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11.1 Sequences
261
THEOREM11.1.12
Ifasequenceisboundedandmonotonicthenitconverges.
Wewillnotprovethis;theproofappearsinmanycalculusbooks. Itisnothardto
believe:supposethatasequenceisincreasingandbounded,soeachtermislargerthanthe
onebefore,yetneverlargerthansomexedvalueN.Thetermsmustthengetcloserand
closertosomevaluebetweena
0
andN.ItneednotbeN,sinceNmaybea\too-generous"
upperbound;thelimitwillbethesmallestnumberthatisaboveallofthetermsa
i
.
EXAMPLE 11.1.13
Alloftheterms(2
i
1)=2
i
arelessthan2,andthesequenceis
increasing. Aswehaveseen,thelimitofthesequenceis1|1isthesmallestnumberthat
isbiggerthanallthetermsinthesequence. Similarly,alloftheterms(n+1)=narebigger
than1=2,andthelimitis1|1isthelargestnumberthatissmallerthanthetermsofthe
sequence.
Wedon’tactuallyneedtoknowthatasequenceismonotonictoapplythistheorem|
itisenoughtoknowthatthesequenceis\eventually"monotonic,thatis,thatat some
pointitbecomesincreasingordecreasing.Forexample,thesequence10,9,8,15,3,21,4,
3=4,7=8,15=16,31=32;:::isnotincreasing,becauseamongtherstfewtermsitisnot.
Butstartingwiththeterm3=4itisincreasing,sothetheoremtellsusthatthesequence
3=4;7=8;15=16;31=32;:::converges. Sinceconvergencedependsonlyonwhathappensas
ngetslarge,addingafewtermsatthebeginningcan’tturnaconvergentsequenceintoa
divergentone.
EXAMPLE11.1.14
Showthatfn
1=n
gconverges.
Werstshowthatthissequenceisdecreasing,thatis,thatn
1=n
>(n+1)
1=(n+1)
.Consider
the real function n f(x) ) = = x
1=x
when x x  1. . We e can n compute e the e derivative, , f
0
(x) =
x
1=x
(1 lnx)=x
2
,andnotethatwhenx3thisisnegative.Sincethefunctionhasnegative
slope,n
1=n
>(n+1)
1=(n+1)
whenn3. Sincealltermsofthesequencearepositive,the
sequenceisdecreasingandboundedwhenn3,andsothe sequence converges. (Asit
happens,we cancomputethe limit inthiscase,but weknowit convergesevenwithout
knowingthelimit;seeexercise1.)
EXAMPLE11.1.15
Showthatfn!=n
n
gconverges.
Againweshowthatthesequenceisdecreasing,andsinceeachtermispositivethesequence
converges. Wecan’ttakethederivativethistime,asx!doesn’tmakesenseforxreal. But
wenotethatifa
n+1
=a
n
<1thena
n+1
<a
n
,whichiswhatwewanttoknow.Sowelook
ata
n+1
=a
n
:
a
n+1
a
n
=
(n+1)!
(n+1)n+1
n
n
n!
=
(n+1)!
n!
n
n
(n+1)n+1
=
n+1
n+1
n
n+1
n
=
n
n+1
n
<1:
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262
Chapter 11 1 SequencesandSeries
(Againitispossibletocomputethelimit;seeexercise2.)
Exercises 11.1.
1. Compute lim
x!1
x
1=x
)
2. Usethesqueezetheoremtoshowthat lim
n!1
n!
nn
=0.
3. Determinewhetherf
p
n+47 
p
ng
1
n=0
convergesordiverges. Ifitconverges,computethe
limit. )
4. Determinewhether
n
2
+1
(n+1)2
1
n=0
convergesordiverges.Ifitconverges,computethelimit.
)
5. Determine e whether
n+47
p
n+3n
1
n=1
converges or r diverges. If f it converges, compute the
limit. )
6. Determinewhether
2
n
n!
1
n=0
convergesordiverges. )
While much more can be said about sequences, , we e nowturn to ourprincipalinterest,
series. Recallthat t aseries,roughlyspeaking,isthe sumofasequence: iffa
n
g
1
n=0
isa
sequencethentheassociatedseriesis
X1
i=0
a
n
=a
0
+a
1
+a
2
+
Associated with a a series s isa second sequence, called the e sequence e of partial sums
fs
n
g
1
n=0
:
s
n
=
Xn
i=0
a
i
:
So
s
0
=a
0
; s
1
=a
0
+a
1
; s
2
=a
0
+a
1
+a
2
; :::
A series s converges if the e sequence e of f partial sums s converges, , and otherwise e the series
diverges.
EXAMPLE11.2.1 Ifa
n
=kx
n
,
X1
n=0
a
n
iscalledageometricseries. Atypicalpartial
sumis
s
n
=k+kx+kx
2
+kx
3
++kx
n
=k(1+x+x
2
+x
3
++x
n
):
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11.2 Series
263
Wenotethat
s
n
(1 x)=k(1+x+x
2
+x
3
++x
n
)(1 x)
=k(1+x+x
2
+x
3
++x
n
)1 k(1+x+x
2
+x
3
++x
n 1
+x
n
)x
=k(1+x+x
2
+x
3
++x
n
x x
2
x
3
 x
n
x
n+1
)
=k(1 x
n+1
)
so
s
n
(1 x)=k(1 x
n+1
)
s
n
=k
1 xn+1
1 x
:
Ifjxj<1, lim
n!1
x
n
=0so
lim
n!1
s
n
= lim
n!1
k
1 x
n+1
1 x
=k
1
1 x
:
Thus,whenjxj<1thegeometricseriesconvergestok=(1 x).When,forexample,k=1
andx=1=2:
s
n
=
1 (1=2)n+1
1 1=2
=
2n+1 1
2n
=2 
1
2n
and
X1
n=0
1
2n
=
1
1 1=2
=2:
Webeganthechapterwiththeseries
X1
n=1
1
2n
;
namely,thegeometricserieswithouttherstterm1. Eachpartialsumofthisseriesis1
lessthanthecorrespondingpartialsumforthegeometricseries,soofcoursethelimit is
alsoonelessthanthevalueofthegeometricseries,thatis,
X1
n=1
1
2n
=1:
Itisnothardtoseethatthefollowingtheoremfollowsfromtheorem11.1.2.
THEOREM 11.2.2
Suppose that
P
a
n
and
P
b
n
are convergent t series, , and c isa
constant. Then
1.
X
ca
n
isconvergentand
X
ca
n
=c
X
a
n
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264
Chapter 11 1 SequencesandSeries
2.
X
(a
n
+b
n
)isconvergentand
X
(a
n
+b
n
)=
X
a
n
+
X
b
n
.
Thetwopartsofthistheoremaresubtlydierent.Supposethat
P
a
n
diverges;does
P
ca
n
alsodivergeifcisnon-zero? Yes: : supposeinsteadthat
P
ca
n
converges;thenby
thetheorem,
P
(1=c)ca
n
converges,but thisisthe same as
P
a
n
, whichbyassumption
diverges. Hence
P
ca
n
also diverges. Note e that we are applying the theorem m with h a
n
replacedbyca
n
andcreplacedby(1=c).
Nowsuppose that
P
a
n
and
P
b
n
diverge;does
P
(a
n
+b
n
) alsodiverge? ? Nowthe
answerisno:Leta
n
=1andb
n
= 1,socertainly
P
a
n
and
P
b
n
diverge. But
P
(a
n
+
b
n
) =
P
(1+ 1) ) =
P
0 = = 0. Of f course, sometimes
P
(a
n
+b
n
) willalsodiverge, , for
example,ifa
n
=b
n
=1,then
P
(a
n
+b
n
)=
P
(1+1)=
P
2diverges.
Ingeneral,thesequenceofpartialsumss
n
ishardertounderstandandanalyzethan
thesequenceoftermsa
n
,anditisdiculttodeterminewhetherseriesconvergeandifso
towhat.Sometimesthingsarerelativelysimple,startingwiththefollowing.
THEOREM11.2.3
If
P
a
n
convergesthen lim
n!1
a
n
=0.
Proof. Since
P
a
n
converges, lim
n!1
s
n
=Land lim
n!1
s
n 1
=L,becausethisreallysays
thesamethingbut\renumbers"theterms. Bytheorem11.1.2,
lim
n!1
(s
n
s
n 1
)= lim
n!1
s
n
lim
n!1
s
n 1
=L L=0:
But
s
n
s
n 1
=(a
0
+a
1
+a
2
++a
n
) (a
0
+a
1
+a
2
++a
n 1
)=a
n
;
soasdesired lim
n!1
a
n
=0.
Thistheorempresentsaneasydivergencetest:ifgivenaseries
P
a
n
thelimit lim
n!1
a
n
does not t exist t or has a a value e otherthan zero, the series diverges. Note e well that the
converseisnot true:If lim
n!1
a
n
=0thentheseriesdoesnotnecessarilyconverge.
EXAMPLE11.2.4
Showthat
X1
n=1
n
n+1
diverges.
Wecomputethelimit:
lim
n!1
n
n+1
=16=0:
Looking at the rst fewterms s perhapsmakesit clearthat the e serieshas s no o chance of
converging:
1
2
+
2
3
+
3
4
+
4
5
+
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11.2 Series
265
willjustgetlargerandlarger;indeed,afterabitlongertheseriesstartstolookverymuch
like+1+1+1+1+,andofcourseifweaddupenough1’swecanmakethesum
aslargeaswedesire.
EXAMPLE11.2.5
Showthat
X1
n=1
1
n
diverges.
Here thetheoremdoesnotapply: lim
n!1
1=n=0,soitlookslikeperhapstheseriescon-
verges. Indeed,ifyouhavethefortitude(orthesoftware)toadduptherst1000terms
youwillndthat
1X000
n=1
1
n
7:49;
soitmightbereasonabletospeculatethattheseriesconvergestosomethingintheneigh-
borhoodof10. Butinfactthepartialsumsdogotoinnity;theyjustgetbigvery,very
slowly. Considerthefollowing:
1+
1
2
+
1
3
+
1
4
>1+
1
2
+
1
4
+
1
4
=1+
1
2
+
1
2
1+
1
2
+
1
3
+
1
4
+
1
5
+
1
6
+
1
7
+
1
8
>1+
1
2
+
1
4
+
1
4
+
1
8
+
1
8
+
1
8
+
1
8
=1+
1
2
+
1
2
+
1
2
1+
1
2
+
1
3
++
1
16
>1+
1
2
+
1
4
+
1
4
+
1
8
++
1
8
+
1
16
++
1
16
=1+
1
2
+
1
2
+
1
2
+
1
2
andsoon. Byswallowingupmoreandmoretermswecanalwaysmanagetoaddatleast
another1=2tothesum,andbyaddingenoughofthesewecanmakethepartialsumsas
bigaswelike.Infact,it’snothardtoseefromthispatternthat
1+
1
2
+
1
3
++
1
2n
>1+
n
2
;
so to o make e sure e the sum m is over 100, forexample, we’d d add up termsuntil we e get t to
around1=2
198
,thatis,about410
59
terms.Thisseries,
P
(1=n),iscalledtheharmonic
series.
Exercises11.2.
1. Explainwhy
X1
n=1
n
2
2n+1
diverges. )
2. Explainwhy
X1
n=1
5
21=n+14
diverges. )
3. Explainwhy
X1
n=1
3
n
diverges. )
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266
Chapter 11 1 SequencesandSeries
4. Compute
X1
n=0
4
( 3)n
3
3n
.)
5. Compute
X1
n=0
3
2n
+
4
5n
.)
6. Compute
X1
n=0
4
n+1
5n
)
7. Compute
X1
n=0
3
n+1
7n+1
)
8. Compute
X1
n=1
3
5
n
)
9. Compute
X1
n=1
3
n
5n+1
)
Itisgenerallyquitedicult,oftenimpossible,todeterminethevalueofaseriesexactly.
Inmanycasesitispossibleatleasttodeterminewhetherornottheseriesconverges,and
sowewillspendmostofourtimeonthisproblem.
Ifallofthetermsa
n
inaseriesarenon-negative,thenclearlythesequenceofpartial
sumss
n
isnon-decreasing. Thismeansthatifwecanshowthat t thesequenceofpartial
sums is bounded, , the e seriesmust t converge. We e know w that t if f the e seriesconverges, , the
termsa
n
approachzero,butthisdoesnotmeanthata
n
a
n+1
foreveryn. Manyuseful
andinterestingseriesdohavethisproperty,however,andtheyareamongtheeasiest to
understand.Let’slookatanexample.
EXAMPLE11.3.1
Showthat
X1
n=1
1
n2
converges.
The terms s 1=n
2
are positive and decreasing, , and since lim
x!1
1=x
2
=0, the e terms1=n
2
approach zero. We e seekan n upperbound d for r all the e partialsums, , that is, we e want t to
nda numberN N sothat t s
n
N foreveryn. Theupperboundisprovidedcourtesyof
integration,andisinherentingure11.3.1.
Thegureshowsthegraphofy=1=x
2
togetherwithsomerectanglesthatliecom-
pletelybelowthe curve and that all l have e base length one. Because e the heights s of f the
rectanglesaredeterminedbytheheightofthecurve,theareasoftherectanglesare1=1
2
,
1=2
2
,1=3
2
,andsoon|inotherwords,exactlythe termsoftheseries. Thepartialsum
s
n
issimplythesumoftheareasoftherstnrectangles. Becausetherectanglesalllie
betweenthecurveandthex-axis,anysumofrectangleareasislessthanthecorresponding
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11.3 The e IntegralTest
267
0
1
2
0
1
2
3
4
5
..
..
..
..
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..
..
..
..
.
..
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..
..
.
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..
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..
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..
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.
..
.
..
.
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..
.
..
.
..
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..
.
..
..
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.
..
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..
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..
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...
..
..
..
...
..
...
..
...
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...
...
..
...
...
...
...
...
....
...
....
....
....
....
....
.....
....
.....
......
......
......
.......
.......
........
.........
..........
...........
............
.............
................
..................
......................
..........................
.................................
..........................................
.............................................................
..............................................................
A=1
A=1=4
Figure11.3.1
Graphofy=1=x
2
withrectangles.
area underthecurve, , andso o ofcourseanysumofrectangle areasislessthanthe area
undertheentirecurve,thatis,allthewaytoinnity.Thereisabitoftroubleattheleft
end,wherethereisanasymptote,butwecanworkaroundthateasily.Hereitis:
s
n
=
1
12
+
1
22
+
1
32
++
1
n2
<1+
Z
n
1
1
x2
dx<1+
Z
1
1
1
x2
dx=1+1=2;
recallingthat we computed thisimproperintegralinsection 9.7. Sincethe e sequenceof
partialsumss
n
isincreasingandboundedaboveby2,weknowthat lim
n!1
s
n
=L<2,and
sotheseriesconvergestosomenumberlessthan2.Infact,itispossible,thoughdicult,
toshowthatL=
2
=61:6.
We already know that
P
1=n diverges. What t goes wrong g if we e try to o apply y this
techniquetoit? Here’sthecalculation:
s
n
=
1
1
+
1
2
+
1
3
++
1
n
<1+
Z
n
1
1
x
dx<1+
Z
1
1
1
x
dx=1+1:
The problem m isthat the e improper r integral doesn’t converge. Note e well that this does
not provethat
P
1=ndiverges,justthat thisparticularcalculationfailstoprovethatit
converges. Aslightmodication,however,allowsustoproveinasecondwaythat
P
1=n
diverges.
EXAMPLE 11.3.2
Considera slightlyalteredversion ofgure 11.3.1, showning-
ure11.3.2.
Therectanglesthistimeareabovethecurve,thatis,eachrectanglecompletelycontains
thecorrespondingareaunderthecurve.Thismeansthat
s
n
=
1
1
+
1
2
+
1
3
++
1
n
>
Z
n+1
1
1
x
dx=lnx
n+1
1
=ln(n+1):
Asngetsbigger,ln(n+1)goestoinnity,sothesequenceofpartialsumss
n
mustalso
gotoinnity,sotheharmonicseriesdiverges.
268
Chapter 11 1 SequencesandSeries
0
1
2
0
1
2
3
4
5
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A=1
A=1=2
A=1=3
Figure11.3.2
Graphofy=1=xwithrectangles.
Theimportantfactthatclinchesthisexampleisthat
lim
n!1
Z
n+1
1
1
x
dx=1;
whichwecanrewriteas
Z
1
1
1
x
dx=1:
Sothesetwoexamplestakentogetherindicatethatwecanprovethataseriesconverges
orprovethatitdivergeswithasinglecalculationofanimproperintegral. Thisisknown
astheintegraltest,whichwestateasatheorem.
THEOREM11.3.3
Suppose thatf(x)>0andisdecreasingonthe innite interval
[k;1)(forsomek1)andthata
n
=f(n). Thentheseries
X1
n=1
a
n
convergesifandonly
iftheimproperintegral
Z
1
1
f(x)dxconverges.
Thetwoexampleswehaveseenarecalledp-series;ap-seriesisanyseriesoftheform
X
1=n
p
. Ifp0, lim
n!1
1=n
p
6=0,sotheseriesdiverges. Forpositivevaluesofpwecan
determinepreciselywhichseriesconverge.
THEOREM11.3.4
Ap-serieswithp>0convergesifandonlyifp>1.
Proof. Weusetheintegraltest;wehavealreadydonep=1,soassumethatp6=1.
Z
1
1
1
xp
dx= lim
D!1
x
1 p
1 p
D
1
= lim
D!1
D
1 p
1 p
1
1 p
:
Ifp>1then1 p<0and lim
D!1
D
1 p
=0,sotheintegralconverges. If0<p<1then
1 p>0and lim
D!1
D
1 p
=1,sotheintegraldiverges.
11.3 The e IntegralTest
269
EXAMPLE11.3.5
Showthat
X1
n=1
1
n3
converges.
Wecouldofcourseusetheintegraltest,butnowthatwehavethetheoremwemaysimply
notethatthisisap-serieswithp>1.
EXAMPLE11.3.6
Showthat
X1
n=1
5
n4
converges.
Weknowthatif
X1
n=1
1=n
4
convergesthen
X1
n=1
5=n
4
alsoconverges,bytheorem11.2.2.Since
X1
n=1
1=n
4
isaconvergentp-series,
X1
n=1
5=n
4
convergesalso.
EXAMPLE11.3.7
Showthat
X1
n=1
5
p
n
diverges.
Thisalsofollowsfromtheorem11.2.2: Since
X1
n=1
1
p
n
isap-serieswithp =1=2<1, , it
diverges,andsodoes
X1
n=1
5
p
n
.
Sinceitistypicallydiculttocomputethevalueofaseriesexactly,agoodapprox-
imationisfrequentlyrequired. Inarealsense,agoodapproximationisonlyasgoodas
weknowitis,thatis,whileanapproximationmayinfactbegood,itisonlyvaluablein
practiceifwecanguaranteeitsaccuracytosomedegree. Thisguaranteeisusuallyeasy
tocomebyforserieswithdecreasingpositiveterms.
EXAMPLE11.3.8
Approximate
X
1=n
2
totwodecimalplaces.
Referringtogure11.3.1,ifweapproximatethesumby
XN
n=1
1=n
2
,theerrorwemakeis
thetotalareaoftheremainingrectangles,allofwhichlieunderthecurve1=x
2
fromx=N
outtoinnity. Soweknowthetruevalueoftheseriesislargerthantheapproximation,
andnobiggerthantheapproximationplustheareaunderthecurvefromN to innity.
Roughly,then,weneedtondNsothat
Z
1
N
1
x2
dx<1=100:
270
Chapter 11 1 SequencesandSeries
Wecancomputetheintegral:
Z
1
N
1
x2
dx=
1
N
;
soN =100isagoodstartingpoint. Addinguptherst100termsgivesapproximately
1:634983900,andthatplus1=100is1:644983900,soapproximatingtheseriesbythevalue
halfwaybetweenthesewillbeatmost1=200=0:005inerror.Themidpointis1:639983900,
butwhilethisiscorrectto0:005,wecan’ttellifthecorrecttwo-decimalapproximation
is1:63or1:64.WeneedtomakeN bigenoughtoreducetheguaranteederror,perhapsto
around0:004tobesafe,sowewouldneed1=N0:008,orN=125.Nowthesumofthe
rst125termsisapproximately1:636965982,andthatplus0:008is1:644965982andthe
pointhalfwaybetweenthemis1:640965982.Thetruevalueisthen1:6409659820:004,and
allnumbersinthisrangeroundto1:64,so1:64iscorrecttotwodecimalplaces.Wehave
mentionedthatthetruevalueofthisseriescanbeshowntobe
2
=61:644934068which
roundsdownto1:64(just barely)andisindeedbelowthe upperboundof1:644965982,
againjust barely. Frequentlyapproximationswillbeevenbetterthanthe\guaranteed"
accuracy,butnotalways,asthisexampledemonstrates.
Exercises 11.3.
Determinewhethereachseriesconvergesordiverges.
1.
X1
n=1
1
n=4
)
2.
X1
n=1
n
n+1
)
3.
X1
n=1
lnn
n2
)
4.
X1
n=1
1
n+1
)
5.
X1
n=1
1
en
)
6.
X1
n=1
n
en
)
7.
X1
n=2
1
nlnn
)
8.
X1
n=2
1
n(lnn)2
)
9. FindanN N sothat
X1
n=1
1
n4
isbetween
XN
n=1
1
n4
and
XN
n=1
1
n4
+0:005. )
10. FindanN N sothat
X1
n=0
1
en
isbetween
XN
n=0
1
en
and
XN
n=0
1
en
+10
4
.)
11. FindanN N sothat
X1
n=1
lnn
n2
isbetween
XN
n=1
lnn
n2
and
XN
n=1
lnn
n2
+0:005.)
12. FindanN N sothat
X1
n=2
1
n(lnn)2
isbetween
XN
n=2
1
n(lnn)2
and
XN
n=2
1
n(lnn)2
+0:005.)
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