is geometric, this ratio of successive terms is always the same number; hence, the phrase “common ratio”).

(1)

Example Exercise 1

Remembering the Rule for Geometric Series

In our lecture introducing infinite series, we learned that geometric series converge if the absolute value of the common ratio is less than one. We define the common ratio as the ratio of any term in the argument of the series divided by the previous term, that is, the ratio of

successive terms (if the sequence a

_n

is geometric, this ratio of successive terms is always the same number; hence, the phrase “common ratio”).

In short, if r is the common ratio of a geometric sequence a

_n

, then we define r as below.

1



ⁿ

n

r a a

Moreover, if r  1 , then the geometric series converges.

  1

^1

1





 

 



ⁿ ^k^aaⁿ_n ^k n k

a a

a r

If a

_n

is geometric, when does

^

 



 ⁿ n k

a converge?

(2)

If the ratio a

_n_₁

a

_n

is constant (always the same number), then the sequence a

_n

is geometric.

Otherwise, the sequence is not geometric. Nevertheless, the Ratio Test extends the idea that a sequence b

_n

converges (even if non-geometric) given that the ratio of successive terms b

_n_₁

b

_n

is less than one eventually.

In words, if the ratio of successive positive terms of a

_n

approaches a (obviously non-negative) limit less than one, then the series





 ⁿ n k

a converges. If the ratio of successive terms of a

n

approaches a limit greater than one, then the series





 ⁿ n k

a diverges. If the ratio of successive terms of a

_n

approaches a limit equal to one, then the series





 ⁿ n k

a either converges or diverges.

Warning! If 0   L 1 and lim 

_1



 _n _n



n

a a L , the series





 ⁿ n k

a converges, but it does not necessarily converge to L.

What is the Ratio Test?

Ratio Test Overall Hypotheses:





 ⁿ n k

a is a series with positive terms.

Part 1 Hypothesis: 0 lim 

_1

 1



 _n _n



n

a a

Conclusion:





 ⁿ n k

a converges Part 2

Hypothesis: lim 

_1

 1

 _n _n



n

a a

Conclusion:





 ⁿ n k

a diverges

*Notice the Ratio Test makes no conclusions if lim 

_1

 1

 _n _n



n

a a .

(3)

Example Exercise 3 Converge or Diverge?

The Ratio Test is especially useful for series containing factorial terms as the following work shows.

 

     

 

1

4 1 1 ! 4 1 !

lim lim

4 ! 4

1 !

4 4 1 ! lim

! 4

lim 4



 



 

      

       

   

  

 



  

   

 



n

n n

n n n

n n

 ^{1 !} 

4 ! 4

 



n

n n

   

 

lim 4 1 ! 1 !

lim 4

1 !



 

 

 

 

          

  

n

n n n

n n ^  ⁿ ^ ^{1 !} 

4 4

lim 0



 

 

 

        



n

The series

 

1

4 1 !





 

 

  

 



ⁿ

n

n converges by the Ratio Test.

Does

 

1

4 1 !





 

 

  

 



ⁿ

n

n converge or diverge?

(4)

Look at the limit of the ratio of successive terms of n n

ⁿ

! .

 

   

 

   

 

   

 

1 1

1

1 1 !

lim lim

1 ! ! 1 !

1 1 !

lim

1 !

1 1

lim

1 !

 

 



     

  

   

     

   

   

         

 

  

n n n

n

n n n n

n n n

n n

 n !

   

 

   

1 1 1

lim

1 1 1

lim



 

 

 

   

        

 



n

n n n

n n

n

n n

 ⁿ ^ ¹ 

 



¹



1 lim 1

lim 1

lim 1 by



 



 

 

  

  

 

 

    

          

 



n

n n n

n n

n n n

n n

n

e definition

The series

1

!





 

 

 



ⁿ

n

n diverges by Ratio Test.

Does

1

!





 

 

 



ⁿ

n

n converge or diverge?

(5)

Example Exercise 5 Converge or Diverge?

Look at the limit of the ratio of successive terms of ² ⁵ 5



n n

.

1 1

1

2 5 2 5 2 5 5

lim lim

5 5 5 2 5

2 5 5

lim

5 5 2 5

2 5

lim 5

 

 









      

  

    

   

  

       

 

n n n n

n n

n n n n

5 5 



n

 

1

5 2 5 2 5 2 5 2

2 5

1 2 5

lim

5 2 5

2 2

1 lim

5 2 1

1 2 lim

5 1





 

 

  

 

  

      

   

 

       

  

     

n

n n n

n

2 5

1 2 0

5 1 0



  





The series

1

2 5 5





  

 

 



ⁿ ⁿ

n

converges by the Ratio Test.

Does

1

2 5 5





  

 

 



ⁿ ⁿ

n

converge or diverge?

(6)

Notice the similarities this theorem has with the Ratio Test What is the Root Test?

Root Test Overall Hypotheses: Given





 ⁿ n k

a , there exists some integer N such that n  N  a

_n

 0 . Part 1

Hypothesis: ⁰ ^ ^lim

_n_

 

ⁿ

^a

ⁿ

^ ¹

Conclusion:





 ⁿ n k

a converges Part 2 Hypothesis: ^lim

_n_

 

ⁿ

^a

ⁿ

^ ¹

Conclusion:





 ⁿ n k

a diverges

*Notice the Root Test makes no conclusions if ^lim

_n_

 

ⁿ

^a

ⁿ

^ ¹ ^.

(7)

Example Exercise 7 Converge or Diverge?

Look at the limit of the nth root of n

²

5

ⁿ

.

 

2 2 2

1

2

lim lim lim lim

5 5 5 5

   

     

   

     

     

n n

n n n

n n n n

n n n

n

Let  ^lim



 

ⁿ ²

N

n

n . Take the natural log of both sides and recall the limit property

   

lim f g  f lim g assuming the outputs of g are in the domain of f.

   

 

2

H

0

2

ln ln lim

ln lim ln

ln 2 lim ln ln 2 lim 1

ln 0

1 lim



  

  

 

 

 

 

   

  

  

   

  

 

  



n

n n

n n n

n

n n

N n

n

N n

N

e N

n

2

Looking at a table of the terms of as

may be just as convincing as this symbolic argument, but, alas, we eschew buttons.

 

n

n n

Thus, ^lim



₅

²

¹ ₅ ^lim



 

²

¹ ₅ ¹ ¹ ₅

 

    

 

 

n n

n n n

n n , and the series

2

1

5

ⁿ

n



n



 

 

 

 converges by the Root Test.

Does

2

1

5





 

 

 



ⁿ

n

n converge or diverge?

(8)

Look at the limit of the nth root of

5 7

7 9

  

  

 

n

n .

 

7 2 7

7 7

1 5 1 7

5

7 1 7 9

7 7

lim lim lim 0

9 9 1

  

          

              

          

 

n

n n n

n

n n n

n n

The series

5 7 1

7 9





    

   

    

 



ⁿ

n

n converges by Root Test.

Does

5 7 1

7 9





    

   

    

 



ⁿ

n

n converge or diverge?

(9)

#1 lim 

_1

 0 converges by Ratio Test

 _n _n

 

n

a a #5 lim 

_1

 diverges by Ratio Test

 _n _n

 

n

a a e

#7 ^lim   ¹ diverges by Root Test

 ⁿ _n

  

_e

n

a

^

#9 ^lim   ² ¹ diverges by Root Test

 ⁿ _n

  

n

a

#11 ^lim   ^{0 1} converges by Root Test

 ⁿ _n

  

n

a

Practice Problems in Calculus: Concepts and Contexts by James Stewart 1 ^st ed. problem set: Not available

2 ^nd ed.& 3 ^rd ed. problem set: Not available

Practice Problems in Calculus: Early Transcendentals by Briggs and Cochran

1st ed. problem set: Section 8.5 #9-25 odd Practice Problems

Determine if the following series converge or diverge.

#1

1

10 !





 

 

 



ⁿ

n

n #2

1

2 !





   

  

n

#3 Explain why the Ratio Test is inconclusive for

₃

1

3





 

 

 



n

n .

#4 Explain why the Ratio Test is inconclusive for

1





 

 

 



n

n .

#5 

²



1







 ⁿ n

n e #6  

²

1

1 4





  

 

  

 



ⁿ

n

n n

#7

2 2 1

1 1





    

    

   

 



ⁿ

n

n en

 #8  

1







 ⁿ n

ne

#9

₂

1

2





 

 

 



ⁿ

n

n #10

1





   

   

 

 

 



ⁿ ⁿ

n

n e

#11

1



1



   

     

 

 



ⁿ

n

n #12

2

 

1 ln





 

 

 



ⁿ

n