3.4 direct comparison and limit comparison tests

7/30/2019 3.4 Direct Comparison and Limit Comparison Tests

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Direct comparison Test and Limit Comparison Test

If an and bn are the nth terms of two infinite series, and if we knowthe behavior of one of them, then the behavior of the other can beobtained using the direct comparison test, provided the followingconditions are satisfied.

If 0 < an bn for all values of n, then

1 1

1. ,n n

n n

If b converges then a converges

1 1

2. ,n n

n n

If a diverges then b diverges

Direct Comparison Test


2/17

Use direct comparison test to determine theconvergence or divergence of the series

an is the given series.

21

1

3 1n n

We have to identify a series bn whose behavior weknow and such that an < bn

Since the degree of the denominator of an

is 2, bn

alsomust be of degree 2.

The obvious choice is2

1n

bn

2 21 1

3 1n n

21

1

is a converging -seriesn pn

2 21 1

1 1converges when compared to

3 1n nn n


3/17


4/17


What is a suitable bn to compare an?


1n

b

n

4 4

1 1

3 1n n

1/ 41

1is a -series with p = 1/4

n

pn

1/441 1

1 1diverges when compared to

3 1n n nn

4

1

3 1na

n

1/ 4

1n

bn

41

1

3 1n n

Since p< 1, this is a diverging p-series.

bn < an and bn diverges


5/17


What is a suitable bn to compare ansuch that an < bn


1n

bn

21 1

for 3

!

n

n n

21

1is a converging -series with p >1

n

pn

21 1

1 1converges when compared to!n nn n

1

!na

n

0

1

!n n

0 1

1 11 also converges

! !n nn n


6/17

If an and bn are the nth terms of two infinite series, and if we know

the behavior of one of them, then the behavior of the other can be

obtained using the limit comparison test.

If an > 0 and bn > 0 for all values of n and

n

n

n

aLim L

b

Limit Comparison Test

where L is finite and positive:

Review:1

What is ?n

Limn

As n increases, the denominatorgrows without bounds.

01

nLim

n

22What is ?

( 1)nn

Limn n

The numerator and the denominatorhave the same degree

The limit is the ratio ofthe leading coefficient.

22 2

(2

1) 1nn

Limn n

n nThe two series a and b with both converge or diverge.


7/17

Use the limit comparison test to determine theconvergence or divergence of the series

What is a suitable bn whose behavior we know.


nb

n

n

nn

aLim

b

2

1

1na

n

22

1

1n n

2

1

1

1n

nLim

n

2 1n

nLim

n

1

is a diverging harmonic

series.

This limit is finiteand positive.

nb

n na and b either both converge or diverge.

nn

Since b diverge a divergs, .es


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nb

n

n

nn

aLim

b

2

5 3

2 5n

na

n n

2

5 3

2 51n

n

n nLim

n

2

(5 3)

2 5n

n nLim

n n

5

is a diverging harmonic

series.

This limit is finite and positive

22

5 3

2 5n

n

n n

nb


nn



9/17




2n n

b

n

nn

aLim

b

1( 1)2n

n

na

n

1( 1)2

1

2

n

n

n

n

nLim

1

2

2( 1)

n

n n

nLim

n

2

12 ( 1)2nn

n

n

2

( 1)n

nLim

n

This is a converging geometric series.



nn

Since b converge a convergs, .es


10/17




nb

n

n

nn

aLim

b

2

1

1na

n n

2

1

11n

n nLim

n

2 1n

nLim

n n

1

2

This is a diverging harmonic series.

21

1

1n n n

2n

nLim

n



nn



11/17

1. nth Term Test

Review of Tests so far

2. Geometric Series Test

3. Telescopic Series Test

4. Integral Test

5. Direct Comparison Test

6. Limit Comparison Test


12/17

Test the convergence or divergence of theseries: of the series

Which of the tests is appropriate in this case?

This is a geometric series with a = 5 and1

5r

1

5r

1

5 1

This is a converging geometric series.

Can you find the sum of this converging

geometric series?

0

15

5

n

n

Answer:25

6


13/17




1n

b

n

n

nn

aLim

b

2

1

3 2 15na n n

2

2

1

3 2 151n

n nLim

n

2

23 2 15n

nLim

n n

1

3

is a converging p-series.

This limit is finite and positive.

24

1

3 2 15n n n

Use limit comparison test

nb


nn

Since b converge a convergs, .es


14/17

Test the convergence or divergence of theseries:

What is an appropriate test in this case?

When you are in doubt, always try the nth term test?

2 3n

nLim n

1

2

Since the limiting value of the nth term is not zero, this seriesdiverges.

1 2 3n

n

n


15/17



Since the corresponding function can beintegrated, we should try integral test.

2 2( )

( 1)

xf x

x

2 21 ( 1)n

n

n

The corresponding function is:

2 21( )x

xdx

x2 + 1 = u

2xdx = du

xdx = du/2

2

1

2

u

u

d

1

2u 2

1

2( 1)x

2 21 ( 1)

xdxx

2 21

lim( 1)

b

bx dx

x 21

1lim2( 1)

b

b x

2

1 1

2( 1) 2(1 1)b

1

4 Since the integral converges,

the series converges.


16/17



Use partial fractions to simplify

3

( 3)n n

3

A B

n n

Equate the numeratorand denominator.

3 = A(n + 3) + Bn Setting n = 0 gives:

( 3)

( 3)

A n bn

n n

A = 1

1

3

( 3)n n n

Setting n = -3 gives: B = -1

1

3( 3)n n n

1

1 1

3n n n

114

1 12 5

1 13 6

1 1 ......4 7

1 11

2 3

This is a converging

telescopic series.


17/17

Given that:

2

2 21 3

1 1,

(2 1) 8 (2 1)n nfind

n n

2

21

1 1 101

(2 1) 9 9n n

21

1

(2 1)n n

2

2 21 3

1 1

(2 1) (2 1)n n

2

2 2 23 1 1

1 1 1

(2 1) (2 1) (2 1)n nn n n

2 10

8 9

3.4 direct comparison and limit comparison tests

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