Section 8.4: Comparison Tests
Direct comparison test
(Baby comparison test)
Suppose 0 ≤ ak ≤ bk for all k
If
 b converges,
If
a
k
k
then
diverges, then
 a converges.
k
b
k
diverges.
Aside:
To use any of the Comparison Tests,
the terms must be positive.
1
1 , but
 2
k
k
1
 k diverges !
ln k
 k
ln k
1
and

k
k
1
 k diverges
Diverges by the Direct Comparison Test.

3

k
4
5
k 1
3
1
 3 
k
4 5
4
k
and

k
1
3   converges
4
Converges by the Direct Comparison Test.


k 2
1
4 k 3
1
1

4 k 3
4 k
and

Diverges by the Direct Comparison Test.
1
4 k
diverges
But
The baby test doesn’t really help much with

1
4 k 3
1
1

4 k 3
4 k
Limit Comparison Test
Suppose bk , ak > 0 and
ak
lim
is not zero or infinity,
k  b
k
then
a
k
and
b
k
either both converge or both diverge

1
k 3

Looks like
1
k
1
lim k  3
k 
1
k
So

1
k 3
The thing


and
1
k
1
k
 lim
k 
k
k 3
 1
do the same thing
does is diverge!
So

1
diverges by the limit comparison test.
k 3
3
 4k  3
Looks like
1
 3  4 
k
3
k
5
4
lim
k
k 
1
3 
4
So
3
 4k  3
1
and  3  
4
4k
 lim k
k  4  5
 1
k
do the same thing
k
1
The thing  3   does is converge !
4
3
So  k
converges by the limit comparison test.
4 3
2k 3  3k
 3k 4  5
Looks like

1
k
2k 3  3k
3
4
2
k
 3k
3
k

5
lim
 lim
k 
k 
1
3k 4  5
k
So
2k 3  3k
 3k 4  5
The thing
and
1
k
1
k
k
1
2

3
do the same thing
does is diverge!
2k 3  3k
So  4
diverges by the limit comparison test.
3k  5
3 k 1
 k 2  3k
Looks like

1
k 3/2
3 k 1
3/2
2
3
k

1
k
lim k  3k  lim 2
k  k  3k
k 
1
1
k 3/ 2
So
3 k 1
 k 2  3k
The thing
k
and
k
1
3/2
1
3/2
 3
do the same thing
does is converge!
So

3 k 1
diverges by the limit comparison test.
k 2  3k
Zero Infinity test
(Common Sense Comparison Test)
Suppose bk, ak > 0
ak
If lim  0 and
k  b
k
b
k
converges, then
ak
  and  bk diverges, then
If lim
k  b
k
a
k
a
k
converges.
diverges.
1
Compare with  23 ?
k
ln k
 k3
ln k
3
k
lim
k  1
k2
ln k k 2
 0
 lim
3
1
k  k
ln k
1
So  3 is smaller than  2
k
k
1
And  2 Converges
k
ln k
So  3 Converges by the Zero-Infinity Test.
k
Log-q Series
ln k
 kq
converges if and only if q>1
(just like p-series)