12. 1 A sequence is…
(a) an ordered list of objects.
1 1 1 1
1, , , ,
...
2 4 8 16
(b) A function whose domain is a set of integers.
Domain: 1, 2, 3, 4, …,n…
Range
a1, a2, a3, a4, … an…
{(1, 1), (2, ½), (3, ¼), (4, 1/8) ….}
Finding patterns
Describe a pattern for each sequence. Write
a formula for the nth term
1 1 1 1
1, , , ,
...
2 4 8 16
1
2
n1
1 1 1 1
1, , ,
,
...
2 6 24 120
1
n!
1 4 9 16 25
, , ,
,
...
4 9 16 25 36
n2
( n  1)
2
Write the first 5 terms for
1 2 3 4 n 1
0, , , , ...
...
2 3 4 5
n
On a number line
n 1
an 
n
As a function
The terms in this sequence get closer and closer
to 1. The sequence CONVERGES to 1.
Write the first 5 terms
(1)n 1  n  1
an 
n
1 2
3 4
n 1
0,  , ,  , .    ...
...
2 3
4 5
n
The terms in this sequence do not get close to
Any single value. The sequence Diverges
Write the terms for an = 3
The terms are 3, 3, 3, …3
The sequence converges to 3.
y= L is a horizontal asymptote when
sequence converges to L.
A sequence that diverges
(1)n 1  n  1
an 
n
Sequences
Write the first 5 terms of the sequence.
Does the sequence converge? If so, find the value.
(1) n 1
an 
2n  1
1 1 1 1
1,  , ,  ,
3 5 7 9
(1)n 1
lim n 
0
2n  1
The sequence converges to 0.
an  (1)
n 1 
1
1 2 3 4
1


 0,  ,  ,  ,
n


2 3 4 5
The sequence diverges.
n
1
limn (1) 1   does not exist
 n
12.2 Infinite Series

1
1 1 1
1
 2n  2  4  8  16  ...
n 1
Represents the sum of the terms in a sequence.
We want to know if the series converges to
a single value i.e. there is a finite sum.

1 1  1  1  1  ...
n 1
The series diverges because sn = n. Note that the
Sequence {1} converges.

1
 n(n  1)
n 1

1
1 1 1
1
1
 n(n  1)  2  6  12  20  30  ...
n 1
Partial sums of
1
1

1 2 2
1
1
2
s2 


1 2 2  3 3
1
1
1
3
s3 



1 2 2  3 3  4 4
s1 

1
1 1 1
1
1
 n(n  1)  2  6  12  20  30  ...
n 1
and
sn 
1
1
1
1
n


 ...

1 2 2  3 3  4
n(n  1) n  1
If the sequence of partial sums converges,
the series converges
1 2 3 4 5
n
, , , , ...
... Converges to 1 so series converges.
2 3 4 5 6 n 1
Finding sums

1
 n(n  1)
n 1
Can use partial fractions to rewrite

1
 n(n  1) 
n 1

1 
1
1 1
1 1
1
1
1
  n  n  1   (1  2 )  ( 2  3)  ( 3  4 )  ... n  n  1  ...
n 1

1
1
 n(n  1)  limn (1  n  1)  1
n 1
. he partial sums of the series
T
Limit

1
 n(n  1)
n 1
Geometric Series

1
 2n
n 1
1 1 1
1
   
 ...
2 4 8 16
Each term is obtained from the preceding
number by multiplying by the same number r.
Find r (the common ratio)
1 1
1
1



 ...
5 25 125 625
2 4 8 16
  
 ...
3 3 3
3

 ar
n 1
n 1
Is a Geometric Series
Where a = first term and r=common ratio
Write using series notation
n 1

11
1 1 1
1
 2  2 
  
 ...
2 4 8 16
n 1
3 12
48 192



 ...
5 25 125 625
2 4 8 16
  
 ...
3 3 3
3

3 4 
 5   5 
n 1

n 1
2 n 1
 3  2
n 1
The sum of a geometric series
sn  a  ar  ar  ar  ...ar
2
3
rsn  ar  ar  ar  ...ar
2
3
n
n 1
Multiply each term by r
sn  rsn  a  ar
n
n
a  ar
a(1  r )
sn 

,r 1
1 r
1 r
n
if | r | 1,
Sum of n terms
subtract
r  0 as n  .
a
Geometric series converges to sn  1  r , | r | 1
If r>1 the geometric series diverges.
n
Find the sum of a Geometric Series
Where a = first term and r=common ratio

 ar
n 1

11
 2  2 
n 1

n 1
3 4 
 5   5 
n 1
n 1
n 1
a

,
1 r
| r | 1
1
2 1
1
1
2
3
5 31
4 9 3
1
5

2 n 1
 3  2
n 1
The series diverges.
Repeating decimals-Geometric Series
0.080808 
a

8
102
10
2

and r 
n 1
ar

n 1
8

8
10
4

8
10
6

8
8
10
 ...
1
102
8  1 
 2  2 
n 110  10 
n 1
8

2
a
8
n 1
10
ar




1 r 1 1
99
n 1
102
The repeating decimal is equivalent to 8/99.
Series known to converge or diverge
1. A geometric series with | r | <1 converges
2. A repeating decimal converges
3. Telescoping series converge
A necessary condition for convergence:
Limit as n goes to infinity for nth term
in sequence is 0.
nth term test for divergence:
If the limit as n goes to infinity for the nth
term is not 0, the series DIVERGES!
Convergence or Divergence?

n  10
 10n  1
n 1

 1.075 
n 1

1
1
n  n2
n 1

n
4
 2n
n 1
A sequence in which each term is less than or
equal to the one before it is called a monotonic
non-increasing sequence. If each term is greater
than or equal to the one before it, it is called
monotonic non-decreasing.
A monotonic sequence that is bounded
Is convergent.
A series of non-negative terms converges
If its partial sums are bounded from above.
12. 3 The Integral Test
Let {an} be a sequence of positive terms.
Suppose that an = f(n) where f is a continuous
positive, decreasing function of x for all xN.
Then the series and the corresponding integral
shown both converge of both diverge.
f(n)

 an
n N
f(x)

 f ( x)dx
N
The series and the integral both converge or both diverge
Area in rectangle corresponds to term in sequence
Exact area under curve is between
If area under curve is finite, so is area in rectangles
If area under curve is infinite, so is area in rectangles
Using the Integral test

n
 n2  1
an  f ( n) 
n 1

n
n 1
2
f ( x) 
x
x2  1
b
b
1 2x
2


dx

lim
dx

lim
ln(
x

1)
b   2
b  
 x2  1

1
2
x

1
1
1
x
limb (ln(b  1)  ln 2)  
2
The improper integral diverges
Thus the series diverges
Using the Integral test

1
 n2  1
n 1

1
an  f ( n) 
b
1
n 1
2
f ( x) 
1
x2  1
1
 x2  1 dx  limb  x2  1 dx  limb arctan x
1
1
  
limb  (arctan b  arctan1)   
2 4 4
The improper integral converges
Thus the series converges
b
1
Harmonic series and p-series

1
 n p Is called a p-series
n 1
A p-series converges if p > 1 and diverges
If p < 1 or p = 1.

1
1 1 1 1
1

1





....
 ....
n
2 3 4 5
n
n 1
Is called the harmonic series and it
diverges since p =1.
Identify which series converge and which diverge.


1
1
n 1 3
n


n 1


n 1
1
n
1

n3

1
n
n 1


n 1

100
n2
3 4 
 5   5 
n 1
n 1
12. 4Direct Comparison test

Let

 an
 an be a series with no negative terms
n 1
Converges if there is a series
n 1

 cn
n 1
Where the terms of an are less than or equal to
the terms of cn for all n>N.

 an Diverges if there is a series
n 1

 dn
n 1
Where the terms of an are greater than or equal
to the terms of dn for all n>N.
Limit
Comparison
test
Limit Comparison test
an
lim x 
 c,
bn
0c


Then the following series  an and  bn
n 1
n 1
both converge or both diverge:
an
lim x 
0
bn


bn Converges then  an Converges
and Amd
n 1
n 1

an
lim x 

bn Diverges then

and
bn
n 1

 an Diverges
n 1
Convergence or divergence?

1
 2  3n
n 1


n 1

1
3n  2
n
 n2  1
n 1
Alternating Series
A series in which terms alternate in sign

 (1)
n 1

 (1)
n
n 1
(

1)
an

n 1
n 1
1
2
n 1

or
an

n
1 1 1
1
   
 ...
2 4 8 16
1
1 1 1
 (1) n   1  2  3  4  ...
n 1
n
Alternating Series Test

n 1
(

1)
an  a1  a2  a3  a4  ...

n 1
Converges if:
an is always positive
an an+1 for all n  N for some integer N.
an0
If any one of the conditions is not met, the
Series diverges.
Absolute and Conditional Convergence

• A series  an is absolutely convergent if the
n N

corresponding series of absolute values  an
n N
converges.
• A series that converges but does not converge
absolutely, converges conditionally.
• Every absolutely convergent series converges.
(Converse is false!!!)
Is the given series convergent or divergent? If it is
convergent, its it absolutely convergent or
conditionally convergent?


(1)
n 1


n 1
n ( n 1) / 2
n
3
(1)
n 1
( n  1)
n

( 1)
 ln(n  1)
n 1
n

( 1)
 n
n 1
n
a) Is the given series convergent or divergent? If
it is convergent, its it absolutely convergent or
conditionally convergent?


(1)
n 1
n ( n 1) / 2
n
3
1 1
1
1
  

 ...
3 9 27 81
This is not an alternating series, but since


n 1
(1) n ( n 1) / 2
n
3


1
 3n
n 1
Is a convergent geometric series, then the given
Series is absolutely convergent.
b) Is the given series convergent or divergent? If
it is convergent, its it absolutely convergent or
conditionally convergent?
n

(1)
1
1
1
 ln(n  1)   ln 2  ln 3  ln 4  ......
n 1
Converges by the Alternating series test.


n 1
(1)
1
1
1



 ......
ln(n  1)
ln 2 ln 3 ln 4
n
Diverges with direct comparison with the harmonic
Series. The given series is conditionally convergent.
c) Is the given series convergent or divergent? If
it is convergent, its it absolutely convergent or
conditionally convergent?


n 1
(1)
n 1
(n  1)
2 3 4 5
    
n
1 2 3 4
By the nth term test for divergence, the series
Diverges.
d) Is the given series convergent or divergent? If
it is convergent, its it absolutely convergent or
conditionally convergent?

(1)
 n
n 1
n
1
1
1
1




1
2
3
4
Converges by the alternating series test.


n 1
(1)
n
n
1
1
1
1




1
2
3
4
Diverges since it is a p-series with p <1. The
Given series is conditionally convergent.
The Ratio Test

Let  an be a series with positive terms and
n N
an 1
lim n

an
Then
• The series converges if ρ < 1
• The series diverges if ρ > 1
• The test is inconclusive if ρ = 1.
The Root Test

Let  an be a series with non-zero terms and
n N
limn n | an |  L
Then
• The series converges if L< 1
• The series diverges if L > 1 or is infinite
• The test is inconclusive if L= 1.
Convergence or divergence?

n
2
 n!
n 1

n
3
 n2 2n 1
n 1


n 1
e
2n
nn
.
Procedure for determining Convergence
Power Series (infinite polynomial in x)

n
2
n
c
x

c

c
x

c
x

.....
c
x
...
 n
0
1
2
n
n 0
Is a power series centered at x = 0.
and

n
2
n
c
(
x

a
)

c

c
(
x

a
)

c
(
x

a
)

.....
c
(
x

a
)
...
 n
0
1
2
n
n 0
Is a power series centered at x = a.
Examples of Power Series

xn
x2
x3
 n!  1  x  2  3!  ...
n 0
Is a power series centered at x = 0.
and

(1) n
n 0
3n

( x  1) n  1 
1
1
1
( x  1)  ( x  1) 2  ..... n ( x  1) n ...
3
9
3
Is a power series centered at x = -1.
Geometric Power Series

x
n
 1  x  x  x  x  ...x
2
3
4
n 0
a  1 and r  x
a
1
S 

,
1 r 1 x
P1  1  x
P2  1  x  x 2
P3  1  x  x 2  x 3
x 1
n
.
The graph of f(x) = 1/(1-x) and four of its
polynomial approximations
Convergence of a Power Series
There are three possibilities
1)There is a positive number R such that the
series diverges for |x-a|>R but converges for
|x-a|<R. The series may or may not converge
at the endpoints, x = a - R and x = a + R.
2)The series converges for every x. (R = .)
3)The series converges at x = a and diverges
elsewhere. (R = 0)
What is the interval of convergence?

x
n
 1  x  x  x  x  ...x
2
3
4
n 0
Since r = x, the series converges |x| <1, or
-1 < x < 1. In interval notation (-1,1).
Test endpoints of –1 and 1.

 (1)
n 0

 (1)
n0
n
n
Series diverges
Series diverges
n
Geometric Power Series
Find the function

(1) n
n 0
3n

1
1
1
2
( x  1)  1  ( x  1)  ( x  1)  ..... n ( x  1) n ...
3
9
3
n
1
a  1 and r   ( x  1)
3
a
1
3
3
S 



1  r 1  1 ( x  1) 3  ( x  1)
4 x
3
Find the radius of convergence
1
r   ( x  1)
3
1
 ( x  1)  1
3
2  x  4
Geometric Power Series

Find
ofof
convergence
Find the
the radius
interval
convergence n
0
(1) n
3n
( x  1) n
For x = -2,


(1) n
n 0
n
3
(2  1) 
n


(1) n ( 1) n
n 0
n
3


1
 3n
n 0
Geometric series with r < 1, converges
For x = 4


n 0
(1) n
n
3
(4  1) n 


n 0
(1) n ( 3) n
n
3


3n
 3n
n 0


1
n 0
By nth term test, the series diverges.
Interval of convergence 2  x  4
Finding interval of convergence
lim n 
and

n 0
Use the ratio test:
xn
un 
n

un 1
n
x
n
x n 1

n 1
n 1
un 1
x
n
 lim n 
 x
n
un
n 1 x
x 1
For x = 1

1
n
n0
Harmonic series
diverges
R=1
(-1, 1)
For x = -1
Interval of convergence

(1) n
 n
n 0
Alternating Harmonic series
converges
[-1, 1)
Differentiation and Integration of Power Series

If the function is given by the series
n
2
n
c
(
x

a
)

c

c
(
x

a
)

c
(
x

a
)

.....
c
(
x

a
)
...
 n
0
1
2
n
n 0
Has a radius of convergence R>0, the on the
interval (c-R, c+R) the function is continuous,
Differentiable and integrable where:
f ( x ) 

n 1
nc
(
x

a
)
 n
n 0


( x  a) n
f ( x)dx  C   cn
n 1
n 0
The radius of convergence is the same but the
interval of convergence may differ at the endpoints.
Constructing Power Series
If a power series exists has a radius of convergence = R
It can be differentiated
f ( x)  c0  c1( x  a)  c2 ( x  a)2  .....cn ( x  a)n ...
f ( x)  c1  2c2 ( x  a)  3c3 ( x  a)2.....ncn ( x  a) n 1...
f ( x)  2c2  2 *3c3 ( x  a)  3* 4( x  a) 2....
f ( x)  1* 2 *3c3  2 *3* 4c4 ( x  a)  3* 4 *5( x  a) 2  ...
So the nth derivative is
f
( n)
( x)  n!cn  terms with factor of
( x  a)
Finding the coefficients for a Power Series
f ( n) ( x)  n!cn  terms with factor of
( x  a)
All derivatives for f(x) must equal the series
Derivatives at x = a.
f ( a )  c1
f ( a )  1* 2c2
f ( a )  1* 2 * 3c3
f
( n)
(a)  n!cn
f
(n)
(a)
 cn
n!
If f has a series representation centered at x=a, the
series must be


f
k 0
(a)
f ( a )
f ( a )
 f ( a )  f ( x  a ) 
( x  a)2 
( x  a )3...
k!
2!
3!
(k )

f
(n)
(a) n
x  ...
n!
If f has a series representation centered at x=0, the
series must be


k 0
f
(0)
f (0)
f (0)
 f ( a )  f (0) 
( x  a)2 
( x  a )3...
k!
2!
3!
(k )

f
(n)
(0) n
x  ...
n!
Form a Taylor Polynomial of order 3 for

sin x at a =
4
(n)
(n)
(n)
n
f (x)
0
sin x
1
cos x
2
-sin x
3
-cos x
f (a)
f (a)/n!
2
2
2
2
2
2
2
2
2

2

2
2 * 2!
2
2

2
2 * 3!

The graph of f(x) = ex and its Taylor
polynomials
Find the derivative and the integral

n
2
3
4
x
x
x
x
 n!  1  x  2  3!  4!  ...
n 0


n 0
 1
n
2 n 1
x
(2n  1)!
x3
x5
x7
 x


...
3!
5!
7!
Taylor polynomials for f(x) = cos (x)
Converges only at x = 0