The Comparison Tests and limit comparison Test

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The Comparison Tests and limit comparison Test Math 102

The Comparison Tests and limit comparison Test

There are two comparison tests. One is called The Comparison Test and the other one is called The Limit Comparison Test. Both tests require one to choose another series to which the given series is

compared. During the course of the class, we will learn about certain well known series that you can use in this regard. So far we have learned about the p-series, which includes the Harmonic series. So these comparison tests work well when the given series looks a little like a p-series.

The Comparison Test:

Let a_nand b_nbe two series of positive terms. Then

If b_nconverges anda_n ≤b_nfor all n sufficiently large, then a_nalso converges.

If b_ndiverges anda_n ≥b_nfor all n sufficiently large, then a_nalso diverges.

The Limit Comparison Test:

Let a_nand b_nbe two series of positive terms. If lim ⁿ

n n

a C

b

→∞ = ,

where C is finite and positive, then either both series converge or both series diverge.

There are several observations that one must keep in mind in applying these Comparison tests.

a. Both tests apply to series of positive terms. Actually, if either of the series have a finite number of negative terms, then the tests still apply because a finite number of terms does not change the convergence or divergence of any series.

b. In general, one will be given a series, say a_n. One then chooses the series b_nfrom your

knowledge of previously studied series. The p-series is often a good choice, but you will need to pick a particular value for p.

c. In applying these tests, one must verify that the conditions of the tests are satisfied. In particular, for the Comparison Test one must verify the appropriate inequality holds.

d. In applying the Limit Comparison Test, if lim ⁿ 0

n n

a b

→∞ = , then no information about convergence or divergence is obtained and you have to choose some other series with which to compare.

e. Neither of these tests provide an estimate of the error as the Integral Test does.

Example 1: Determine whether or not the series ₃

12 3 2

n

n n

∞

= + + converges or diverges.

Solution: Applying the Comparison Test:

The series ₂

1

1 2

n n

∞

=

converges since it is a constant multiple of the p-series with p=2.

3 3 2

1

2 3 2 2 2

n n

n n ≤ n = n

+ + for all n≥1. So by the Comparison Test, the series ₃

12 3 2

n

n n

∞

= + + converges.

- Applying the Limit Comparison Test:

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The series ₂

1

n 2n

∞

=

3 3

3

2 2 3

2 2

2 3 2

lim 1 lim2 3 2 lim2 3 2 1 0

2

n n n

n n

n n n

→∞ →∞ →∞

+ + = = = ≠

+ + + + .

So by the Limit Comparison Test, the series ₃

12 3 2

n

n n

∞

= + + converges.

Example 2: Determine whether or not the series ₃

12 3 2

n

n n

∞

= − + converges or diverges.

Solution: The series ₂

1

1 2

n n

∞

=

3 3

3

2 2 3

2 2

2 3 2

lim 1 lim2 3 2 lim2 3 2 1 0

2

n n n

n n

n n n

→∞ − + = →∞ = →∞ = ≠

− + − + .

So by the Limit Comparison Test, the series ₃

12 3 2

n

n n

∞

= − + converges.

Example 3: Determine whether or not the series

2

ln

n

n n

∞

=

converges or diverges.

Solution: Consider

2

1

n n

∞

=

and note that this is the Harmonic series which diverges. Furthermore lnn 1 n ≥n for 3

n≥ . So by the Comparison Test, the series

2

ln

n

n n

∞

=

diverges. (If one tries to use the Limit Comparison Test, then limln / lim ln

1/

n n

n n n

n

→∞ = →∞ = ∞. So the Limit Comparison Test does not provide an answer for this problem. The Integral Test can be used to show that this series diverges.)

Problems:

1-20. Use the comparison test to establish the convergence or divergence of

1

( )

n

f n for each given ( )

f n :

1. ( ) ₂ 6

2 4 2

f n = n n

+ + 2. ( ) 6

2ⁿ 2 f n =

+ 3. f n( ) lnn

= n 4. ( ) ₂ 1

f n 3

n n

= −

5.

( )

²

( ) 1 f n 1

= n

− 6. ( )

2ⁿ f n = n

7. 1

( ) ln 1

f n = n

+ 8. ( )

4 f n n

=n +

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9. 2

( ) 1

1 f n n

n

= −

+ 10. ₁

2

( ) 6 f n =n

−

11. ( ) 6

f n 2

= n

+ 12. ( ) 3

5ⁿ 2 f n =

+ 13.

43

( ) 2

2 1

f n n

n n

= + + 14. ( ) 3

5 2

n

f n = n

+

15. ( ) 6

5 2

n

f n = n

+ 16. ( ) ₃

2 f n n

=n +

17. ( ) 1

5ⁿ

f n = n 18. 1 1

( ) 2 1 4

n

f n = n +

19. 1

( )

n n

f n e e + −

= 20. ( ) 1 1

f n 3

n n

= −

− 21. f n( ) 3 cosn

n

= + 22. f n( ) 1_n

=n