1

1 dx

x x

Checking off homework:1) 8.8-10.5 Test - Form A - Review Packet 2) 9.1-9.2 Combined Assignment Part 2: 9.1 ( 75, 81, 84, 89-97 eoo ) and 9.2 ( 1-4, 12-15, 25-29 odd, 37, 41, 47, 49 )3) 9.1-9.2 Combined Assignment Part 3 : 9.1 ( 51-63 eoo, 79, 119-124 ) and 9.2 ( 61-69 eoo, 83, 85)

Bellwork:

The 8.8-10.5 Retest has been postponed until Friday.

Homework: Read 9.3.9.3 (7,9,13,21,29,36,39,40,75,79,87-97 odd)

Bellwork Solutions – Improper Integration with a convergent integral:

1

1 dx

x x

3

2

1lim

b

bx dx

1

21

lim 2

b

b x

22lim

b b

2

9.3 Notes - The Integral Test and p-series

Petrified Forest National Park,Arizona

Greg Kelly, Hanford High School, Richland, WashingtonPhoto by Vickie Kelly, 2007

Yet more ways for us to determine convergence and divergence of series!

9.3 Notes - The Integral Test and p-series

Petrified Forest National Park,Arizona

Greg Kelly, Hanford High School, Richland, WashingtonPhoto by Vickie Kelly, 2007

Yet more ways for us to determine convergence and divergence of series!

Aren’t you excited???

Remember that when we first studied integrals, we used a summation of rectangles to approximate the area under a curve:

This leads to:

Theorem 9.10 The Integral Test

If is a positive sequence and where

is a continuous, positive decreasing function, then:

na na f n f n

and both converge or both diverge.nn N

a

Nf dxx

Example 1: Does converge?1

1

n n n

1

1 dx

x x

3

2

1lim

b

bx dx

1

21

lim 2

b

b x

22lim

b b

2

Since the integral converges, the series must converge.

(but not necessarily to 2.)

If you showed this amount of work on the AP test, you would havea point deducted for not showing that an is positive and decreasing.

Theorem 9.11p-series Test

1

1 1 1 1

1 2 3p p p pn n

converges if , diverges if .1p 1p

We could show this with the integral test.

If this test seems backward after the ratio and nth root

tests (which we will cover soon), remember that larger

values of p would make the denominators increase

faster and the terms decrease faster.

the harmonic series:

1

1 1 1 1 1

1 2 3 4n n

diverges.

(It is a p-series with p=1.)

It diverges very slowly, but it diverges.

Because the p-series is so easy to evaluate, we use it to compare to other series.

Integral Test and p-series

Very often comparing the nth term of a series an with the corresponding function f(x) will yield useful information about an.

If f(x) is continuous, positive and decreasing for x 1 and an = f(n), then

The integral test uses such a comparison.

1 1

( ) will both either converge or diverge.nn

a and f x dx

Test the convergence or divergence of the series 2

1 1n

n

n

The corresponding function is:

2 1n

na

n

2( )

1

xf x

x

For x 1, this function is positive and decreasing.

Therefore, we can use the integral test.

21 1

xdx

x

x2 + 1 = u

2xdx = du

xdx = du/2

21 1

b

b xLim

xdx

When x = 1, u = 2

When x = , u =

2

1

2

b

b u

dLim

u

2

1ln

2 bLim u

=

The integral diverges and hence the series diverges.

Test the convergence or divergence of the series

The corresponding function is:n

na ne ( ) xf x xe

For x 1, this function is positive and decreasing.

Therefore, we can use the integral test.

1

xxe dx

Use the table method for integration.

1

n

n

ne

u dv

x e-x

1 -e-x

0 e-x

(+)

(-)

1

bx

bLim xe dx

1

bx x

bLim xe e

1( 1) (1 1)b

bLim e b e

1( 1)

bx

bLim e x

20e

2

e

Since the integral converges, the series also converges.

Test the convergence or divergence of the series

The corresponding function is:2

1

1na n

2

1( )

1f x

x

For x 1, this function is positive and decreasing.

Therefore, we can use the integral test.

21

1

1dx

x

21

1

1

b

bLim dx

x

1

arctanb

bLim x

arctan arctan1b

bLim

2 4

Since the integral converges, the series also converges.

1 1 1 1 1..........

2 5 10 17 26

21

1

1n n

4

Test the convergence or divergence of the series

The corresponding function is:2 3n

na

n

2( )

3

xf x

x

For x 1, this function is positive and decreasing.

Therefore, we can use the integral test.

21 3

xdx

x

x2 + 3 = u

2xdx = du

xdx = du/2

21 3

b

b xLim

xdx

When x = 1, u = 4

When x = , u =

4

1

2

b

b u

dLim

u

2

1ln

2 bLim u

=

The integral diverges and hence the series diverges.

2

1 2 3....... ......

4 7 12 3

n

n

p-series and the Harmonic series

A series of the form:1

1p

n n

is called a p-series.

Using integral theorem, it can be shown that a p-series converges for p > 1

A p-series diverges for 0 < p 1

When p = 1, the series is called a harmonic series.

A harmonic series is a diverging series.

This can be shown using integral theorem.

1

is a harmonic ser e .1

i sn n

Determine the convergence or divergence of:

This is a p-series with p = 4/3

Since p > 1, this is a converging p-series.

4/31

1

n n

Determine the convergence or divergence of: 3 3 3 3

1 1 1 11 .........

4 9 16 25

3 3 3 3

1 1 1 11 .........

4 9 16 25

2 2 23 3 33

1 1 1 1.........

1 2 3 4

2/3 2/3 2/3 2/3

1 1 1 1.........

1 2 3 4 2

1 3

1

n n

This is a p-series with 0 < p < 1. Therefore, the series diverges.

Partial sum SN as an approximation of S

Very often the sum of an infinite series can be approximated taking the sum of the first n terms.

Of course there will be a small error in this approximation.

The error is a measure of the remainder which is the sum of all the remaining terms.

We can use the following theorem to obtain the remainder RN in the approximation.

If f is a positive and decreasing function such that an = f(n), then the error involved in approximating S, the sum of the series, as SN, the sum of the first n terms of the series, will be less than or equal to:

( )N

f x dx

In other words, If RN = S – SN , then: 0 ( )NN

R f x dx

Approximate the sum of the series

using the first 10 terms.

1/((n + 1)(ln(n + 1))^3), n, 1, 10, 1,), ENTER 1.9821

31

1

( 1) ln( 1)n n n

10

3101

1

( 1) ln( 1)n

Sn n

2nd, STAT, right arrow twice, 5, 2nd, STAT, right arrow,5 and type in

The error in this approximation is given by:

310

1

( 1) ln( 1)dx

x x

310 ln( 1

1

( ) )1

b

b xLi dx

xm

ln(x + 1) = u

1

1dx du

x

When x = 10, u = ln 11

When x = b, u = ln(b + 1)

ln( 1)

3ln11

1b

bLim

udu

Approximate the sum of the series

using the first 10 terms. 3

1

1

( 1) ln( 1)n n n

ln( 1)

3ln11

1b

bLim

udu

ln( 1)3

ln11

b

bLim u du

ln( 1)

2ln11

1

2

b

bLim

u

2 2

1 1

2[ln( 1)] 2(ln11)bLim

b

2

10

2(ln11) 0.087

Approximate the sum of the series

using the first 4 terms.

= 0.5713S4 = e-1 + e-2 + e-3+ e-4

The error in this approximation is given by:4

xe dx

1

n

n

e

4

xe dx

4

limb

x

be dx

4lim

bx

be

4lim ( )b

be e

40 e = 0.0183

Find N such that the error in approximating S using SN is less than 0.001 (RN 0.001) for the series:

If we use N terms to approximate the sum of the series, the remainder RN is given by 2

1

1N

dxx

21

1

1n n

2

1

1N

dxx

2

1

1

b

bN

Lim dxx

arctan

b

NbLim x

arctan arctanb

b NLim

arctan2

N

We need to find the value of N such that this remainder is less than 0.001

arctan 0.0012

N

arctan 0.0012

N

arctan 0.0012

N

tan 0.0012

N

1000N

Determine the convergence or divergence of the series:

Here we use the integral test.

2

1( )

1f x

x x

22

1

1n n n

The corresponding function is:

22

1

1dx

x x

2sec

b

bLim arc x

22

1

1

b

bLim dx

x x

sec se 2cbLim arc arb c

2 3

6

Since the integral converges, the series also converges.

For x > 1, this function is positive and decreasing.

Determine the convergence or divergence of the series:

Do this on your own.

Answer:

This is a geometric series with r = 1.075

0

(1.075)nn

Since r > 1, this geometric series diverges.

Determine the convergence or divergence of the series: 2 3

1

1 1

n n n

2 31

1 1

n n n

2 31 1

1 1

n nn n

Each of these is a p-series with p > 1

Therefore, the series converges.

Now try using the integral test to show that the series converges.

Determine the convergence or divergence of the series:

Here we use the integral test.

3

ln( )

xf x

xThe corresponding function is:

32

ln xdx

x

32

lnb

dxx

Lx

im

Use integration by parts

For x 2, this function is positive and decreasing.

32

ln

n

n

n

lnu x1

du dxx

3

1dv dx

x

2

1

2v

x

22

2

1

2 2

n 1lbL d

x

xxim

xx

2 32

ln 1 1

2 2b

xLim dx

x x

Determine the convergence or divergence of the series: 3

2

ln

n

n

n

2 22

ln 1

2 4

b

b

xLim

x x

2 2 2 2

ln 1 ln 1

2 4 2 4

2

2 2

b

b b

ln 2 10

8 16 1

2ln 2 116

Since the integral converges, the series also converges.

2 32

ln 1 1

2 2b

xLim dx

x x