Chapter 10 Infinite Series

7/25/2019 Chapter 10 Infinite Series

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CHAPTER10 INFINITESEQUENCESANDSERIES

10.1 Sequences

10.2 InfiniteSeries

10.3

TheIntegral

Tests

10.4 ComparisonTests

10.5 TheRatioandRootTests

10.6 AlternatingSeries:AbsoluteandConditionalConvergence

10.7 PowerSeries

10.8 TaylorandMacLaurinSeries

Calculus

&

Analytic

Geometry

II

(MATF

144) 2

10.1 Sequences

Definition

Aninfinitesequenceormoresimplyasequenceisanunendingsuccessionof

numbers,calledterms.

Itisunderstoodthatthetermshaveadefiniteorder,thatis,thereisafirstterm

a1,

a

second

term

a2,

a

third

term

a3,

and

so

forth.

Suchasequencewouldtypicallybewrittenas

a1,a2,a3,a4,.an

wherethedotsareusedtoindicatethatthesequencecontinuesindefinitely.

Somespecificexampleare

1,2,3,4,..

1 1 11, , , ,....

2 3 4

2,4,

6,

8,

..

1,

1,

1,

1,.

Thenumberaniscalledthenthtermorgeneralterm,ofthesequence.


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Calculus&AnalyticGeometryII(MATF144) 3

Example10.1:

Ineachpart,findthegeneraltermofthesequence.

(a)

1 2 3 4, , , ....

2 3 4 5 (b)1 1 1 1

, , , ....2 4 8 1 6

(c)

1 2 3 4, , , ....

2 3 4 5

(d) 1,3,5,7,..


Whenthegeneraltermofsequenceisknown,thereisnoneedtowriteouttheinitial

terms,anditiscommontowritethegeneraltermenclosedinbraces.

Sequence BraceNotation

1 2 3 4, , , .... ....

2 3 4 5 1

n

n 11

n

n

n

n

1 1 1 1 1, , , .... ...

2 4 8 1 6 2n

1

1

2

n

nn

11 2 3 4, , , ....,( 1) ,....2 3 4 5 1

n n

n

1

1

( 1)1

nn

n

n

n

1,3,5,7,..,2n1, 1

2 1 n

nn


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ConvergenceandDivergence

Sometimesthenumbersinasequenceapproachasinglevalueastheindexnincreases.

Thishappensinthesequence

1,

1

2 ,

1

3 ,

1

4 , ,

1

,

whosetermsapproach0asngetslarger,andinthesequence

0, 12 ,23 ,

34 ,

45 , ,1

1 ,

whosetermsapproach1.

Ontheotherhand,sequenceslike

1, 2, 3 , . . , , .

havetermsthatgetlargerthananynumberasnincreases,andsequenceslike

1,1,1,1.,1, . bouncebackandforthbetween1and1,neverconvergingtoasinglevalue.


Example

10.2:

Findthelimitofeachofthesesequences.

(a)

1

n

(b) ( 1)

n (c) 8 2n


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CalculatingLimitsofSequences


Example10.3:

Determinewhetherthesequenceconvergesordiverges.Ifitconverges,findthelimit.

(a)

100

n

(b)

1n

n

(c)

4

4 2

3 1

5 2 1

n n

n n

(e)

2

3

2 5 7n n

n

(b)

5 3

4 2

2

7 3

n n

n n

(b)1( 1)

2 1

n n

n


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TheSandwichTheorem

Example10.4:

Findthelimitofthesequence:

(a)

2

sinn

n

(b)

2cos

3n

n

(c)

1( 1)

n

n


UsingLHpitalsrule.

ThenexttheoremenablesustouseLHpitalsruletofindthelimitsofsomesequences.It

formalizestheconnectionbetween lim nn

a

and lim ( )x

f x .

Example10.5:(EvaluatingalimitusingLHpitalsrule).

(a)

lnlimn

n

n

(b)

2

lim2nn

n

(c)n

n

n/1

lim

(d)

1lim

1

n

n

n

n


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CommonlyOccuringLimits

Thenexttheoremgivessomelimitsthatarisefrequently.


Example10.6:

Findthelimitofeachconvergentsequence.

(a)

lnn

na

n

(b)

3n

na n

(c)

25n

na n

(d)

1

3

n

na

(e)

3 n

n

na

n

(f)

55

!

n

na

n


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10.2 InfiniteSeries


i) ThenthpartialsumSnoftheseries na is1 2 ...n kS a a a

ii) Thesequenceofpartialsumsoftheseries na

is

1 2 3, , ,..... ,...nS S S S

Example10.7:

Giventheseries

1 1 1 1... ...

1 2 2 3 3 4 ( 1)n n

(a)

FindS1,S2,S3,S4,S5andS6.

(b)

FindSn.

(c) Showthattheseriesconvergesandfinditssum.


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Example10.8:

Giventheseries,

1 1

1

( 1) 1 ( 1) 1 ( 1) ... ( 1) ...n n

n

(a)

Find

S1,

S2,S3,

S4,

S5

and

S6.

(b)FindSn.

(c) Showthattheseriesdiverges.


GeometricSeries

Example10.9:

Determinewhethereachofthefollowinggeometricseriesconvergesordiverges.If

theseriesconverges,finditssum.

(a) 01 37 2

n

n

(b) 0135

n

n


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TelescopingSeries

Atelescopingseriesdoesnothaveasetform,likethegeometricseriesdo.Atelescoping

seriesisanyserieswherenearlyeverytermcancelswithaprecedingorfollowingterm.For

instance,theseries

21

1

k k k

Usingpartialfractions,wefindthat

2

1 1 1 1

( 1) 1k k k k k k

Thus,thenthpartialsumofthegivenseriescanberepresentedasfollows:


21 1

1 1 1

1

1 1 1 1 1 1 11 ..

2 2 3 3 4 1

1 1 1 1 1 1 11 ..

2 2 3 4 1

11

1

n n

nk k

Sk kk k

n n

n n n

n

Thelimitofthesequenceofpartialsumsis

1 1lim lim 1

1n

n nS

n n

sotheseriesconverges,withsumS=1.


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Example10.10:

Ineachcase,expressthenthpartialsumsSnintermsofnanddeterminewhether

theseriesconvergesordiverges.

(a)

0

1

( 1)( 2)n n n

(b)

0

1 1

2 1 2 3n n n


DivergentSeries

Onereasonthataseriesmayfailtoconvergeisthatitstermsdontbecomesmall.For

example,theseries

2 2

1

1 4 9 .... ...n

n n

ThisseriesisdivergesbecausethepartialsumsgrowbeyondeverynumberL.Aftern=1,

thepartialsums21 4 9 ....ns n isgreaterthan

2n .

Theoremabovestatesthatifaseriesconverges,thenthelimitsofitsnthtermanas n

is0.Sometimes,itispossibleforaseriestobecomediverges.


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lim 0nn

a

Sothistheoremleadstoatestfordetectingthekindofdivergencethatoccurredinsomeof

theseries.

ThenthTermTest

If, thenfurtherinvestigationisnecessarytodeterminewhethertheseries1

nn

a

isconvergeordiverge.


Example10.11:

ApplyingthenthTermTest

(a)1 2 1n

n

n

(b)

2

1n

n

(c)1

1( 1)n

n


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CombiningSeries


Example10.12:

(a) 11

7 2

( 1) 3n

n n n

(b)

1 11

1 1

2 6n nn

(c)

2

1

1 1

2 3n nn


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10.3 TheIntegralTests


Example10.13:

Determinewhetherthefollowingseriesconverge.

(a) 21 1n

n

n

(b) 1

1

2 5n n

(c)0

ln

n

n

n

ThepSeries

TheIntegralTestisusedtoanalyzetheconvergenceofanentirefamilyofinfiniteseries

knownasthepseries.Forwhatvaluesoftherealnumberspdoesthepseries

converge?

1

1p

n n


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Example10.14:

Testeachofthefollowingseriesforconvergence.

(a) 101

1

n n

(b) 31

1

n n

(c) 24

1

( 1)n n


10.4 ComparisonTests

Wehaveseenhowtodeterminetheconvergenceofgeometricseries,pseriesandafew

others.Wecantesttheconvergenceofmanymoreseriesbycomparingtheirtermsto

thoseofaserieswhoseconvergenceisknown.

Note:Let na , nc and nd beserieswithpositiveterms.Theseries na convergesifitissmallerthan(dominatedby)a

knownconvergentseries nc anddivergesifitislargerthan(dominates)aknowndivergentseries nd .Thatis,smallerthanconvergentisconvergent,andbiggerthandivergentisdivergent.


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Example10.15:

Testthefollowingseriesforconvergence.

(a)1

1

3 1nn

(b)2

11n n

(c) 32

ln

n

n

n


Example10.16:

Testthefollowingseriesforconvergencebyusingthelimitcomparisontest.

(a)1

1

2 5nn

(b) 1

3 2

(3 5)n

n

n n

(c)1

100

70nn

n

ne


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10.5 TheRatioandRootTests

Intuitively,aseriesofpositiveterms na convergesifandonlyifthesequence na

decreaserapidlytoward0.Onewaytomeasuretherateatwhichthesequence na is

decreasing

is

to

examine

the

ratio nn

aa /1 as

n

grows

large.

This

approach

leads

to

the

followingtheorem:


Note:Youwillfindtheratiotestmostusefulwithseriesinvolvingfactorialsor

exponentials.

Example10.17:

Usetheratiotesttodeterminewhetherthefollowingseriesconvergeordiverge.

(a)

1 !2

n

n

n

(b)

1 !n

n

nn

(c)1

n

n

ne

(d)1

( 1)( 2)

!n

n n

n


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Thefollowingtestisoftenusefulifancontainspowersofn.


Example10.18:

Determinewhethertheseriesisconvergentordivergent.

(a)

1 )(ln

1

n

nn

(b) 21

2n

k n

(c) 1 2 1

n

n

n

n


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GuidelinestoUsetheConvergenceTest.

Hereisareasonablecourseofactionwhentestingaseriesofapositiveterms

1nn

afor

convergence.

1.BeginwiththeDivergenceTest.Ifyoushowthat lim 0nna

,thentheseriesdiverges

andyourworkisfinished.

2.Istheseriesaspecialseries?Recalltheconvergencepropertiesforthefollowing

series.

GeometricSeries: convergesif 1anddivergesfor 1.n

ar r r

pseries:1

convergesfor 1anddivergesfor 1.p

p pn

Checkalsoforatelescopingseries.


3.

Ifthegeneralnthtermoftheserieslookslikeafunctionyoucanintegrate,thentry

theIntegralTest.

4.

Ifthegeneralnthtermoftheseriesinvolves,whereaisaconstant,theRatioTestis

advisable.SerieswithninanexponentmayyieldtotheRootTest.

5.

Ifthegeneralnthtermoftheseriesisarationalfunctionof n(orarootorarational

function),use

the

Direct

Comparison

or

the

Limit

Comparison

Test.

Use

the

families

of

seriesgiveninStep2asacomparisonseries.


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10.6 AlternatingSeries,AbsoluteandConditional

Convergence

AlternatingSeries

Weconsider

alternating

series

in

which

signs

strictly

alternate,

as

in

the

series,

1

1

1 1 1 1 1 1 1 11 ...

2 3 4 5 6 7 8

n

n n

Thefactor 1

1 n

hasthepattern{,1,1,1,1,}andprovidesthealternatingsigns.

WeprovetheconvergenceofthealternatingseriesbyapplyingtheAlternatingSeriesTest.


Example10.19:

(a)

1

21

( 1)n

n n

(b)1

21

2( 1)

4 3

n

n

n

n

(c)

1

1

2( 1)

4 3

n

n

n

n


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AbsoluteandConditionalConvergence

Theconvergencetestwehavedevelopedcannotbeappliedtoaseriesthathasmixed

termsordoesnotstrictlyalternate.Insuchcases,itisoftenusefultoapplythefollowing

theorem.

Theseries

1

21

( 1)n

n n

isanexampleofanabsolutelyconvergentseriesbecausetheseries

ofabsolutevalues,

1

2 21 1

( 1) 1n

n nn n

isaconvergentpseries.Inthiscase,removingthealternatingsignsintheseriesdoesnot

affectitsconvergence.


Ontheotherhand,theconvergentalternatingharmonicseries

1

1

( 1)n

n n

hasthe

propertythatthecorrespondingseriesofabsolutevalues,

1

1 1

( 1) 1n

n nn n

doesnotconverge.Inthiscase,removingthealternatingsignsintheseriesdoeseffect

convergence.


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Example10.20: Determinewhetherthefollowingseriesdiverge,convergeabsolutely,or

convergeconditionally.

a)

1

1

( 1)n

n n

b)

1

31

( 1)n

n n

c)1

sin

n

n

n

d)1

1

( 1)2

n

n

n

n



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10.7 PowerSeries

Themostimportantreasonfordevelopingthetheoryintheprevioussectionsisto

representfunctionsaspowerseriesthatis,asserieswhosetermscontainpowersofa

variablex.

Thegoodwaytobecomefamiliarwithpowerseriesistoreturntogeometricseries.

Recallthatforafixednumberr,

2

0

11 ... ,provided 1

1

n

n

r r r r r

Itsasmallchangetoreplacetherealnumberrbythevariablex.Indoingso,thegeometric

seriesbecomesanewrepresentationofafamiliarfunction:

2

0

1

1 ...

,provided

11

n

nx x x xx


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ConvergenceofPowerSeries

Webeginbyestablishingtheterminologyofpowerseries.



25/33


Example:

Findtheintervalandradiusofconvergenceforeachpowerseries:

a)

0 !

n

n

x

n

Solution:

Thecenterofthepowerseriesis0andthetermsoftheseriesare.Wetesttheseries

forabsoluteconvergenceusingRatioTest:

1

1

/( 1)!lim (RatioTest)

/ !

! lim (Invertandmultiply)

( 1)!

1 lim 0 (Simplifyandtakethelimitwith fixed)

( 1)

n

nn

n

nn

n

x n

x n

x n

nx

x xn


Noticethatintakingthelimitas n ,xisheldfixed.Therefore, 0 forallvaluesofx,whichimpliesthattheintervalofconvergenceofthepowerseriesis

x andtheradiusofconvergenceis R


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b)0

( 1) ( 2)

4

n n

nn

x

Solution:

WetestforabsoluteconvergenceusingtheRootTest:

(1) ( 2)lim (RootTest)

4

2 1

4

n n

nnn

x

x

Inthiscase, dependsonthevalueofx.Forabsoluteconvergence,xmustsatisfy

21

4

x

whichimplies

that

2 4x .

Using

standard

techniques

for

solving

inequalities,

the

solutionsetis 4 2 4, or 2 6x x .Thus,theintervalofconvergence

includes(2,6).


TheRootTestdoesnotgiveinformationaboutconvergenceattheendpoints,x=2

andx=6,becauseatthesepoints,theRootTestresultsin 1 .Totestfor

convergenceattheendpoints,wemustsubstituteeachendpointintotheseriesand

carryoutseparatetests.

Atx=2,thepowerseriesbecomes

0 0

0

( 1) ( 2) 4 Substitute 2andsimplify

4 4

1 Divergesbynthtermtest/divergencetest

n n n

n nn n

n

xx

Theseriesclearlydivergesattheleftendpoint.Atx=6,thepowerseriesis

0 0

0

( 1) ( 2) 41 Substitute 6andsimplify

4 4

1 Divergesbynthtermtest/divergencetest

n n nn

n n

n n

n

n

xx


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Thisseriesalsodivergesattherightendpoint.Therefore,theintervalofconvergence

is(2,6),excludingtheendpointsandtheradiusofconvergenceisR=4.

(Whentheconvergencesetistheentirexaxis).

Showthat

the

power

series

1 !

n

n

x

n

convergesforallx.


Example10.21:

(Convergenceonlyatthepointx=0).

Showthatthepowerseries 1!

n

n

n x

convergesonlywhenx=0.


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Example10.22:

(Convergencesetisaboundedinterval).

Findtheconvergencesetforthepowerseries 1

n

n

x

n

.


AccordingtotheTheorem11.23,thesetofnumbersforwhichthepowerseries 0

nn

n

a x

convergesisanintervalcenteredatx=0.Wecallthistheintervalofconvergenceofthe

powerseries.Ifthisintervalhaslength2R,thenRiscalledtheradiusofconvergenceofthe

series.IftheserieshasradiusofconvergenceR=0,andifitconvergesforallx,wesaythat

R .


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Example10.23:

Findtheintervalofconvergenceforthepowerseries 0

2n n

n

x

n

.

Whatistheradiusoftheconvergence?

Example10.24:

Findtheintervalofconvergenceofthepowerseries 0

( 1)

3

n

nn

x

.


TermbytermDifferentiationandIntegration.

Supposethatapowerseries 0

nn

n

a x

hasaradiusofconvergencer>0,andletfbedefined

by

2 30 1 2 3

0

( ) ... ...n n

n n

n

f x a x a a x a x a x a x

foreveryxintheintervalofconvergence.Ifr


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Example10.25:

Findafunctionfthatisrepresentedbythepowerseries

2 31 ... ( 1) ...n n

x x x x

Example10.26:

Findapowerseriesrepresentationfor 2

1

(1 )x if 1x .


Example10.27:

Findapowerseriesrepresentationfor ln(1 )x if 1x .

Example10.28:

Findapowerseriesrepresentationfor 1tan x

.


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10.8 TaylorandMacLaurinSeries

Inthepreviouslecture,weconsideredpowerseriesrepresentationforseveralspecial

functions,includingthosewheref(x)hastheform

1

(1 )x , ln(1 )x and1tan x

,

providedxissuitablyrestricted.Wenowwishtoconsiderthefollowingquestion.

Ifafunctionfhasapowerseriesrepresentation

0

( ) n

nn

f x a x

or 0( ) ( )

nn

n

f x a x c

whatistheformof na ?

Supposethat,

2 3 40 1 2 3 4

0

( ) ...n nn

f x a x a a x a x a x a x


andtheradiusofconvergenceoftheseriesisr>0.Apowerseriesrepresentationfor ( )f x

maybeobtainedbydifferentiatingeachtermoftheseriesforf(x).Wemaythenfinda

seriesfor ( )f x bydifferentiatingthetermsoftheseriesfor ( )f x .Seriesfor ( )f x ,(4)( )f x

andsoon,canbefoundinsimilarfashion.Thus,

2 3 1

1 2 3 41( ) 2 3 4 ...

n

nnf x a a x a x a x na x

2 22 3 4

2

( ) 2 (3 2) (4 3) ... ( 1) n

nn

f x a a x a x n n a x

33 4

3

( ) (3 2) (4 3 2) ... ( 1)( 2) n

nn

f x a a x n n n a x

andforeverypositiveintegerk,

( )( ) ( 1)( 2)...( 1)k n knn k

f x n n n n k a x


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Eachseriesobtainedbydifferentiationhasthesameradiusofconvergencerastheseries

forf(x).Substituting0forxineachoftheserepresentations,weobtain

0 1 2 3(0) , (0) , (0) 2 , (0) (3 2)f a f a f a f a

andforeverypositiveintegerk,

( )( ) ( 1)( 2)...(1)k kf x k k k a

Ifweletk=n,then

( )( ) !

nnf x n a

Solvingtheprecedingequationsfor ,....,,, 210 aaa weseethat

0 1 2 3

(0) (0)(0), (0), ,

2 (3 2)

f fa f a f a a

And,ingeneral,

( )(0)

!

n

n

fa

n


MacLaurinSeriesforf(x)


0

( ) n

nn

f x a x

withradiusofconvergencer>0,then( )(0)k

f existsforeverypositiveintegerkand

( )(0)

!

n

n

fa

n

.Thus,

( )2

( )

0

(0) (0)( ) (0) (0) ... ...

2! !

(0)

!

nn

nn

n

f ff x f f x x x

n

fx

n


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TaylorSeriesforf(x)


withradiusofconvergencer>0,then( )

( )k

f c existsforeverypositiveintegerkand

( )( )

!

n

n

f ca

n

.Thus,

2

( )

( )

0

( )( ) ( ) ( )( ) ( ) ...

2!

(0) ( ) ...!

( )( )

!

n

n

nn

n

f cf x f c f c x c x c

f x cn

f cx c

n

0

( ) ( )n

nn

f x a x c

Example10.29:

FindtheMacLaurinSeriesforf(x)=cosx.

Example10.30:

FindtheTaylorSeriesforf(x)=lnxatc=1.

Documents

Chapter 10 Infinite Series