01-Series-Basic Concepts.ppt

Series: Guide to Investigating Convergence

Understanding the Convergence of a Series

A series converges to λ if the limit of the sequence of the partial sums

of the series is equal to λ

Example (1)

numbersrealfixedarerandaWhere

ar

SeriesGeometricThe

n

n

1

1

11

:;111

)1(

)1()1(

,)2()1(

)2(

)1(

)1(

:

,

132

122

0

1

0

r

a

r

arT

isseriestheofsumpartialthntheofsequenceTheThusr

a

r

ar

r

raT

rarT

aarTrT

getwefromgSubtractin

arararararrT

equationresultingthefromitgsubtractinandrby

equationgmultiplyinbyatarrivedisTforformulaA

ararararaT

where

TisseriestheofsumpartialthntheofsequenceThe

ars

seriesgeometrictheConsider

n

n

nn

n

nn

nnn

nnn

n

nnn

n

n

n

nn

divergesseriesthesoand

existnotdoesT

getwerIf

rr

aarwriteWe

r

atoconvergesseriesthesoand

r

a

r

a

r

a

r

arT

getwerIf

rondependssequencethisofTThe

nn

n

n

n

nn

n

nn

lim

:,1.1

1;1

1

1)1

(0)11

(limlim

:,1.1

:lim

0

1

Warning

aasWhileTpositiveaandrFor

arsWhiler

aTrFor

thatNotice

ars

seriestheoftermthntheofssequenceThe

r

a

r

arT

seriestheofsumspartialthntheofTsequenceThe

betweenhdistinguisPlease

nn

nn

n

n

nn

nn

n

nn

n

n

n

n

limlim,lim:1.2

0limlim,1

lim:1.1

:

.2

&11

.1

:,

1

1

The Sum of the series

nn

knn

n

Tt

Thus

seriestheofsumspartialthntheofTsequencetheoftimilThe

toequalisexistswhenseriestheofsumthedefinitionBy

lim

,

:,,,,

QuestionsCheck, whether the given series is convergent, and if convergent find

its sum

113

22

02

13

012

2

3

2.3

7

2.2

7

5.1

nn

n

nn

n

nn

n

Example (2) Telescoping Series

knsst

thatsuchssequenceaistherethatAssume

tseriestheConsider

Examples

nnn

n

knn

;

,

,

)1(

1

knn

nkn

nk

nnnnkkkkkk

nnnkkkkn

n

kinn

stoconvergesTthennumberrealatoconvergessifsoand

ssT

Thus

ss

ssssssssss

tttttttT

So

tseriestheofsumpartialthntheofsequencethebeTLet

,

,

)()()()()(

,

1

1

1132211

12321

Warning

0lim,limlim

:,

:

.2

&

.1

:,

1

nn

nn

knn

n

n

n

nkn

n

tWhilessT

thenconvergessIf

thatNotice

t

seriestheoftermthntheoftsequenceThe

ssT

seriestheofsumspartialthntheofTsequenceThe


Examples of this type of telescoping seriesA Convergent Telescoping Series

6

70

24

7

lim65

7

2

7

)!(3

7

2

7

)3)(2(

7

65

7,

65

7

.1

44

2

1

2

42

4

nn

n

nnn

n

nnn

ssnn

Thus

sstn

sLet

thatshownn

nnnnhaveWe

nntseriestheConsider

Solutions

202lim)1(

2

Thus,

0toconverges><

Then

>2

<><

)1(

22

1

2

)1(

2

:

)1(

2.2

11

1

11

nn

n

n

nnn

n

n

nnn

ssnn

sBut

sstn

s

Let

nnnnnnt

havewe

nntLet

knsst

thatsuchssequenceaistherethatAssume

tseriestheConsider

Examples

nnn

n

knn

;

,

,

)2(

1

knn

knn

nk

nnnnkkkkkk

nnnkkkkn

n

kinn

stoconvergesTthennumberrealatoconvergessifsoand

ssT

Thus

ss

ssssssssss

tttttttT

So

tseriestheofsumpartialthntheofsequencethebeTLet

,

,

)()()()()(

,

1

1

1123121

12321

Examples of this type of telescoping seriesA Divergent Telescoping Series

)1ln(lim

)1ln()1ln(0

:

,

ln

ln)1ln()1

ln()1

1ln(,

)1

1ln(

11

1

11

nT

nnssT

isseriestheofsumspartialthntheofTsequencetheThus

sst

Then

nsLet

nnn

n

nhaveWe

ntseriestheConsider

nn

nn

n

nnn

n

nnn

Questions ICheck, whether the given series is

convergent, and if convergent find its sum

])1arctan([arctan4

)!1(.3

1.2

)1(

2.1

1

22

1

n

n

n

nn

n

n

n

nn

Questions IIShow that the following series is a telescoping series, and then determine

whether it is convergent

nn

nn

n

n

ssnThus

nn

sthen

n

s

Let

nnnn

nnnn

tniH

n

nn

1

21

211

21

21

21

21

21

21

21

21

21

2

1

sin,

sin2

)2

12cos(

sin2

)2

1)1(2cos(

,sin2

)2

12cos(

:

sin2

)2

12cos(

sin2

)2

12cos(

sin2

)2

12cos()

2

12cos(

sin2

)cos()cos(

sin2

sinsin2sin

sin.2

)1(.1

nn

nn

n

n

ssnThus

nn

sthen

n

s

Let

nnnn

nnnn

tniH

n

nn

1

21

211

21

21

21

21

21

21

21

21

21

2

1

sin,

sin2

)212

cos(

sin2

)2

1)1(2cos(

,sin2

)212

cos(

:

sin2

)212

cos(

sin2

)212

cos(

sin2

)212

cos()212

cos(

sin2

)cos()cos(

sin2

sinsin2sin

sin.2

)1(.1

12

50

12

50

)3)(2(

5

Thus,

0toconverges><

Then

>2

5<><

)3)(2(

5

3

5

2

5

)3)(2(

5

:

)3)(2(

5.2

11

1

11

snn

sBut

sstn

s

Let

nnnnnnt

havewe

nntLet

n

n

nnn

n

n

nnn

The Integral Test

divergebothorconvergeboth

dxxfegralimpropertheandsseriestheeitherThen

Knumbernatural

someforKnnfssatisfyingsequenceabeslet

KoncontinuousandngdecreasipositivebeffunctionaLet

KKnn

nn

)(int

;;)(:&

).,[

Example (3)

0

1

)(

1

pWhere

n

seriespThe

SeriesnicHyperharmoThe

np

111

111

,

1int

1

:int

sin,

),1[;1

)(

;

;1

)(

:

1

0

1

10

01

pifdivergesandpifconvergesn

Thus

pifdivergesandpifconvergesdxx

thatknowweBut

convergesdxx

egralimpropertheiffconvergesn

testegraltheBy

gdecreaandcontinuouspositiveisfhaveWe

xx

xf

functiontheConsider

Nnn

snf

Write

ns

seriesptheConsider

np

p

pn

p

p

pn

np

nn

Warning

01

lim1

limlim

1,.1

:

1.2

&

1.1

:,

33

1

1131

11

nn

sWhile

divergesn

sseriesthepFor

thatNotice

seriestheoftermthntheofn

ssequencetheofeconvergencThe

nsseriestheofeconvergencThe


nnn

n

np

nn

pn

np

nn

QuestionsCheck, whether the given series is convergent.

1

5 2

1

5 2

15 3

13 5

.4.3

1.2

1.1

nn

nn

n

n

n

n

nn

Algebra of Series Convergence

)(

)(

.1

)(

.2

)(

.1

11

1

11

11

1

11

11

1

11

convergentisitthatfollowsitnor

divergentistcscthatfollowsnotdoesitThen

divergentaresandsbothIf

divergentistcscThen

divergentaretandconvergentissIf

tcscissoThen

convergentaretandsbothIf

nn

n

nn

nn

nn

n

nn

nn

nn

n

nn

nn

Questions

seriesdivergentab

seriesconvergentaa

istsseriesthethatsuchtands

seriesdivergenttwoofexampleanGiven

b

na

seriesfollowingtheofeachofeconvergenctheeInvestigat

nn

nn

nn

n

nn

n

nn

n

.

.

:)(,

.2

]2

3

7[.

]5

3

4[.

:.1

111

21

23

12

123

12

Divergence Test

convergessthatfollownotdoesitsIf

Notice

divergessThen

sIf

nnn

n

nn

nn

1

1

,0lim

:

0lim

QuestionsCheck, whether the given series is convergent.

1

158

38

1

13

1.3

163

325.2

])1(1

[.1

nn

n

nn

n

n

nn

nn

n

Convergence Tests

Convergence Tests for Series of Positive Terms

1. Comparison Test

2. Limit Comparison Test

3. Ratio Test

4. Root Test

The Comparison test

divergestthendivergessIf

convergessthenconvergestIf

Then

Nntsand

termspositiveofseriesbetandsLet

nn

nn

nn

nn

nn

nn

nn

11

11

11

,.1

,.1

:

.;

,

Examples

Example (1)

18 5

15 8

23

4.2

23

4.1

n

n

n

n

convergesseriesfollowingtheofeachwhethereInvestigat

Solution

.23

4

:,

?)11

3

4

3

4(

3

4

1

3

4

3

4

23

4.1

15 8

1 5

81 5

81

5 8

15 8

5

85 85 8

convergesn

termspositiveofseriesThe

testcomparisonthebysoand

Whyconverges

n

and

nnbecause

convergesn


nnn

n

nnn

n

.23

4

:,

?)11

3

4

3

4(

3

4

1

3

4

3

4

23

4.2

18 5

1 8

51 8

51

8 5

18 5

8

58 58 5

divergesn


testcomparisonthebysoand

Whydiverges

n

and

nnbecause

divergesn


nnn

n

nnn

n

DefinitionOrder of Magnitude of a Series

11

11

11

11

,lim.3

,0lim.2

,lim.1

:

nn

nn

n

n

n

nn

nn

n

n

n

nn

nn

n

n

n

nn

nn

tthanmagnitudeofordergreaterahass

seriesthethatsaywethent

sIf

tthanmagnitudeoforderlesserahass

seriesthethatsaywethent

sIf

magnitudeofordersamethehavetands

seriesthethatsaywethennumberpositiveaist

sIf

Then


Question

?

lim

lim

,

notWhy

t

s

Or

numbernegativeaist

s

wherecasesthereAre

n

n

n

n

n

n

The Limit Comparison test

11

11

11

11

11

11

.

.3

.

.2

.

,.1

:

nn

nn

nn

nn

nn

nn

nn

nn

nn

nn

nn

nn

tofdivergencethetoleadssofdivergencethe

thentthanmagnitudeofordergreaterahassIf

sofeconvergencthetoleadstofeconvergencthe

thentthanmagnitudeoforderlesserahassIf

divergebothorconvergeboththeyeither

thenmagnitudeofordersamethehavetandsIf

Then


Examples

1512

311

1512

37

163

325.2

163

325.1

n

n

nn

nn

nn

nn

Solution

convergesnn

nntermspositiveofseriesThe

Therefore

Whyconvergeswhichn

seriesthe

asmagnitudeofordersamethehasnn

nnThus

nn

nnn

n

nn

nn

t

s

haveWe

nt

termspositiveofseriestheConsider

nn

nnsLet

n

n

n

nnn

n

n

nnn

nnn

1512

37

15

1512

37

512

5812

5

512

37

15

1

1512

37

1

163

325

;

)?(1

163

325

),0[3

5

163

325lim

1163

325

limlim

1

:

163

325.1

divergesnn

nntermspositiveofseriesThe

Therefore

Whydivergeswhichn

seriesthe

asmagnitudeofordersamethehasnn

nnThus

nn

nnn

n

nn

nn

t

s

haveWe

nt

termspositiveofseriestheConsider

nn

nnsLet

n

n

n

nnn

n

n

nnn

nnn

1512

311

1

1512

311

512

412512

311

11

1512

311

1

163

325

;

)?(1

163

325

),0[3

5

163

325lim

1163

325

limlim

1

:

163

325.2

The Ratio test

divergessthenors

sIf

convergessthens

sIf

Then

termspositiveofseriesabesLet

nn

n

n

n

nn

n

n

n

nn

1

1

1

1

1

,1lim.2

1lim.1

:

Examples

1

1

1

5.3

5

)3(.2

!.1

nn

nn

n

n

n

n

n

n

n

The Root test

divergessthenorsIf

convergessthensIf

Then

termspositiveofseriesabesLet

nn

nn

n

nn

nn

n

nn

1

1

1

1lim.2

1lim.1

:

Examples

1

1

52ln

3.2

25

32.1

n

n

n

n

n

n

n

DefinitionAlternating Series

termspositiveofsequenceaisswhere

s

Or

s

formsfollowing

theofeitherofofseriesaisseriesgalternatinAn

n

nn

n

nn

n

1

1

1

)1(

)1(

Alternating Series Convergence Test

convergestThen

toconvergentandgdecreaissIf

termspositiveofsequenceaisswhere

storstLet

nn

n

n

nn

n

nnn

n

n

nn

1

1

1

111

0sin

)1()1(

Example

testeconvergencseriesgalternatinthebyconvergesn

Thereforen

sand

termspositiveofsequencegdecreaaisn

s

haveWen

s

seriestheConsider

n

n

nn

n

n

n

n

nn

1)1(

;

1limlim

sin1

:

1)1(

1

11

DefinitionAbsolute and Conditional Convergence

divergessbutconvergesitif

llyconditionaconvergessseriesthethatsayWe

convergessif

absolutelyconvergessseriesthethatsayWe

nn

nn

nn

nn

1

1

1

1

,

.2

.

.1

Example (1)

llyconditionaconvergesseriesthesoand

whydivergesnn

seriesthehandothertheOn

convergesn

thatshownhaveWe

ns

seriestheConsider

n

n

n

n

n

n

n

nn

?11

)1(

1)1(

1)1(

11

1

11

Example (2)

absolutelyconvergesn

nseriestheThus

Whyconvergesn

nseriesthesoand

nn

ns

haveWe

n

ns

seriestheConsider

n

n

n

n

nn

n

n

nn

31

13

33

311

2sin)1(

?)(2sin

)1(

12sin)1(

:

2sin)1(

The Ratio test for Absolute Convergence

divergessthenors

sIf

absolutelyconvergessthens

sIf

nn

n

n

n

nn

n

n

n

1

1

1

1

,1lim.2

1lim.1

Examples:Investigate the absolute convergence of

the following series

)!1(

)!12()1(.5

67

5)1(.4

)3cos(5)1(.3

!

)!12()1(.2

!

5)1(.1

1

1

3 41

1

5 91

11

n

n

nn

n

n

n

n

n

n

n

n

n

n

n

n

nn

n

n

More Examples on the Integral Test

Example (1)

convergesesseriesthe

thatShowconvergesdxedxxfIegralimproperThe

nnfsand

oncontinuousandWhypositiveWhypositiveisf

haveWe

exfLet

Solution

esseriestheofeconvergenctheeInvestigat

n

n

nn

x

n

x

n

n

nn

11

11

11

)!()(int

&

1;)(

),1[?)(?),(

:

)(

:

Example (2)

convergesn

sseriesthe

thatShowconvergesdxx

dxxfIegralimproperThe

nnfsand

oncontinuousandWhyingdecreasWhypositiveisf

haveWex

xfLet

Solution

nsseriestheofeconvergenctheeInvestigat

nnn

n

nnn

12

1

12

1

2

12

1

1

1

)!(1

1)(int

&

1;)(

),1[?)(?),(

:1

1)(

:

1

1

Example (3)

2ln

1

2ln

1

ln

1]

ln

1[

lnln

1

ln

1

:

ln

1int

ln

1

2;)(

),2[?)(?),(

:ln

1)(

:

ln

1

22222

2

3

23

23

23

23

2

23

2

limlim

limlim

tx

x

dxxdx

xxdx

xx

havewe

convergesdxxx

egralimproperthe

thatshowingbyconvergesnn

thatshowwillWe

nnfsand

oncontinuousandWhyingdecreasWhypositiveisf

haveWexx

xfLet

Solution

nnsseriestheofeconvergenctheeInvestigat

tt

t

t

t

t

n

n

nnn

01-Series-Basic Concepts.ppt

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Transcript of 01-Series-Basic Concepts.ppt