微积分(下)英文教材

Chapter1InfiniteSeries

Generally,forthegivensequence

a1,a2,a3......an,.......,

theexpression

formedbythesequencea1,a

2,

a3......an,.......,

a1

a2

a3

.....

an

.......,

iscalledtheinfiniteseriesoftheconstantsterm,denotedbyan,thatis

n1

an=a1a2a3.....an.......,

n1

Wherethenthtermissaidtobethegeneraltermoftheseries,moreover,thenthpartialsumoftheseriesisgivenby

Sna1a2a3.....an.

1.1Determinewhethertheinfiniteseriesconvergesordiverges.

Whileit’possibletoaddtwonumbers,threenumbers,ahundrednumbers,orevenamillionnumbers,it’impossibletoaddaninfinitenumberofnumbers.

Toformaninfiniteserieswebeginwithaninfinitesequenceofrealnumbers:a0,a1,a2,a3.....,wecannotformthesumofalltheak(thereis

aninfinitenumberoftheterm),butwecanformthepartialsums

0

S0

a0

ak

k

0

1

S1

a0

a1

ak

k0

2

S2

a0

a1

a2

ak

k

0

3

S3a0a1a2a3ak

k0

.

n

Sna0a1a2a3.......anak

k0

Ifthesequence{Sn}ofpartialsumshasafinitelimitL,

Wewrite

L

ak

k

0

andsaythattheseriesakconvergestoL.wecallLthesumof

k0

theseries.

Ifthelimitofthesequence{Sn}ofpartialsumsdon’texists,wesay

thattheseriesakdiverges.

k0

Remarkitisimportanttonotethatthesumofaseriesisnotasumintheorderingsense.Itisalimit.

EX1.1.1provethefollowingproposition:

Proposition1.1.1:

If

(2)If

1

1,

thenthe

thenthe

ak

k0

xk

k0

converges,and

diverges.

xk

1

;

k0

1x

Proof:thenthpartialsumofthegeometricseriesaktakesthe

k0

formSn

1x1

x2

x3.......

xn1Multiplicationbyxgives

xSn

x(1x1

x2

x3.......

xn1)=x1

x2

x3.......

xn1

xn

Subtractingthesecondequationfromthefirst,wefindthat

(1

x)Sn

1

xn.For

x

1,

thisgives

Sn

1

xn

1

x

Ifx

1,then

xn

0,andthisbyequation.

limSn

lim

1

xn

1

1

x

1

x

n

0

n0

Thisproves(1).

Nowletusprove(2).Forx=1,weuseequationanddevicethat

Snn,

Obviously,limSn,akdiverges.

nk0

Forx=-1weuseequationandwededuce

Ifnisodd,thenSn0,

Ifniseven,thenSn1.

ThesequenceofpartialsumSnlikethis0,-1,0,-1,0,-1..

Becausethelimitofsequence{Sn}ofpartialsumdoesnotexist.By

definition1.1.1,wehavetheseriesxKdiverges.(x=-1).

k0

Forx1withx1,weuseequation.Sinceinthisinstance,we

havelimSn

lim1

xn

n

n

1

x

.Thelimitofsequenceofpartialsumnotexist,

theseriesxkdiverges.

k0

Remarktheaboveseriesiscalledthegeometricseries.Itarisesin

somanydifferentcontextsthatitmeritsspecialattention.

geometricseriesisoneofthefewserieswherewecanactuallygiveanexplicitformulaforSn;acollapsingseriesisanother.

1

k0(k1)(k2)

Solutioninordertodeterminewhetherornotthisseriesconvergeswemustexaminethepartialsum.Since

1

1

1

(k1)(k2)

k1k

2

Weusepartialfractiondecompositiontowrite

1

1

1

1

Sn

2

2.

3..........

....

n(n1)

(n1)(n

2)

1.

1

1

)(

1

1

1

1

1

1

1

1

(

)

(

)

..............

(

)(

)

12

23

34

nn1

n1n2

1

1

1

1

1

1

1

1

1

1

2

2

3

34

........................

nn1n1

n2

Sinceallbutthefirstandlastoccurinpairswithoppositesigns,thesumcollapsestogive

Sn

1

1

n2

Obviously,asn

,Sn

1.thismeansthattheseriesconvergesto1.

limSn

lim(1

1

)

1

therefore

1

1

n

n

n

2

n0(k1)(k

2)

EX.1.1.3provesthefollowingtheorem:

Theorem1.1.1thekthtermofaconvergentseriestendsto0;namelyif

akConverges,bydefinitionwehavethelimitofthesequence{Sn}of

k0

partialsumsexists.Namely

n

limSn

lim

akl

n

n

0

k

Obviously

n

since

ansnsn1,wehave

limSn1

lim

ak

l.

n

n

0

k

liman

lim(Sn

Sn1)

limSn

limSn1

ll

0

n

n

n

n

Achangeinnotationgivesliman

0.

k

Thenextresultisanobviously,butimportant,consequenceof

(Adivergestest)iflimak0,orif

k

limandoesnotexist,thentheseriesakdiverges.

k

k0

limak0,andthen

k

akconverge.Infact,therearedivergentseriesforwhichlimak

0.For

k0

k

example,theseries

1

1

1

.............

1

.Since

itis

k1k

1

2

.....

n

sequence{Sn}ofpartialsum

{Sn}

1

1

1

n

is

unbounded.

So

1

2

.............

n

n

n

limSn

lim

n

,thereforetheseriesdiverges.

n

n

But

limak

lim

1

0

k

k

k

EX.1.1.3determinewhetherornottheseries:

k

0

1

2

3

4

..........Converges.

k0k

2

3

4

5

1

Solutionsincelimak

lim

k

lim

1

1

0,thisseriesdiverges.

1

k

k

k

1

k

1

k

1

k02k

Solution

1

the

given

series

is

a

geometric

series.

xk

(1)k,

and

x

1

1,byproposition

k0

k0

2

2

converges.

Solution2

Sn

1

1

1

.........

1

1,①

2

4

2

n

1

1

1

1

.........

1

1

,②

Sn

2

2

2

3

2

n1

2

n

2

2

①-②(1-

1

)Sn

1

1

,

Sn

2(1

1

n).

2

n

2

1n)

2

limSn

lim2(1

2

n

n

2

Bydefinitionofconvergesofseries,thisseriesconverges.

EX.1.1.5proofsthefollowingtheorem:

Theorem

If

the

series

k

0

akand

k0

bk

converges,then(1)

(akbk)alsoconverges,andisequalthesumofthetwoseries.

k0

(2)IfCisarealnumber,then

Cak

alsoconverges.Moreoverif

k0

aklthen

Cak

Cl.

k0

k0

Prooflet

n

n

n

n

Sn(1)

ak,Sn2

bkSn(3)

(akbk),Sn4

Cak

k0

k

0

k0

k0

NotethatSn(3)Sn(1)Sn(2)andSn(4)CSn(1)

Since

Then

limSn1

n

limSn(3)

n

l,limSn(2)

m,

n

lim(Sn(1)

Sn(2))limSn(1)

limSn(2)

lm

n

n

n

limSn(4)

n

limCSn(1)

n

ClimSn(1)Cl.

n

Theorem1.1.4(squeezetheorem)

Supposethat{an}and{cn}bothconvergetolandthatanbncnfor

nk,(kisafixedinteger),then{bn}

alsoconvergestol.

limsin3n

0

.

n

n

Solution

Forn

1

sin3n

1

1

1

1,

(

)

,since

lim()

0,andlim()0,

n

n

n

n

n

n

n

theresultfollowsbythesqueezetheorem.

Forsequenceof

variablesign,itishelpfultohavethefollowing

result.

provethatthefollowingtheoremholds.

Ifliman

0,thenliman

0,

n

n

Proof

since

an

an

an,

Namelythesqueezetheorem,weknowtheresultistrue.

Exercise1.1

(1)

Anexpressionoftheform

a1a2

a3

iscalled

(2)

A

seriesa1a2a3is

said

to

convergeifthesequence

Sn

converges,whereSn=

1.

Thegeometricseriesa

ar

ar2

convergesif

;inthiscase

thesumoftheseriesis

2.

Ifliman

0

,wecanbesurethattheseriesan

n

n1

3.

Evaluate

r(1

r)k,

0r

2.

k

0

4.

Evaluate

(1)kxk,

1

x

1.

k

0

5.

Showthat

ln

k

diverges.

k

1

k1

Findthesumsoftheseries6-11

6.

10.

1

7.

1

8.

1

9.

3

k

3(k

1)(k

2)

k

12k(k1)

k1k(k3)

k010k

3k

4k

11.

2k

3

k0

5k

k

03k

12.

Derive

the

following

results

from

the

geometric

series

(1)kx2k

1

2,

|x|1.

k

0

1x

Testthefollowingseriesforconvergence:

13.

n

14.

1

n11n

k02k3

1.2SeriesWithPositiveTerms

ThecomparisonTest

Throughoutthissection,weshallassumethatournumbersanarex0,

thenthepartialsum

Sna1a2anareincreasing,i.e.

S1S2S3SnSn1

Iftheyaretoapproachalimitatall,theycannotbecomearbitrarily

large.ThusinthatcasethereisanumberBsuchthatSnBforalln.

SuchanumberBiscalledanupperbound.ByaleastupperboundwemeananumberSwhichisanupperbound,andsuchthateveryupperboundBisS.Wetakeforgrantedthataleastupperboundexists.ThecollectionofnumbersShasthereforealeastupperbound,i.e.,theren

isasmallestnumberssuchthat

Sn

Sforalln.Inthatcase,thepartial

sums

Sn

approach

S

asalimit.

In

otherwords,givenanypositive

number

0,wehave

S

Sn

Sforallnsufficientlylarge.

S1

S2

Sn

S

ThissimplyexpressesthefactSistheleastofallupperboundsforourcollectionofnumbersSn.Weexpressthisasatheorem.

Theorem

Letan(n1,2,)beasequenceofnumbers

0andletSn

a1a2

an.IfthesequenceofnumbersSnisbounded,

thenitapproachesalimitS,whichisitsleastupperbound.

Theorem1.2.2Aserieswithnonnegativetermsconvergesifandonlyifthesequenceofpartialsumsisboundedabove.

Theorem1.2.1and1.2.2giveusaveryusefulcriteriontodeterminewhenaserieswithpositivetermsconverges.

Theconvergenceordivergenceofaserieswithnonnegativetermsis

usuallydeducedbycomparisonwithaseriesofknownbehavior.

Theorem1.2.3(TheOrdinaryComparisonTest)Let

an

and

n1

bnbetwoseries,with

an

0forallnand

bn

0foralln.Assumethat

n1

thereisanumbersc

0,

suchthat

an

cbn

for

all

n,andthat

bn

n1

converges,then

anconverges,and

an

cbn.

n1

n1

n1

Proof:

Wehave

a1a2

ancb1

cb2cbnc(b1

b2

bn)cbn.

n1

Thismeansthat

c

n

isaboundforthepartialsums1

a

2

n.

b

a

a

n1

Theleastupperboundofthesesumsisthereforecbn,thusprovingour

n1

theorem.

Theorem1.2.3hasananaloguetoshowthataseriesdoesnotconverge.

Theorem1.2.4(Ordinary

ComparisonTest)Let

an

and

bn

be

n1

n1

twoseries,withanandbn0foralln.Assumethatthereisanumber

c0suchthat

an

cbn

forallnsufficientlylarge,and

bn

doesnot

n1

converge,then

an

diverges.

n1

Proof.

Assume

an

cbn

for

n

n0

,

since

bn

diverges,we

can

n1

makethepartialsum

N

bn

bn

bn

0

1

bN

n

n0

0

N

N

N

arbitrarilylargeasNbecomesarbitrarilylarge.But

an

cbncbn.

nn0

nn0

nn0

N

aNarearbitrarily

largeasN

Hencethepartialsum

an

a1

a2

n

1

becomesarbitrarilylarge,arehenceandiverges,aswastobeshown.

n1

Remarkonnotationyouhaveeasilyseenthatforeachj0,ak

k0

convergesiff

akconverges.Thistellsusthat,indeterminingwhether

kj1

ornotaseriesconverges,itdoesnotmatterwherewebeginthe

summation,wheredetailedindexingwould

contributenothing,wewill

omit

itandwrite

withoutspecifyingwherethesummationbegins.

For

instance,it

makessensetoyouthat

12convergesand

1

k

k

divergeswithoutspecifyingwherewebeginthesummation.Butintheconvergentcaseitdoes,however,affectthesum.Thusforexample

1

2,

1

1,

1

1,andsoforth.

k02k

k12k

k22k

2

12

converges.

n1n

Solution

Letuslookattheseries:

1

1

1

1

1

1

1

1

1

12

22

32

42

52

72

82

152

162

Welookatthegroupsoftermsasindicated.Ineachgroupofterms,ifwedecreasethedenominatorineachterm,thenweincreasethefraction.Wereplace3by2,then4,5,6,7by4,thenwereplacethenumbersfrom8to15by8,andsoforth.Ourpartialsumsthereforelessthanorequalto

1

1

1

1

1

1

1

1

andwenotethat2occurs

12

22

22

42

42

42

82

82

twice,4occursfourtimes,8occurseighttimes,andsoforth.Ourpartial

sumarethereforelessthanorequalto

1

1

1

1

1

1

1

1

12

22

22

42

42

42

82

82

andwenotethat2occurstwice,4occursfourtimes,8occurseighttimes,andsoforth.Hencethepartialsumsarelessthanorequalto

1

2

4

8

1

1

1

?

=1+

12

22

42

82

2

4

8

Thusourpartialsumsarelessthanorequaltothoseofthegeometricseriesandarebounded.Henceourseriesconverges.

Generallywehavethefollowingresult:

1

1

1

1

1

1

,wherepisaconstant,

Theseriesn1np

2p

3p

4p

np

iscalledap-series.

p

1,thep-seriesconverges;andifp1,then

thep-seriesdiverges.

n2

converges.

3

n1n

1

SolutionWewrite

n2

1

1

1(1

1

)

.Thenweseethat

n3

n

1

n

1

1

n2

n3

n2

1

11

1

n3

1

2n

2n.Since

n1n

doesnotconverge,itfollowsthattheseries

n2

doesnotconvergeeither.Namelythisseriesdiverges.

n1n3

1

n2

7

converges.

n12n4

n3

Proof:

Indeedwecanwrite

n

2

7

n2(1

7

2)

1

1

7

2

n

n

2n4

4(2(1

3

4)n2

2(1

3

n3n

)3

n

n

)3

4

n

n

Fornsufficientlylarge,thefactor

1

7

2

iscertainlybounded,

2(1

n

3

)3

4

n

n

andinfactisnear1/2.Hencewecancompareourserieswith

1

n2to

seeconverges,because

12

convergesandthefactorisbounded.

n

1

diverges.

ln(k

b)

Solution

1Weknow

that

ask

,

lnk

0

.

It

followsthat

k

ln(k

b)

0

,and

thus

that

ln(k

b)

ln(k

b)k

b

0.

Thus

for

k

k

b

k

k

b

k

sufficiently

large,

ln(k

b)

k

and

1

1

.

Since

1

diverges,

k

ln(k

b)

k

wecanconcludethat

1

diverges.

ln(k

b)

Solution

2Another

way

to

showthat

ln(k

b)

k

forsufficiently

largek

is

toexaminethe

function

f(x)

xln(x

b)

.At

x

3

the

functionispositive:

f(3)3ln932.1970

Sincef'(x)

1

1

b

0for

allx

0,

f(x)

0

forall

x

3

.It

follows

x

that

ln(x

b)x

forall

x

3.

Wecomenowtoasomewhatmorecomparisontheorem.Ourproof

reliesonthebasiccomparisontheorem.

Theorem1.2.5(TheLimitComparisonTest)Let

ak

and

bkbe

serieswith

positive

terms.Iflim(ak

)

l,

wherel

is

somepositive

k

bk

number,then

ak

and

bk

convergeordivergetogether.

Proof

Choose

between0andl,since

ak

l,weknowforall

bk

ksufficientlylarge(forall

k

greaterthansomek0)|ak

l|.

bk

Forsuchk

wehavel

ak

l

,andthus(l

)bk

ak

(l

)bk

bk

thislastinequalityiswhatweneeded.

(1)If

ak

converges,

then

(l

)bk

converges,

and

thus

bkconverges.

(2)Ifbkconverges,then(l)bkconverges,andthus

akconverges.

Toapplythelimitcomparisontheoremtoaseriesak,wemustfirst

findaseriesbkofknownbehaviorforwhichakconvergestoa

bk

positivenumber.

ExDeterminewhethertheseriessinconvergesor

k

diverges.

Solution

Recallthatas

x

0,sinx

1

As

k

,

0

andthus

x

.

k

sin

k

1

.Since

k

diverges,so

sin(

k)diverges.

k

5

k

100

convergesor

2k2

k

9k

diverges.

Solution

For

largevalueofk,

5kdominatesthenumeratorand

2k2

kdominatesthedenominator,thus,for

suchk,

5

k

100

differs

2k2

k

9

k

littlefrom

5

k

5.Since

2k2

k

2k2

5

k

100

5

10k

2

k

200k

2

1

20

k

1

2k2

k

9k

2k2

100k2

k

45

k

1

9

2

2k

And

52

5

12converges,thisseriesconverges.

2k

2

k

Theorem

ak

and

bk

beserieswith

positiveterms

andsupposethusak

0

,then

bk

If

If

If

If

bk

ak

ak

bk

converges,thenakconverges.

diverges,thenbkdiverges.

converges,thenbkmayconvergeordiverge.

diverges,thenakmayconvergeordiverge.

[Parts(3)and(4)explainwhywestipulatedl0intheorem1.2.5]

Theorem1.2.7(theroottest,Cauchytest)letakbeaserieswith

nonnegativetermsandsupposethat

1

limkak

limakk

,if

<1,

ak

converges,

if

>1,

k

k

ak

diverges,if=1,thetestisinconclusive.

Proofwesupposefirst

<1andchoose

sothat

u

1.Since

1

1

(ak)k

,wehaveakk

,forallksufficientlylargethusak

kforall

ksufficiently

large

since

kconverges(a

geometric

serieswith

0<

1

akconverges.

Wesupposenowthat

11

(ak)k,wehave(ak)k

forallksufficientlylarge.

1andchooseso

thatu1

.since

forallksufficiently

large.Thus

k

ak

Since

k

diverges(ageometricserieswith

1

)thetheorem

akdiverges.

Toseetheinconclusivenessof

theroottestwhen

1

,notethat

1

1forboth

(ak)k

1and

1

1

(1

1

(1

1

1

:(ak)k

)k

)2

12

1,(ak)k

(

1)k

1

1

k2

k

k2

1

k

kk

kk

Thefirstseriesconverges,buttheseconddiverges.

1

convergesor

(lnk)k

diverges.

Solution

Fortheseries

1

,applyingtheroottestwehave

(lnk)k

1

lim(ak)k

k

lim

1

,theseriesconverges.

0

lnk

Determinewhetherseries

2k3

convergesordiverges.

(k)

Solution

Fortheseries

2kk,applyingtheroottest,wehave

(3)

1

3

2[1k]3

(ak)k

2(.

1)k

2

13

21.Sotheseriesdiverges.

k

k

Determineswhether

theseries

1

k

(1

)convergesor

k

diverges.

1)k

1

Solutioninthecaseof(1

,wehave(ak)k

k

applyingtheroottest,itisinconclusive.Butsinceak

(1

to1andnotto0,theseriesdiverges.

e

1

1

.If

1

k

)kconverges

k

Wecontinuetoconsideronlyserieswithterms0.Tocompare

suchaserieswithageometricseries,thesimplesttestisgivenbythe

ratiotesttheorem

Theorem1.2.8(Theratiotest,DAlemberttest)letakbeaseries

withpositivetermsandsupposethat

ak1

lim,

ak

If1,akconverges,if1,akdiverges.

Ifthe1,thetestisinconclusive.

Proof

wesupposefirstthat

1,sincelim

ak1

1

kak

SothereexistssomeintegerNsuchthatifnN

an1

C

Then

aN1

CaN,

aN

2

CaN1

C2aN

and

ingeneralby

an

induction

aNkCkaN,

Thus

N

k

c2aN......ckaN

an

aN

caN

n

N

aN(1

c

c2

c3

........ck)

aN

1

c

1

Thusineffect,wehavecomparedourserieswithageometricseries,andweknowthatthepartialsumsarebounded.Thisimpliesthatourseriesconverges.

Theratiotestisusuallyusedinthecaseofaserieswithpositivetermsan

suchthat

an1

1.

lim

n

an

nn

converges.

n13

Solution

welet

an

nn,then

an1

n

n11.3n

n

1.1

,thisratio

3

an

3

n

n

3

1

,andhencetheratiotestisapplicable:theseries

approaches

asn

3

converges.

kk

diverges.

k!

Solutionwehave

an1

(k1)k1k!

k1

k

1

k

an

(k1)!kk

(

k

)

(1

k

)

Solim

an1

lim(1

1

)k

e

n

an

k

k

Sincee1,theseriesdiverges.

1

diverges.

2k

1

1

Solutionsince

ak1

1

.2k

12k1

2

k

ak

2(k1)11

2k32

3

1

k

lim

ak1

lim

2

1.

2

k

k

ak

k

3

k

Thereforetheratiotestisinconclusive.Wehavetolookfurther.

Comparisonwiththeharmonicseriesshowsthattheseriesdiverges:

1

1

1

.

1

1

dverges.

2k1

2(k1)

2

,

2(k1)

k

1

Exercise1.2

1.Theordinarycomparisontestsaysthatif____andif

biconverges.

Thenakalsoconverges.

2.Assumethatak

0andbk

0.ThelimitcomparisonTestsaysthatif

0<____<+

then

ak

and

bkconvergesordivergetogether.

3.Let

lim

an1.TheratioTestsaysthataseries

akofpositiveterms

n

an

convergesif___,divergesif____andmaydoeitherif___.

Determinewhethertheseriesconvergesordiverges

4.

k

5.

1

6.

1

17.

1

k3

1

(2k

1)2

k

2k2

k

8.

tan1k

9.

1

10.

(

3

)

k

11.

ln

k

12.

10k

k2

1

2

4

k

k!

k3

1

k!

k2

2

k(2)

k

1

13.

14.

15.

16.

17.

kk

100k

k3

6k

3

1k.

18.

k!

104k

19.

Let

{an}beasequenceofpositivenumberandassumethat

an1

1

1foralln.showthattheseriesandiverges.

an

n

1.3Alternatingseries,Absoluteconvergenceandconditionalconvergence

Inthissectionweconsiderseriesthathavebothpositiveandnegativeterms.

Alternatingseriesandthetestsforconvergence

Theseriesof

theform

u1

u2

u3u4.......iscalledthealternating

series,whereun

0

foralln,heretwoexample:

1

1

1

1

1

1

....

(

1)n1

2

3

4

5

6

,

n1

n

1

2

3

4

5

....

(

1

2

3

4

5

6

1)n

1

n1

n

Weseefromtheseexamplesthatthenthtermofanalternatingseriesis

theforman

(1)n1unoran

(1)nun,whereunisapositivenumber(in

factun

an

.)

The

followingtestsaysthatifthetermsofanalternatingseries

decreasetoward0inabsolutevalue,thentheseriesconverges.

Theorem1.3.1(LeibnizTheorem)

Ifthealternatingseries

(1)nunsatisfy:

n1

(1)unun1(n=1,2);(2)limun0,

n

thentheseriesconverges.Moreover,itissumu1,andtheerrorrnmake

byusingsnofthefirstntermstoapproximatethesumsoftheseriesis

notmorethanun1,thatis,rnun1namelyrnssnun1.

oftheideabehindtheproof.Wefirstplots1u1onanumberline.

Tofinds2wesubtractu2,sos2istheleftofs1.Thentofinds3we

addu3,sos3istotherightofs2.But,sinceu3<u2,s3istotheleftofs1.

Continuinginthismanner,weseethatthepartialsumsoscillatebackandforth.Sinceun0,thesuccessivestepsarebecomingsmallerandsmaller.Theevenpartialsumss2,s4,s6,........areincreasingandtheoddpartial

sumss1,s3,s5,........aredecreasing.Thusitseemsplausiblethatbothare

convergingtosomenumbers,whichisthesumoftheseries.Therefore,inthefollowingproofweconsidertheevenandoddpartialsumsseparately

Wegivethefollowingproofofthealternatingseriestest.Wefirstconsidertheevenpartialsums:

s2

u1

u2

0,

Since

s4

s2

(u3

u4)s2,

since

u2u1

u4u

Ingeneral,

s2n

s2n2

(u2n1u2n)s2n2sinceu2nu2n1

Thus0s2

s4

s6

............s2n.............

Butwecanalsowrite

s2nu1(u2u3)(u4

u5)....(u2n2

u2n1)

u2n

Everyterminbracketsis

positive,sos2n

u1

foralln.therefore,the

sequence{s2n}ofevenpartialsumsisincreasingandboundedabove.Itis

thereforeconvergentbythemonotonicsequencetheorem.Let

’scallit

limits,thatis,

lims2n

s

Nowwecomputethelimitoftheoddpartial

n

sums:

lims2n1

lim(s2n

u2n1)

n

n

lims2n

limu2n

1

s

0(bycondition(2))

s

n

n

Since

both

the

even

and

oddpartial

sumsconverge

tos,

we

havelimsn

s,andsotheseriesisconvergent.

n

showsthat

the

following

alternating

harmonic

series

is

convergent:

1

1

1

..........

(

1)n1

1

3

4

n

.

2

n1

Solution

thealternatingharmonicseriessatisfies

(1)un

1

1

un

1;

(2)

limun

lim1

0

n

1

n

n

n

n

SotheseriesisconvergentbyalternatingseriesTest.

Testtheseries

(1)n3nforconvergenceanddivergence.

n1

4n1

Solutionthegivenseriesisalternatingbut

limun

lim

3n

lim

3

3

4n

1

4

1

0

n

n

n

4

n

Socondition(2)isnotsatisfied.Instead,welookatthelimitofthenth

termoftheseries:

liman

lim(

1)3nThislimit

doesnotexist,sotheseriesdivergesby

n

n

4n

1

thetestfordivergence.

Testtheseries

(

2

1)n

forconvergenceordivergence.

n1n

1

Solutionthegivenseriesisalternatingsowetrytoverifyconditions(1)and(2)ofthealternatingseriestest.

Unlikethesituationinexample1.3.1,itisnotobviousthesequence

givenby

un

n

isdecreasing.Ifweconsidertherelatedfunction

n2

1

x

f(x)

，we

easilyfindthat

x2

1

f'(x)

x2

1

2x2

1

x2

0whenverx2

1

.

(x2

1)2

(x2

1)2

Thusf

isdecreasingon[1,)

andsof(n)

f(n1).Therefore,{un}is

decreasing

Wemayalsoshowdirectlythatun1un,thatis

n

1

n

(n1)2

1

n2

1

Thisinequalityitequivalenttotheonewegetbycrossmultiplication:

(n

n

1

n

(n

1)(n2

1)n[(n

1)2

1]

1)2

1

n2

1

n3

n2

n1n3

2n2

2n1n2

n

Sincen

1,

we

know

thatthe

inequality

n2

n1is

true.Therefore,

un1

unand

{un}isdecreasing.

Condition(2)isreadilyverified:

n

1

limun

lim

lim

n

0,thusthegivenseriesisconvergentby

n

2

1

n

n

n

n

1

n

theAlternatingseriesTest.

Absoluteandconditionalconvergence

Inthissectionweconsiderseriesthathavebothpositiveandnegativeterms.Absoluteandconditionalconvergence.

akisnotserieswithpositive

k1

terms,iftheseriesakformedwiththeabsolutevalueoftheterms

k1

anconverges,theseriesakiscalledabsolutelyconvergent.Theseries

k1

akiscalledconditionallyconvergent,iftheseriesakconverges

k1k1

butakdiverges.

k1

ifakconverges,theakconverges.

Proofforeachk,akakak,andtherefore0akak2ak.if

akconverges,then2ak2akconverges,andtherefore,by

theorem1.2.3(theordinarycomparisontheorem),(akak)converges.

Sinceak(akak)akbythetheorem1.1.2(1),wecanconcludethat

akisconvergence.

TheabovetheoremwejustprovedsaysthatAbsolutelyconvergentseriesareconvergent.

Aswellshowpresently,theconverseisfalse.Thereareconvergent

series

that

are

notabsolutely

convergent;suchseriesarecalled

conditionallyconvergent.

Provethefollowingseriesisabsolutelyconvergent

1

1

1

1

1

.............

22

32

42

52

Proof

If

wereplacetermbyit

’absolutevalue,weobtaintheseries

1

1

1

.....

1

32

42

22

ThisisaPserieswithP=2.Itisthereforeconvergent.Thismeansthattheinitialseriesisabsolutelyconvergent.

provesthatthefollowingseriesisabsolutelyconvergent:

1

1

1

1

1

1

1

1

1

22

23

24

25

26

27

28.............

2

Proofifwereplaceeachtermbyitsabsolutevalue,weobtaintheseries:

1

1

1

1

1

1

1

1

1

22

23

24

25

26

27

28.............

2

Thisisaconvergentgeometricseries.Theinitialseriesisthereforeabsolutelyconvergent.

Ex.1.3.6provesthatthefollowingseriesisonlyconditionallyconvergent:

1

1

1

1

1

(

1)n

1

3

4

5

6.............

n

2

n1

Proof

thegivenseriesisconvergent.

Since(1)

un1

1

un

1

,(2)limun

lim

1

0,

n

1

n

n

n

n

Sothisseriesisconvergentbythealternatingseriestest,butitisnotabsolutely.

Convergent:ifwereplaceeachtermbyitisabsolutevalue,weobtainthedivergentharmonicseries:

111

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

234

Sothisseriesisonlyconditionallyconvergent.

Exercise1.3

(a)whatisthealternatingseries.

Underwhatconditionsdoesanalternatingseriesconverge?

Iftheseconditionaresatisfied,whatcanyousayaboutthe

remainderafternterms?

2.

if

un0forall

n,thealternatingseries

u1u2u3...........will

convergeprovidedthatthetermsaredecreasinginsizeand_____.

3.

If

akconverges,wesaythatseries

akconverges___;if

ak

converges,but

akdiverges,wesaythat

akconverges____.

Determinewhetherthefollowingseriesconvergeabsolutely

4.

sinn

5.

(

1)n

6.

(1)n

cos3n

n3

n2

1

n2

n

Determinewhetherthefollowingseriesconvergeandwhetherthey

convergeabsolutely.

7.

(

1)n1

8.

(

1)n

9.

(

1)

n

nl

n

n

n

n3

l

10.(

n1

n

11.

(

n

n

1)

2

1)

2

1

n

n

Testtheseseriesfor(1)absoluteconvergence(2)conditionalconvergence

12.

(1)k1

13.

k

k)

(1)(k1

2k1

14.

sin(

2)15.

k

k

(1)

2

k

4k

1.4powerseries

Theformula

u1

(x)u2(x)................

un(x)....

IscalledtheinfiniteseriesoffunctionontheintervalI,where

u1

(x),u2(x),................un(x)

.......areallfunctionsdefinedontheintervalI

Fortheseriesoffunction(1.4.1),thepointx0iscalledtheconvergent

pointifu1(x0)u2(x0),...............un(x0).......isconvergent.Wecalltheset

onwhichtheinfiniteseries(1.4.1)offunctionconvergesitsconvergentset.

powerseriesisaseriesinwhicheachtermisapowerfunction,namely,apowerseriesisaseriesoftheform

xn...

n0

）

Wherexisavariableandcnarcconstantscalledthecoefficientsof

theseries.

Theseries(1.4.2)isaseriesofconstantsthatwecantestforconvergenceordivergence.Apowerseriesmayconvergeforsomevaluesofxanddivergeforthevaluesofx.Thesumoftheseriesisafunction.

f(x)

c0

c1x

c2x2

...

cnxn

...

Whosedomainisthesetofallxinwhichtheseriesconverges.Notice

thatf.resemblesapolynomial,theonlydifferenceisthatfhas

infinitelymanyterms

Moregenerally,aseriesoftheform

cn(x

a)n

c0

c1(x

a)

c2

(x

a)2

........

n0

iscalledapowerseriesin(xa)orapowerseriescenteredataorapowerseriesabouta.Noticethatinwritingoutthetermcorrespondington0in（）and(1.4.3).Wehaveadoptedtheconventionthat

(xa)01evenwhenxa.Noticealsothatwhenxaallofthetermsare0forn1andsothepowerseries(1.4.3)alwaysconverges

whenxa.

Sinceasimpletranslationconverts

ak(xa)k

into

akxk,wecan

k0

k

0

focusourattentiononpowerseriesoftheform

akxk

.

k

0

Whendetailedindexingisunnecessary,wewillomititand

writeakxk.Webegindiscussionwithadefinition.

Apowerseries

akxk

issaidtoconverge

(1)At

x1,iff

akx1kconverges

(2)Onthesetsiff

akxk

convergesfor

each

x

ins.Thefollowing

resultisfundamental

Theorem1.4.1AbelTheorem(1)Ifthepowerseriesakxkconverges

at

x1

0

thenitconvergesabsolutely.Forany

xthatx

x1(2)Ifthe

powerseries

akxk

divergesat

x1

0,thenitdivergesforany

xthat

x

x1.

Proof

If

akx1k

converges,

then

akx1k

0

In

particular,

fork

k

k

sufficientlylarge,akx1k

1Andthusakxk

akx1k

x

x

Forx

x1,

x1

x1

wehave

x

1.

x1

Theconvergenceof

akxk

followsbycomparisonwiththegeometric

series.This

proves

thefirst

statement,Supposenow

that

akx1k

diverges.By

the

previonsargument,There

can

not

exist

x

with

x

x1

suchthat

akxkconverge