9.5
Testing Convergence at
Endpoints
Quick Review
Determine whether the improper integral converges or diverges.
1
1. 
dx
x
x
2. 
dx
x 1
ln x
3. 
dx
x
1  cos x
4. 
dx
x

4
3
1
2

1
3

1

1
2
Quick Review
Determine whether the function is both positive and decreasing
on some interval ( N , ).
5. f ( x)  3 / x
7x
6. f ( x) 
x 8
3 x
7. f ( x) 
3 x
sin x
8. f ( x) 
x
2
2
2
Quick Review Solutions
Determine whether the improper integral converges or diverges.
1
1. 
dx converges
x
x
2. 
dx diverges
x 1
ln x
3. 
dx diverges
x
1  cos x
4. 
dx converges
x

4
3
1
2

1
3

1

1
2
Quick Review Solutions
Determine whether the function is both positive and decreasing
on some interval ( N , ).
5. f ( x )  3 / x
Yes
7x
6. f ( x) 
Yes
x 8
3 x
7. f ( x) 
No
3 x
sin x
8. f ( x) 
No
x
2
2
2
What you’ll learn about







Integral Test
Harmonic Series and p-series
Comparison Tests
Alternating Series
Absolute and Conditional Convergence
Intervals of Convergence
A Word of Caution
… and why
Additional tests for convergence of series are introduced in
this
section.
The Integral Test
Let a be a sequence of positive terms. Suppose that a   f (n),
n
n
where f is a continuous, positive, decreasing function of x for

all x  N (N a positive integer). Then the series  a and the
n N

n
integral  f ( x)dx either both converge or both diverge.
N
Example Applying the Integral
Test
1
Does 
converge?

n 1
n n
The Integral Test applies because f ( x) 
1
x x
is a continuous, positive,
decreasing function of x for x  1.
1
dx  lim  x dx  lim  2 x 

x x
 2

 lim  
 2  2
k


Since the integral converges, so must the series.

1
k
k 
k 
1
3 / 2
1/ 2
k 
k
1
Harmonic Series p-series

1
 is called a p - series, where p is a real constant.
n
 

n 1
p
A harmonic series is the p -series where p =1.
The Limit Comparison Test
(LCT)
Suppose that a  0 and b  0 for all n  N (N a positive integer).
n
n
a
 c, 0  c   , then  a and  b both converge or both diverge.
b
1. If lim
n
n 
n
n
n
2. If lim
n 
a
 0 and  b converges, then  a converges.
b
n
n
n
n
3. If lim
n 
a
  and  b diverges, then  a diverges.
b
n
n
n
n
Example Using the Limit
Comparison
Test
Determine whether the series converges or diverges.
3 5 7
2n  1
   ...  
4 9 16
 n  1

2
n 1
For n large,
2n  1
2n 2
behaves like
 , so compare it to the terms of  (1/ n).
n
n
 n  1
2
 2n  1
a
lim  lim
b
n
n 
n 
n
1
2
 n  1
2
 lim
n 
n
2n  1 n
 2
 n  1 1
ˆ
(Use l'Hopital's
Rule)
2

Since the limit is positive and  (1/ n) diverges, 
n 1
2n  1
also diverges.
 n  1
2
The Alternating Series Test
(Leibniz’s Theorem)
The series   -1 u  u  u  u  ...

n 1
n 1
n
1
2
3
converges if all three of the following conditions are satisfied:
1. each u is positive;
n
2. u  u for all n  N , for some integer N ;
n 1
n
3. lim u  0.
n 
n
The Alternating Series
Estimation Theorem
If the alternating series   1 u satisfies the conditions of

n 1
n
n 1
The Alternating Series Test, then the truncation error for the
nth partial sum is less than u and has the same sign as the
n 1
first unused term.
Rearrangement of Absolutely
Convergent Series
If  a converges absolutely, and if b , b , ..., b ,... is any rearrangement
n
1
2
n
of the sequence a  , then  b converges absolutely and  b   a

n
n
n 1

n
n 1
n
Rearrangement of Conditionally
Convergent Series
If  a converges conditionally, then the terms can be rearranged to
form a divergent series. The terms can also be rearranged to form a
series that converges to any preassigned sum.
n
How to Test a Power Series for
Convergence
1. Use the Ratio Test for find the values of x for which the series converges
absolutely. Ordinarily, this is an open interval a - R  x  a  R.
In some instances, the series converges for all values of x. In rare cases,
the series converges only at x  a.
2. If the interval of absolute convergence is finite, test for convergence or
divergence at each endpoint. The Ratio Test fails at these points. Use a
comparison test, the Integral test, or the Alternating Series Test.
3. If the interval of absolute convergence is a - R  x  a  R, conclude that
the series diverges for x - a  R, because for those values of x the nth term
does not approach zero.
Example Finding Intervals of
Convergence
For
what values of x does the following series converge?
  -1

2n
n 1
n 1
2
4
6
x
x x x
    ...
2n 2 4 6
Apply the Ratio Test to find the interval of absolute convergence, then check
the endpoints if they exist.
2 n2
u
x
2n
lim
 lim

u
2n  2 x
n 1
n 
n 
2n
n
2n
 lim
x  x
2n  2
The series converges absolutely for x  1 or on the interval  -1,1 .
2
2
n 
2

At x  1 and x  -1, the series is 
 -1
n 1
. This converges by the
2n
Alternating Series Test. The interval of convergence is  -1,1 .
n 1
Procedure for Determining
Convergence
Quick Quiz Sections 9.4 and
9.5
You may use a graphing calculator to solve the following problems.
1. Which of the following series converge?
2
I. 
n 1

n 0
2
2 1
II. 
3 1
(A) I only
(B) II only
(C) III only
(D) II and III
(E) I and II

n 1
n
n

III. 
n 1
4
n
n
Quick Quiz Sections 9.4 and
9.5
You may use a graphing calculator to solve the following problems.
1. Which of the following series converge?
2
I. 
n 1

n 0
2
2 1
II. 
3 1
(A) I only
(B) II only
(C) III only
(D) II and III
(E) I and II

n 1
n
n

III. 
n 1
4
n
n
Quick Quiz Sections 9.4 and
9.5
2. Which of the following is the sum of the telescoping series


n 1
2
 n  1 n  2 
(A) 1/3
(B) 1/2
(C) 3/5
(D) 2/3
(E) 1
?
Quick Quiz Sections 9.4 and
9.5
2. Which of the following is the sum of the telescoping series
2
?

 n  1 n  2 

n 1
(A) 1/3
(B) 1/2
(C) 3/5
(D) 2/3
(E) 1
Quick Quiz Sections 9.4 and
9.5
3. Which of the following describes the behavior of the series
ln n
?
  1
n
I. converges II. diverges III. converges conditionally

n 1
(A) I only
(B) II only
(C) III only
(D) I and III only
(E) II and III only
n
Quick Quiz Sections 9.4 and
9.5
3. Which of the following describes the behavior of the series
ln n
?
  1
n
I. converges II. diverges III. converges conditionally

n 1
(A) I only
(B) II only
(C) III only
(D) I and III only
(E) II and III only
n
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