The Ratio Test and Root Test ∑ ∑ ∑

The Ratio Test and Root Test Math 102

© Dr. S. Almezel & Dr. H. Alsolumi Page 1

The Ratio Test and Root Test

The Ratio Test provides a method to verify a given series converges absolutely. Unlike most of our previous tests, it applies to any series. Since absolute convergence implies convergence, the Ratio Test may be useful in showing a series converges. The drawback is that it sometimes fails to provide an answer.

The Ratio Test:

Let a_n be a given series. Then 1- If lim ⁿ¹ 1

n n

a L

a

+

→∞ = < , then a_nconverges absolutely (and therefore converges);

2- If lim ⁿ¹ 1

n n

a L

a

+

→∞ = > , orlim ^{n 1}

n n

a a

+

→∞ = ∞, then a_ndiverges;

3- If lim ^{n 1} 1

n n

a a

+

→∞ = , no information for a_nis provided by this test.

The Root Test:

Let a_nbe a given series. Then 1- If limⁿ _n 1

n a L

→∞ = < , then a_nconverges absolutely (and therefore converges);

2- If limⁿ _n 1

n a L

→∞ = > , or limⁿ _n

n a

→∞ = ∞, then a_ndiverges;

3- If limⁿ _n 1

n a

→∞ = , no information for a_nis provided by this test.

Example 1: Determine whether the series

( )

0

1 3

!

n n

n n

∞

=

− is absolutely convergent.

Solution:

( ) ( )

( ) ( )

1 1

1

1 3

`1 ! 3 ! 3

lim lim lim 0 1

3 `1 ! 1

1 3

!

n n

n

n n n

n n n

n n

n n

n

+ +

+

→∞ →∞ →∞

−

+ = = = <

+ +

− .

So by the Ration Test the series

( )

0

1 3

!

n n

n n

∞

=

− converges absolutely.

Example 2: Determine the whether the series

2 3 4 5 6

3 5 7 9 11

0

3 3 3 3 3 3

1 2 2 2 2 2 2

n n

∞ a

=

= − − + + − − +

is convergent.

Solution:

1- Find a_n: 3_{2 1} 2

n

n n

a = ₋ forn≥1.

The Ratio Test and Root Test Math 102

© Dr. S. Almezel & Dr. H. Alsolumi Page 2

2- Compute ^{n 1}

n

a a

+ :

1

1 2 1

2 1

1 1

2 1 2 1

3 3 2 3

lim 2 .

3 2 3 4

2

n

n n

n n n

n n n

n n n

n

a a

a a

+

+ −

+ + +

→∞ +

−

= = = ⋅ =

3- compute lim ^{n 1}

n n

a a

+

→∞ : lim ¹ 3. 4

n

n n

a a

+

→∞ =

4- The series converges since 3 1.

L = <4

Example 3: Determine whether or not the series

1

3 2

n

n

∞ n

=

converges.

Solution:

( ) ( )

1

1

3 3

1 2 1 2

lim lim lim

3 2

n

n

n n n n

n

n n

a

a n

n

+

+

→∞ →∞ →∞

+ + ⋅

= =

3lim 1 3lim 1 1 3 1

2ⁿ 2ⁿ 2

n

n n

→∞ →∞

= + = + = > .

So the series

1

3 2

n

n

∞ n

=

diverges.

Example 4: Determine whether or not the series

1 2 1

n

n

n n

∞

=

−

+ converges.

Solution:

lim lim lim 1 1.

2 1 2 1 2

n

n n n

n n n

n n

a n n

→∞ →∞ →∞

= − = = <

+ +

So by the Root Test, the given series converges absolutely and therefore it converges.

Example 5: Determine whether or not the series ²

1 2 1

n

n

n n

∞

= + converges.

Solution:

2 2

lim lim lim .

2 1 2 1

n

n n n

n n n

n n

a n n

→∞ = →∞ = →∞ = ∞

+ +

So the series ²

1 2 1

n

n

n n

∞

= + diverges.

Problem:

1-14. Use the Ratio test to conclude what you can about the convergence or divergence of

1

( )

n

f n for each given ( )f n :

The Ratio Test and Root Test Math 102

© Dr. S. Almezel & Dr. H. Alsolumi Page 3

1. f n( ) 3ⁿ₃

=n 2. ( ) ( 1) 3 ²

2

n n

f n = − n ⁻

3. ( )

( 1) ⁿ f n n

n e

= + 4. ( ) 7

! n n

f n = n 5.

102

( ) (2 1)!

n

f n = n

− 6.

( 1) 3 1

( ) !

n n

f n n

− −

= 7.

3 2

( ) 2 3

n

f n = n 8. ( ) 2

3

n

f n =n 9. ( ) ( 5 1)₂

1

n

f n n

= −

+ 10. ( ) ² ₂7 1

!

n n

f n n n

+ +

= 11. ( )

3 !

n n

f n n

= n 12. ( ) 5 !

(2 )

n n

f n n

= n

13. ( ) !

1 3 5 (2 1) f n n

= n

⋅ ⋅ ⋅⋅⋅ − 14. ( ) ( 2)

( 1)!

f n n n n

= + +

15-22. Use the Root test to conclude what you can about the convergence or divergence of

1

( )

n

f n for each given ( )f n :

15. f n( ) 1_n

=n 16. f n( ) 2ⁿ₂

=n

17. ( ) 1 1

n

f n = +n 18.

[

^{( 1)}

]

( ) ln( 1)

n

f n n

n

= −

+ 19. ( ) ( 1) ¹

(3 1)

n

n n

f n n

= − +

+ 20. ( )

(2 1/ )

n n

f n n

n n

= +

21. ( ) ( 1) , 2

(ln )

n n n

f n n n

n

= − ≥ 22.

2

( ) 1

n n

f n = n +