Ratio and Root Tests
Ratio test
ak 1
Suppose ak  0 and lim
L
k  a
k
If L  1, then
a
If L  1, then
a
k
k
converges.
diverges.
If L  1, then try another test.

1

 
k 2  3 
ak 1
lim
L
k  a
k
k
k 1
lim
k 
1
 
3
k
1
 
3
1

3
So Converges by the Ratio Test
What’s Really Happening?
The ratio test checks to see if the terms
are dropping off by some factor.
This is great for series with an rk or a k!

Ratio Test is no help,
but we know it diverges.
1

k 1 k  4
lim
k 
1
k 5
1
k 4

1

2
k
4
k 1
lim
k 

k 4
lim
k  k  5
1
Ratio Test is no help,
but we know it converges.
1
(k  1)2  4
1
k2  4
k2  4
1
 lim
2
k  (k  1)  4

3k

k 1  k  2  !
3k 1
(k  2)!
lim
k   k  3  !
3k
3k 1
(k  2)...1
lim
k   k  3  ( k  2)...1
3k
3
lim
k  k  3
0
So converges by the Ratio Test
 k  1
 k  1!
k 1
k
k
 k!
lim
k 
 k  1
k!
k
k
k 1
k (k  1)(k  2)...1
lim
k   k  1 k ( k  1)( k  2)...1
kk
lim
k 
 k  1
k
k
k
 k 1 
 1
 lim 
 lim 1  

k 
k 
 k 
 k
k
Diverges by the Ratio Test
k
 e
3 k
k
Find all x>0 so that  x converges.
lim
k 
 k  1
k
3
3
x
xk
k 1
 lim
k 
 k  1
k
3
Try Ratio Test
3
x
 x
So the series converges if 0 ≤ x < 1
and diverges if x > 1.
We must check x=1 separately.
3
k
 diverges, so our answer is [0,1).
3
k
k
|
x

5
|
A better example for future purposes is 
lim
k 
 k  1
k
3
3
| x  5 |k 1
| x  5 |k
 | x 5|
Solve | x  5 |  1
4  x 6
We also have to check x=4 and x=6.
3
k
 Diverges
So our answer is (4,6).
Root test
Suppose ak  0 and lim
k 
If L  1, then
a
If L  1, then
a
k
k
k
ak  L
converges.
diverges.
If L  1, then try another test.
Aside:
The root test is little harder to apply than the ratio test,
but it’s more powerful.
k k
3
 k
lim
k 
k
k
3 k
k
k
 
 lim
k  3
So our series diverges by the root test.
k2
 2k
2
k
lim k k
k 
2
 k
k
 lim
k 
2
2
1

2
Remember lim
k 
k
k  1
So our series converges by the root test.