9.5
Testing Convergence at Endpoints
Petrified Forest National Park,
Arizona
Photo by Vickie Kelly, 2007
Greg Kelly, Hanford High School, Richland, Washington
Remember:
The Ratio Test:
an 1
L
If  an is a series with positive terms and lim
n  a
n
then:
The series converges if
The series diverges if
L 1 .
L 1 .
The test is inconclusive if
L  1.

This section in the book presents several other tests or
techniques to test for convergence, and discusses
some specific convergent and divergent series.

Nth Root Test:
If
a
n
is a series with positive terms and lim n an  L
n 
then:
The series converges if
The series diverges if
L 1 .
L 1 .
The test is inconclusive if
L  1.
Note that the
rules are the
same as for
the Ratio Test.

example:
2
n

n2

n
n 1 2
n
n
n
n 
2
2
2

n
lim
lim n  n  n
n
n 
2

2
?

n
lim n
n 
lim n
formula #104
1
n
n 
e
lim ln
n
e
e
1
lim n
n 1
0
e
1
 
1
nn
1
lim ln n
n n
e
formula #103
ln n
lim
n n
Indeterminate, so we use L’Hôpital’s Rule

example:

n2

n
n 1 2
2
n
n
n
n
n 
2
2
2
n
lim
n  2
1

2
2
n 
1
2
n

n
lim
lim n  n  n
n
2

1

2
?
it converges


n
2

2
n
n 1
another example:
2  2
n 2
2
n
n
n
n
lim
n  n
2
n
2
2

2
1

it diverges

Remember that when we first studied integrals, we used a
summation of rectangles to approximate the area under a
curve:
3
2
This leads to:
1
0
1
If
2
3
4
The Integral Test
an  is a positive sequence and an  f  n  where
f  n  is a continuous, positive decreasing function, then:

a
n N
n
and


N
f  x  dx
both converge or both diverge.


Example 1:
Does
n
1
n
n 1


1
1
x x
dx
converge?
b
 lim  x
b  1

3
2
dx
 lim
b 
2 x
1 b

2
1
 2

 2  2
 lim 
b 
 b

Since the integral converges,
the series must converge.
(but not necessarily to 2.)

p-series Test

1
1 1
1
 p  p  p  

p
1 2
3
n 1 n
converges if
p  1 , diverges if p  1 .
We could show this with the integral test.
If this test seems backward after the ratio and nth root
tests, remember that larger values of p would make the
denominators increase faster and the terms decrease
faster.

the harmonic series:

1 1 1 1 1
     

1 2 3 4
n 1 n
diverges.
(It is a p-series with p=1.)
It diverges very slowly, but it diverges.
Because the p-series is so easy to evaluate, we use it
to compare to other series.

Limit Comparison Test
If
an  0
and
bn  0
for all
n  N (N a positive integer)
If
an
lim  c 0  c   , then both  an and  bn
n  b
n
converge or both diverge.
If
an
lim  0 , then
n  b
n
a
converges if
an
If lim
  , then
n  b
n
a
diverges if
n
n
b
n
converges.
b diverges.
n

Example 3a:

3 5 7 9
2n  1
      
2
4 9 16 25


n 1 n  1
2n 2
When n is large, the function behaves like:

2
n
n
2n  1
2
2 1
an
 n  1
 2n  1 n

lim
 lim
 lim
2
n  b
n 
n  
n n
1
n  1
n
n
2n 2  n
2
 lim 2
n  n  2n  1
harmonic series
1
Since 
diverges, the
n
series diverges.

Example 3b:

1 1 1 1
1
       n
1 3 7 15
n 1 2  1
When n is large, the function behaves like:
an
lim
n  b
n
1
2n
1
n
n
2
1
 lim 2  1  lim n
n 
n 2  1
1
n
2
geometric series
1
Since  n converges, the series converges.
2

Alternating Series
Alternating Series Test
Good news!

example:
  1
n 1
The signs of the terms alternate.
If the absolute values of the terms
approach zero, then an alternating
series will always converge!
n 1
1 1 1 1 1 1 1
       
n 1 2 3 4 5 6
This series converges (by the Alternating Series Test.)
This series is convergent, but not absolutely convergent.
Therefore we say that it is conditionally convergent.

Since each term of a convergent alternating series
moves the partial sum a little closer to the limit:
Alternating Series Estimation Theorem
For a convergent alternating series, the truncation
error is less than the first missing term, and is the
same sign as that term.
This is a good tool to remember, because it is easier
than the LaGrange Error Bound.

There is a flow chart on page 505 that might be helpful
for deciding in what order to do which test. Mostly this
just takes practice.
To do summations on the TI-89:
1

8 

2
n1
5
n
becomes
 8*(1/ 2 ^ n, n,1,5)
F3

1

8 

2
n 1
n
becomes
31
4
4
 8*(1/ 2 ^ n, n,1, )
8

To graph the partial sums, we can use sequence mode.
MODE
Y=
Graph…….
4
ENTER
u1   (8*(3/ 4) ^ k , k ,1, n)
ENTER
WINDOW
GRAPH

To graph the partial sums, we can use sequence mode.
Graph…….
MODE
Y=
4
ENTER
u1   (8*(3/ 4) ^ k , k ,1, n)
ENTER
WINDOW
GRAPH
Table

To graph the partial sums, we can use sequence mode.
Graph…….
MODE
Y=
4
ENTER
u1   (8*(3/ 4) ^ k , k ,1, n)
ENTER
WINDOW
GRAPH
Table
p