounded.
1. 𝑎𝑛 = 𝑛
Since 𝒏 = 𝟏, 𝟐, 𝟑, … : 𝒂𝒏 ≥1
Therefore, 𝒂𝒏 is
bounded below.
Since lim 𝒏 = ∞ : The sequence is not bounded above.
𝒏→∞
2. 𝑎𝑛 =
𝑛
𝑛+1
Since 𝒏 = 𝟏, 𝟐, 𝟑, … : 𝒂𝒏 >0
Since
𝒏
lim
𝒏→∞ 𝒏+𝟏
= 𝟏 : 𝒂𝒏 <1.
Therefore, 𝒂𝒏 is
bounded.
Limit of a Sequence
A sequence {𝑎𝑛 } has the limit 𝐿 and we write:
lim 𝑎𝑛 = 𝐿
𝑛→∞
or 𝑎𝑛 → 𝐿 as 𝑛 → ∞
if we can make the terms 𝑎𝑛 as close to 𝐿 as
we like by taking 𝑛 sufficiently large.
If lim 𝑎𝑛 exists, we say the sequence
𝑛→∞
converges (or is convergent). Otherwise, we
say the sequence diverges (or is divergent).
Reminder: Properties of Limits
Let b and c be real numbers, let n be a positive integer, and
let f and g be functions with the following limits:
l
i
m
f()
x
L
x

c
l
i
m
g
()
x
K
x

c
Constant Function
lim
bb
Limit of x
lim
xc
Limit of a Power of x
Scalar Multiple
x
c
x
c
lim
x
c
n
n
x

c
l
i
m
b
f
(
x
)

b

L


x

c
Reminder: Properties of Limits
Let b and c be real numbers, let n be a positive integer, and
let f and g be functions with the following limits:
l
i
m
f()
x
L
x

c
Sum/Difference
Product
Quotient
Power
l
i
m
g
()
x
K
x

c
l
i
m
f
(
x
)(

g
x
)

L

K


x

c
l
i
m
f
(
x
)

g
(
xL
)


K


x

c
fx
() L
l
i
m , K

0
xc

g
(
x
) K
l
i
m
()
L
fx

n
x

c
n
Example
Determine if the sequences below converge or diverge. If
the sequence converges, find its limit.
1. 𝑎𝑛 =
2. 𝑎𝑛 =
𝑛
𝑛+1
lim
n 
𝑛
10+𝑛 n 
lim
n
n 1
n
10  n
L. H .
 lim  1
n
1
1
L. H .
 lim 1 101n 1/2
n 
2


 lim 2 10  n  
3. 𝑎𝑛 =
ln 𝑛
𝑛
lim
n
n 
L. H .
ln n
n
Converges
to 1
1
n
 lim 1  0
n
Diverges
Converges
to 0
White Board Challenge
Determine whether the sequence converges or
diverges. If it converges, find its limit.
22 −2 32 −2 42 −2 52 −2
,
,
,
,
…
2
2
2
2
2
3
4
5
2
1) −2
(𝑛 +
𝑎𝑛 =
(𝑛 + 1)2
Converges to
1
Absolute Value Theorem
It is not always possible to easily find
the limit of a sequence. Consider:
1
1 1
−1 𝑛
−1, 2 , −3 , 4 , … , 𝑛 , …
The Absolute Value Theorem states:
If lim 𝑎𝑛 = 0, then lim 𝑎𝑛 = 0.
𝑛→∞
𝑛→∞
Example
Determine if the sequences below converge or diverge.
1. 𝑎𝑛 =
lim
n 
2. 𝑎𝑛 =
−1 𝑛
𝑛
 1n
Because of the Absolute Value
Theorem, Converges to 0
 lim 1n  0
n
n
−1 𝑛
Since the limit does not equal 0, we can
not apply the Absolute Value Theorem. It
does not mean it diverges. Another test is
needed.
lim  1  lim1

1
n
n
n
The sequence diverges since it
does not have a limit: -1,1,-1,1,-1,…
Theorem:
Bounded, Monotonic Sequences
Every bounded, monotonic sequence is
convergent.
Example: Investigate the sequence below.
𝑎1 = 2
𝑎𝑛+1 = 12(𝑎𝑛 + 6)
a1  2
a2  4
a3  5
a4  5.5
a5  5.75
a5  5.85
The sequence appears to be
monotonic: It is increasing.
The sequence appears to be
bounded: 2 ≤ 𝑎𝑛 ≤ 6
The limit of the sequence appears
to be 6.
Since the
sequence
appears to be
monotonic and
bounded, it
appears to
converge to 6.