Sequence and Its Convergence

8/15/2019 Sequence and Its Convergence

1/52

UNIT III

nfinite Series andDifferential Calculus

LERNING OBECVES: The ntutve concep of sequence of nmers nvolves not ony a se o nme ut als a ode n whch these nus have en paced. Ou am n ths chapte is to study the convegence behavor o varos nnte sees ln ths chapter we have dscssed

• Sequences and convegence of sqences.

• nnte sees and the convegence

• Comparso rato ntea and cauchy s oot tests the convergence of postvee sees

Lentz test the convegence f alteatng sres

ondtonal and asole convegence of a nnte sees

3. INFINTE SERIES

31 1 equencesAncton/ : N - 1whose doman s the setN of all natural nmes and ange a st of al nmers s called a seqece of real number or smpy a real seqece

lf/ EN, then() s genely denoted y X o U and s caed the th te thesequece {x}, {} o {}· The set of all dstnct tes ofa seqence s called ts rage.

Fo example, consde he sqence

{xn} = (-tr}= -1, 1 1, 1, 1. . }.

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3.2 UNIT Ill

Its th term is x = (-1)" and th range is { 1 } Simiay r the sequencexn = {!} , the th erm s x =!. All the elements in this sequence are dstc

ne nhus the range r ths sequence is an innte t.

A sequence {x deed by x =c, where c is a xed eal number r all n E N, iscalld a osant equene

A sequence {x, is sid to e bounded aboe i here exiss a real numer K such thatx, � K r all 1 E N. he numbe K is led n upper bound of he sequence {x.

A sequence {x} is id to be bounded below if thee eists a ea number k, such thatx, �kr all n E N. Te number k is called a lower bound o the sequence {x}

A sequence whch is boundd both ave and below is clled a bounded equence.hus a sequence {x} is sid to e bounded i there exist wo real numrskandK such

that k �

� K r all n E N.[ we choose M =ma { k I IK} hen {x} s bounded f x I :M r al n there exiss no real number M such tha Ix I $ M r al / EN then he sequence

{x} is sd to unboundd { 1}For example consder te sequence ; nE Since 0 0 thee exiss a ostiveineger n0 ( such hat

Ix, - A I < E r a � n{E}

I the sequence {x} converges o the lt), hen we wrte Jm x =A.o

A sequence {x,} is said o be dergen i l x, is not nte, that s im x is + oor -. -o n

For eample consider he sequence {x=

{2 } hen limX

n=

im �=0

n n2"{I} nE(nte Hence te seqence2,

is convergen

3/52

Innite Ses ad Dlfern Cacuus 3.3

On te other and fwe consider te seence {xn} = {2, then Jim = o Hencehe seuence {} is divergent.

�oA sequence ·{xn} which nether conerges to a nte number nor dverges to o o -,

is called an oscillator sequece.

For exampe consider the sequence = -1)" Her the ven elemens al +I ad s x I wheeas the d elements a -1 and sxi - 1 1 Hence theseuence -I" oscilates fitely betwee and IOn the other and if = (" then the sequence oscilate fe

between o and + o.

EXPLE 3.1 I

.2

If a - 3 5 ,show ta m On = 3n

Slutin: Let > 0. Then

, + 21 I -7 I 70" - = 3 + 5 33 + 5) = 33 + 5) < eimplies

or

3 5 > 3

5 > 9-3·

h

7 5Te 0 a pos1tve nteger eater t an 9e en,

l Hence m O = 3 \ O �

<

er all >

Theorem Every coergnt qun ha uie limi.

eorem 2 ,·ry eee is und

4/52

3.4 UNT Il

Rmark .1 The convee o heoe 3. ay not b te hee nddqences which do not converge Fo eampe, the equence (- 1 )"} i od b itds not converge Ln ct it d no have a unique imit

Cauchy's Pnciple of Converencehe fllowing ndaental theoe is ue f eteining whethe a qence convege o not

Theorm .: (Cauchy's prnciple of covgc). A ecssay and sufcntdin he comrgc o a qn {x,} ha r each 1 > 0 ther its ao ti\c nge 0 suh ha

\ .\ - , I < f all 1 >/1

no.

Theoem 34: A squn o a nb- 1 o nt an ny i t i y seqne

EXPL 32

Uing auchy pinciple o convegence show that the sequence x} whee x = +1 1 I - + + · ·+ not convegent2 3 n

' ·

Soluon: Spse on the conay tha { is convergent. Then takg B = !, theCauchy's pncipal yels

2

(3. 1 )B

1 1 I I m Ix X m =+-+·+->-+-+ · · ·+- (m ters) == ,m I m + 2 2m 2m 2m 2m 2m 2

that is1

I X 2m - X m I > 2 1

which s a contaicton to (32) Hence the qence is not convegent

5/52

Infiite Seies ad Dlerena Cacul 3.

Mootoic Sequence

A squnc {n i aid o b monotonic increasing or seadiy increasing ifXni � X n rall n.

A equence {n i d o e monoonic decreasing o seadily decreasing if+�5 r ll n

A squnc x} i id o b monoonic if i is hr monotonc incrsng omonoonic deceasing.

A squnc {xn} i sicly inreasing V +i > Xn r al n nd sricly decresing f. X 1 < r ll n

Fo xmple,

(I') 1l l 1 . d

, 2 ,3 4 · s a monooc ceag squnc

(ii {n 1 } is monoonc ncrasing enc.(iii {n} is a monoonic incring sequence

Theorem 3.5: A onoonc qc awa) nd o a lii i o ii)

Ths a oooic ·qunc i ih co o din it cao osc)

Theorem 3.6: A cr and uin codiion h on oa

monoon qun is h bodd

EXPLE 33

Show ha he qc {} whee

i convegn.

Soluon: W hav

l 1 1x, = 1 + - + - + -

I 2! ,

1- =( l ) > O

n + r n

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3.6 UNIT r

Hene {xn} is monotonic nreasing Fther1 I I 1 1 'x = l +-+-+ +- I + I +-+ · +< I + = 3

n l! 2! n! 2 2 n-l

I

-

�2

and s x is nded Hene y theoem 3.6, the gven seuence converges to d hih is less than 3. In ftXn e = 271 ... , he base o the natual logathm.

EXPLE 3.4

Show that te seqene {xn, whee

Xn = (t + � )"

,

is onvegent

Soluton: We ve

By Bnomial theoem f a posiive ntege

n n( - 1) n(n 1 n 2) 1 n(n ) ... l X 1 + - + + + -n 21 n2 ! n3 n! n

Theefe, hanging n to n + 1, we get

All the ftors in the nmeratos o X nd Xni ae sitive uer each tor n henmerator o x+ 1 s eate tan he oesnding ctor n e nmerator owhe

7/52

IftSre ad DleretaJ Cacus 3.7

e denomiaors oXn ad Xn+i are same Therefre Xni > Xn ad so {x} is monoonicicreasing Aso

l l I I l

<

l +J +-

+· . .

+- < ) +

1 +-++

+. . ·

+ <

32 ! 2 2 2 2

and n is bounded. Hence {x converges o a fnie posive limi whose value is lessthan 3. In c e im = e =271

n-o

EXPLE 35

Show ha iml +x)1/ = e.o

Solu S b 1

on: u sug x = - we ge y

im (1 + !)")0 y

and so he resul fllows fom Example 3.4

EXPLE 3.6

Sow ha he suence whose h e s

27X = 3 +2

(i is monoonc ncreasng, is unded above, i is bounded below, and (iv hasa mi

Solon F e e quece

Ten

2 - 7Xn =

3n +2'

2 + 1) - 7 2 - 5X nI 3(n + l) + 2

= 3n + 5 ·

2 5 27Xn+ I - Xn = 3n + 5

-3n + 2 > 0 fr all .

8/52

3.8 UNIT Ill

Hence the equence i monotoni inceaing The sequence ounded ave and blow.

he up and lowe und ing � and -1 Fu er 2 - 2

m X E --1=3·3+

EXPLE 3.7

Show that

Um (n)/ = l. -

Soluton: Let an= (1 )1 = I + h, whee h > 0. Thenn

( ) I (

) 2 I ( )

1 ( - l) 2=an= I +h = I +h+ 2 - I hn+ · ·+ 1- l . . . h > 2 h. •

Thu,

and s

h2 2 < -!

n-

( 2

) 1/2 0

9/52

lnnteSre and Dlren1I C1cuJus 3.9

Theorem 3.7: (Cauy). Ifa, i iive r a rlu · ofn

po' idd he ate limit xi�ts

I. I n Li On I1m = m - 1- n-"

\ hilc tudying onegene of innite seie . we ha obseve ha ths theem ofauchy mplies that wenever D' Aembe atio te appabe auhys ot tis ao applicble t e· he eree of a ee of po itivc tem

3.12 Series

An expression of e r m a1 a 2+ a0 ··denoted by La is caed iisris. Te te a s caed te n r m of e sees Furer1

S = a 1 + a · · · a,

is caled te partial sum of sris a =I

Te denition of convergence or divergence of e series a dends on e=I

convergence or divergence o te seqence {S}:1 of parial sus0

Let L a b a seres of real nuber wit paa sms S = a1+ a 2 a f hen= o

sequence {S}_ coveges to/ we say at e series L an coveges o te s and -�

we writeLa1 = If S}:, dverges hen te series a aso diverge

nI nn case o nnte series, te Cac's cieria r te convergece takes te

ta is

m < e m > n > o()

an+ +a+2 + · · · +al < m > > o().

If �e tae m

ten e as expression taes te rm

lan+d o(


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3.10 UNIT Il

orim a 0.

whch is a neca cnditin fr the cnveence fthe ee an=I

However lim = 0 is not a ucient cnn fr the cnvergene of the ere0

L a n of sitive tes. For exampe,n=I

I�,

n

o I o

�j' n lgnare divege series ise of the ct tha i O=0. But the codion is sometiesuse lft, if lim a# 0, we an t on/ythat the ere s not onvergen.

-

er

3 8:

psi.tv

t n

s

e

iterconv r

o dv g to +o.

Theorem 3. 9: Covergee diegee o oscillatio o e psthc , ot atd b} the addi o oi

ion

o a i ub1r o it tns.

EXPLE 38

Test he onvergence of the seres

f+ + +

·

V V6 Vs

2 n

+

Soluton: Te nh em of the gie series is

T err,

u ,=J

2 n

: I )=

2 1

�

Jim U= �# 0n-

y2

Tu neessary odiion r convergee is ot stsed HeeLu is dver en

11/52

EXPLE 3.9

Inie Sre ad Dfereta Caculs 3.11

Dscuss the covergece o the geometric seres

+ a + a +a3+···

Solution: he paal sum r the gve series s

If lal I, the

2

-I a- 1S n = 1 + a + + · + a = - , a = 1a-I

1 -aS n= d -a

m S = sce a 0 as n n- 1-a

Thus, f la < l, theIa s coveget ad ts sum s If a I the. 1-a

S= I 1 +I (n erms) = n

and s Lm S n= o HeceI

s dverget n

fa> , the S n> and so m S = o ece the seres s dverget. Ifa 1 theseies becomes

no

and the

1+1+11+·

S2n= 0 a -1= 1

Hee he seres osclates bewee 0 ad Ifa -1, the successve tems crease mgude The

a - S2 n I

(ve)a-

s moooc creasg whle S t s etve ad umecly cresg wth n. Heeth seres oscates bewee+o d o

12/52

3. 12 UNIT Ill

EXPLE 3.10

Show by direct smmtion of n terms tht the series

is convergent.

I 1 1t+ .3+34+···

Soluon: The th te o the given seres is

heree,I

a1 =I2;Th the ptil sum is

1 I Io = =-

( + I I n+I1 1

o nn

n

+l

IS n = a , + 02+ .+On

I --1 . n+

Therreim 1 0 In-o

Hence the given series converges toI

EXRCISE 3.1

Sequences

1 Show tht t c {} con iI< r � 1.2 So that 3+� i=2

l

3 .Give n exmpe of monotoniclly incresing seqence tht is i) convegent ii) no

convergent { t}Ans. i) I; , ii {n}.4 f O = ( l + !) r ll ntr nmber n, show tht im O n

I. I "m O= .{ 3 4 5 -6 }HJ: The seqence s 2 2T, 4,S, · · · The s o inrior nm a

srior nmsa

{ � }n

{�}

herereI

3 1 2

4 • >

im lub{�}I ima glb{ -� ·}- ' 3' ' 2 ' 4

13/52

If S d Dierential Culs 3.13

5 Show at he sequence {} where

canno conerge.

1 1 1= 1+-++ · +4 7 n - 2

6 Using Caucy' geneal pincipal of convegence, ow ha he sequence { 1} isconergen

7 Sow ha lm [v] = 0.

2 TSTS FOR CONVRGNC OF OSIIV RM SRIS

2 C Tet

Te conergence o diegence behaio of gien erie i geneally deemned by comparng s ens w e es o anoer series whoe onergence haor s nownSuch ompasons ae called pn

omariso s (a f�Un and Lv, ae wo ere of posiie em suc a U <v and Lv i conegen hen u, ao conergen

omariso s ): IfLu, and, ae wo eie of poiie ems uc ha U>v and2v1 i diegen en_1 als deges

omariso s D: lf Lun and Lv ae wo eies of siie em en

() if lmU= nie and non-ze) en L" and Lv, h conege o diee-o v,

ogee

EXPLE 3. 1

Sow a e seies f �· ale e Hnc Series oneges if p > I and degesifP· $ n=

Solution: ake p > , Gop e ems of he seies a ow:

14/52

3.14 UNJT Ill

Bu

and so on

Hence,

The right hand side is a geometric progression wth common ratio 'l� < I, snce > 1ence by comparison test the seies is convergent

If = l, then the series is

nd

I Il +-= l +-

2 2,

and so on.

1 1 1I--+ .

2 3 4

1 1 1 1 1 1 1 1+-+-->----=-5 6 7 8 8 8 8 8 2

Theree, the gven series s less han he series

l I

I++++ . 2 2 2

Leaving aside the t te, his series is a geometric pogresson whose common atio s 1.Therere, this series is divergent and so is the given series.

fO


15/52

EXPLE 3.2

est the convergence o the sers

1

Ifnte Seres and Dierea Calculs 3.15

t+.3+3

4+·

Solution: We noe that

But L� is convergent (onic Series fr p = ) Hence the given seres isnconvergent.

EXPLE 3.1 3

est he convegnce o the sies

n + ln +n

2

+ l)(n 2 + f

Souton: The h tem o the given seres s

(n + ln + u - -n-n2+ ln2+-, 2( 1 +)( +�)n nTae Vn 2. Then,n

i+) i+), � }� ( ( � ) = l, f nite and non-zero.

+ I+,2 1

Therefre, un and Lvn converge or dverge together ut the series Vn =Lconverges. Therefre, Lu s convergent


16/52

3.16 UNIT Ill

EXPE 3.1 4

Examine the convergene of the sees

L [(n3 + t)i-].

Solution: The th term of the seies isu,=(n 3+1)Ln= '1+ ,> -nnI+ :-nn[ 1 ,) -1]

= [I+ � � G-1) (_) 2+··- Jj ! [ ._ · · ·] =. [!�_ · · 3 3 2! 3 3 3 9 3 n2 3 9 n

ICoose v 2 enIflm U= lim [ � � - ·] =�, nie and non-zero1-0 v, 3 3 3

Hence, Lu, and21 convere or divere toether ButL v,=L�

is conveent

TherereLu is also converen

EXPLE 3 1 5Eamine the converence of the series

Solution: he th te of the iven series is

u=v-.=� [v +.n 1

lake 1 = r:hen

y

I Un 1m = tn- V n n-o

If. d

+In

R

- 2

1te a noneo

TherereLu andLv1 convere or divere toethe tLJdive erru1 is also diverent.


17/52

nnite Sries nd Dferenlal Caculs 3.17

EXPLE 3.16Exmine the convergence of the sees0

I

+i·I

lon: he th ten o te seres is-

i - , I"• - 2 I - (I +�) !( I �

"Take V =�.Ten lim u, lim = I, ite and non-zero Terere, u and

V n-

n

l lv converge or diverge togeer But he r v = 1 i conveent Henceu

i alo convergentEXPE 3.1 7Exmine he convergence of the ee

P 2 P ()

. 2 3q 4q(i) L�sin � Soluon: i e t ten oe series s

akehen

nL +lq=

lq =

lq·q I+; n- P l +;

PI' U r J 1n .

m V = q = +

herere Lu nd Lv convee or derge together t L =L converges if q -p > l and diverges iq p $ l Hence Lu converges if q -p > 1 anddiverges q p

18/52

3.18 UNIT I

(ii) Here

I

1 IU= - s.

n n

sn- l 1Since lim-1 n I, it llows that sin and so U We therere, take V= 2, n 1 1 nand then

1

I

21 sm-

limL

Jim �si- Jim I. 1

Theere Iu and I v, converge or dverge together Bt Lv= L:

converges.

Hence :u is alo convergent by Comparison est

EXPE 318

IfIu is a convergent series of sitve term show that Lt� s als convergenGive n emple to show tht the onve ned not be ue.

Soluton: Since u is convergent "n - 0 a - c. herere, there exiss apositve integer m such that 0 $ U $ I r all n � m and s� $ u, r all n � m. Hence,by Comparion Test,L� is convergen 1

However the convese need not b te Fo example if we takeL U L-, then' u; = �· The seesL� is convergent butL � does not converge L. n 1

EXPLE 319Sh h

1 I 1 1

ow t t t e sees2 2 + 23 + 4 + · 1

s conver gent.

Solution Since 2 < 3, each erm o he given series in ess han or eqal to thecoeponding tem o the sees

which is a geometc sees with common rtio ! and so converges. Hence, by Comparsontest the given ries converges 2


19/52

nfnJe Series and Dierenta Caclu 3.19

XPLE 320Sow tht the sees

is conveget

I 3 5-

+ 2 .3 4+

3 4 5 ..

Soluton: The th tem of the given sees is

2 - 2-�

ang v . we note that Lim u = 2 (ite) Thereeu

nd L v converge or v 1diverge ogethe BuL� is convegent ence u is convegent

EXPLE 3.2

th th I2 sExme e convegence o e sees 4 5 1 + Solon: The th te of the given es is

ITke v = 2 Then

23 5 (2 �) (2 �)"• = 41 I '

(4

)

+ .

22+ 1 U . 13 un m = -1- y1 14 l 2 1+nite and nonzero

I

Hence, u1 and Lv1 converge or dverge together But L Vn=

L12 convergeshereren converges


20/52

3.20 UNIT l

EXPLE 322

Tst th convrgnc o th sis

I I I1 ·23+2.34+35

Soluton: Th th tn o th sris isI

"• = n(n I)(n+ ) n' ( 1 +D ( 1+;) ·1

Tak v = 3 Th1lim u,

(1

)

I

( )= I, fit and no-zron- v,

l +- 1 n .Thrfr, Lun nd Iv, convg o dvg togth But th ssL n =L� isconvrgnt Hnc th giv sies Eu is also convrgnt

EXPLE 3.23

Exami th covgnc o th sis

I (# -V).Soluton: Th th trm o h sis is

.n4 + I -JUn

v

;

-n l

9

1·

v

n4

I

J]

J

n-J

ak V n 2 Thn

fl

I , I I,

:

_

= :� ( 1 ! ( 1 i= 2'

I+ + l+1

fi d ozro

Thefe t se E and Ev conveg o dive ogether Bt h E= is covgnt. H, t givn ri is covgnt�n-

21/52

Infnite Sries and DUTee1 Cclus 32

Cachy's Itegra Test

Let /be a non-negaive mononic deeasig nton of

� I and e

f(1) = in al

sitive integal values on. hen, the seiesLn and the integral j f(x)d convege odiege together. . 1EXPLE 3.24

1Apply Cauhy's ntega es o show ha he seies L- onveges p > I, andiveges 0 < p � I. nP

Soluon: he th te of the seies is

U = .

As r Cauchys integra es, we have

hee

1n =n=J(x) = _

xWe obsee hat/is positive and monotoni eeasing � l Theee, by Cauhysinegal test Lln and j x) convege o divege togeher.I p = , then

When p > L hen

J J l /0 [xl-p = xd [Pd = 1 _ p

fx)d 0-

-

l] = fnit . p - 1 p I p -

Thus Jf)d onveges and so by Cauhys inegal test Lun also conveges p > lI

22/52

3.22 UNIT m

Wen 0 < p < I , ten

0

I IJ(x)=1

!-P�=1

[o- I ] = o.-p -p

I

husJ(x) diveges and so, by Cauchys integal est, Lu n dveges p < I.

lfp=I, then

0x) = � = [lgx� = o - lg I = o o0

I

hus, Jx) diverges in tis case and so by Cauchy's integl test Lun dvegesI

p = I.

Hene L Un L conveges p > l and dveges p $ 1

EXPLE 3.25 o 1Apply Cachys inel test t show that the es L (l Y conveg fp > I adiveges if < p $ 1 n2 n o n

Soluton: Fo he gven seies the th tem is

IU n= lg ·

Taking Un

f (n), one gets 1/(x) = (I x gx

Fo x � 2 and sitive p, is stive and mnotonially deceasing Hence, Cauchy'sintel test is applicable.

When p - l

o/(x) = J

( lgx) =lo x)-

p

+t

·x -p+ 1 2

23/52

Infnite Sr and DUJrea Caculus 3.3

Now ifp > 1, hen p 1 s siive and so

0

J o

f ( x)d = _ _ [ 1 lo _ _ [ o- I

p (lox 1 2

p - 1 lo2- 1

- 1 f ) )(lo2 1

e ·

ThusJ(x) convres and hence by Caucy's integral test the iven sees Lun

converes r p > 1.

fp < 1, then 1 p s ositve and so

(

J(x) = [( lox)P]� = ! [ o (lo2)1-P] =op p

2

ence y Caucys interal test, Lu dvees p < 1.I

lfp= 1, en/(x) = . Thererex ox

oj(x) =j � = [lo Jx�=o - o lo2 = o.0

2

Thusj(x) dveres and so Lun also divees p = I. Hence te seres ( 1 Y convees ifp 1 and diveres ifp $ 1 .

n

o

EXPLE 3.26

Eamine the converence of e series L�Soluon:

We ave 1Un=-=f(n).y

24/52

3.24 UNIT Ill

Theere,/ x) and t i potive and monotonic decreaing herere auchyinteal te i applicable. We have

0

JJx) =Jd= [�x] oI Thu the integral Jf(x)d i divergent Hence by Cauchy integral tet the eie

d'. 1vergent

3.2.3 Ratio Test

Let Lun b a ere o itve tem. Then

(). 'f 1

Un+IIun is convrgn 1 1m< .

n- Un

(1 1 d' ' ' lnI 1" IS 1vegent 1 m> .n Ln

(I li"nI = I, thi tet give no inrmation about the convegence o divegence o

n- U

When Ratio tet f then we appy the fllowng two tets:

I. Raabes test In the poitive tem ee Lum i li (�- 1) =k then then ln+ee converge f k> I and diverge f k< 1, bt tet il k= I

2 Laithmi test: In the tive te ee um m (nlg�) =kn UnI

then the ere converge fk> l and divege fk< l bu tet il fk= I.


25/52

nOnte Sees d Dlferea Calcus 3.25

EXPLE 327Examne the convegence o the series

12 .2 1 . 2 31

l·3 5

1 3 5

79

+·.

Souon: he th tem of the gven eris is

The

2 .2 32 ..2u, =

I·3·5·4- 5)4-3)·

Un= )( ) )l 3·5 45 43) 4 I 4 I . 2� . 3 . • • + 1)2

ad soI

u 4)4+) 162 16-;

Un+I=

(n + 1)2

( + 1)2= ( ). l+

Further6-_ 16

u m "2 - Im -- > n-ou ,+1 ,_ ( 1) 2

l+

Hence y D Aebe's atio tes he serieLu, coverges

EXE 3.28Examine the convegence o the series

Soluon: Neglecig he rst tem, we ave

May 07 08

3·6 9 ···3u, = x _ 3 6·9·3)(3+ 3) and+I

-

)x 7·I0 3· 34) 7 0 13·34) 3 7

26/52

3.6 UNlT m

Therere,

,(3+�)lm � = m 3"+

7

!

= m ". !

-

Un+ I 3 + 3 X no ( 3) X Xl 3 +nherere by Aembe's rato test the gven seres converges f! > 1 o i x < 1 and diverges if! I. I x 1 then embets rato test ves no nmaton.

x .

EXPLE 3.29

Test te onvergene of te sees· x4 x62+ 3J+4J+sv+ · ' x 0·

Soluton: Te n th ter of he gven sees s;-2

t n

n+)j and s

herere,

n(t +) v( t+!)Jim = n+ .�= " n 1o l (1+ J) x o ( I) = x2 ·n 1 +-

hus by DAembet's ato test, the gve seres overges f > I hat s f < I, ad dverges f� I When = I the'em� ets ao test s and we have x

akng v n =l' we haven

I 1 = n+ 1) j = n + ·

I Un Lm I f dm - -1 = I , te a non-zero-o V n I +_1


27/52

lnJt Sr 1nd Difeenia Clcus 3.7

Hence, by comparison test 2n andLv converge or dvege togethe. ButL v =

L conve� ges HenceLu also converes. Theree, the gven series converges r

� 1 an diverges r x2 > 1

EXPLE 3.30Test te convergence of the llowing serie:

(i) 2�P J 4

(i)

+ - + + + ...

!3

4

Soluton: (i) The th term of the seies s

Theee

and so

xU =,.

n.

r+•UI = +I)!

JimUn+

lim = O ess than 1, r all nte values of x. ln -on + ·

Hence by D'Alembe's ratio test the given series converges r all nite values ofx(ii) We have

and so

ru =

!1 + 1U (n + l)

lim _ = lim (n + 1)' = o r all values of pn U+I 1-0 l +

Hence the given series converges r all vaues op.

28/52

3.28 UNIT Ill

EXPLE 3.3

Exae the covergece of the sees

. x>0

L

I

Soluon: For the gve ee

Un = xn V J+I UnI =V

n

+ 1)2]ad so

U )'m= m- +I

.

221• ! = Jm

1 I l x - -ln

2 2l+ - + n 2 n1 n 2

· - = -x x

Hece, by D'Alem' rio et Lu, covrg �> I, h , < l ad dveg

! < 1 that >

1 Whex

= l , ths test gve o ato. Bu tht cx n I l- . u - -Ji ( l2

=V.

l +nl2.

Takg n = we ote tha

. u 1 l f dm - mR

= 1 te a ozo. V ,_ l

1 +n2

Ther, Lun ad Lv, covege or dverge ogether ButL V =LdvgThe, gve eis dveges rx = 1 Hece, the gve e coveg x< 1adive x � 1.


29/52

Ifite See ad Dlferental Calcuus 3.29

EXPLE 3.32

et r the covegence of te erie

in2(n + I )2

!

(ii (1)2 (I ) 2 (1 ·3) (I . .3 .4) 23

+ 35

3 57

35.7.9

+ ...

L

(iiin

Souto: For te given erie

2( 1u, =

'

.ad o

1) )2Un+

( l

I. l ( 1)1m -= m ( )

1 3 I -(1 + 1) . = im

=o.n- U+I n nl)(n+2 n- 2

( )2 I-

Hece the ive sees overes by D' Ales ratio est.

(ii For the given ries

( 1·234n n = 35.79 (2n+l) J .3 4 +

1

)

2

l =( )3579 3

Terere,

() 2

2

im

�= lim

+3 = im

1 = oun (12 10 ( ))2-

ece the give ee covege by D'Aember' rtio tet

iii) Te tem o e given erie i

Therere

( )! (l! !L - - -n+I - (n I)n - ( 1)"+

-(n + l) "

30/52

3.30 UNT Ill

nd I. U (n l)" (n + ( l) nun- = l

n= = l -n = e = 2.7 eater ha .-oU+I -o

Hence, the given series onverges by D Alems ratioet

EXPLE 3.33

Test r the convergece o the seres(i) x2 +3x + 4x · · ·

() : x(2n)

()v)

x>O� Ix x I 15 + ·· + n 1 Solution: (i) The gve series isL. Therere

nd sUn = n, U+I = n l )x

I U

n1.

I1m- m =1m

-

=

n-o Un+I n (n + 1 )xn+1 n ( I + �) X

Hence, by 'Alem's ratio test the series converges i> I , tht is x < 1 andiverges i! < 1 hat is x > I . I x = I the to test�ves no inrmtion abutxconvergence. But r x = I the seres ecomes whch s divergen Hence the gvesees converges r x < I nd dverges x

) We ave

Therefre

.U = (2n)!'

l+I = (2n 2)

lim � = Jim �. (2 + 2)(2n + I )(2n)

im (2n + 2){2n I )-o UI o 2n) x+t -o X

31/52

lte Sa and Difrnal CalcuJus 3.31

Hence, by D'Alem's raio test he given series convergesiii) The th te of the series is

Thee re,

and so

I. U 1m= uno U+

I

U = X > 0.n +

U - ( + 1) :+nI -(n+ 1)2 + I

Hence, by D'Aemr's rtio test, the e converges if!

> I orx < and diveges if1

< 1 orx > I. Ifx = I, he Ratio est give no in�ion But in hat casex

Iake = Then

im U =1,

nite and nonzeo. V ence andLV conveges o diveges togeer ButL Vn =L� dverges

TereeL also diverges ence, he ven series conveges r < 1 and divegesr x � 1.iv) Negecng the rst em the t term o the seies is

�U=

n2 + 1' UI =

n + 1)2 + 1.

32/52

3.32 UNIT Il

Theen2( 1 + +

�) lim Lin r X 2 + 2 + 2 = im n2 n .

! = !

-L

l

I

n

n2

+ I

x+I n!

(I

)X X

n2 I +-12

Terefe the seies conveges i!> l orx < l and diverges f

!< l orx > 1 . Forx= 1 ,

e D'Alembr's ratio test give/

no nfnaton. But in suc a�

ase

akng v, =- we get1

Un =,2 1 =

(1 _ .

2

im "• = Jim (1

1 ) I nite and non - zero Vn n I +_2HenceLun and Lvn converge or diverge together ButLvn =L

�

i covegenn :n nvg x = I . Tus, h gvn nvgent x � l ndiverge rx > l

EXPLE 3.34

Examne the convergence f the llowing eie o pitive te.i 1 + l { + l 2 l ) ( + l { 2 + l 3 + I )+ P + I + (P+ 1 (2P + 1) + (P+ 1 )(2P + 1)(3P + l ) +

ii 4 4 2 4 12 20

+ 18 . 27 + 18 27 36 + . . .

I x 1 3 · 5 x4 1 3 5 7 9 (iii) l + 2 4 + 2 4 6 . 8 + 2 4 6 8 10

1 + . .

Solution: i Witout taking noice o the fst term the th tem o the gvense is

(+ (2+ l ) · · ( + l)u - .n - ( + (2P + l ( + I )

33/52

Terere,

and so

Innite Sre and Diferntial Calcus 333

(a + 1) (2a + l ) · · · (na l)[(n + l )a + t

(P+ 1)(2P + l ) (n/ l )(n + ) / 1( l II + - P+ m . m (n + 1 ) 1 = l n n = -0 -0 (n + l )a 1 ( 1 al - a + -n n

Hence by D'Alemts rto test h seis s convergent f � > 1 or p

> a > 0and

divergen f� P >0. f a p e lim = 1 antso the rato test gves no n- Un+Iinomaton. But hen the seres ecome1 + 1 + ,

whch s divergent ence te gven seies onverges / > a > 0 and dverges if ; p > 0(ii) The given seres s

whose th tem s gven by

heere,

ad s

4 1 (Sn 4) .l S 27 6 · (9n + 9)

4 12 20 (Sn 4) (Sn + 4)l = 18 27 36 (9n + 9)(9n + 1)

I.u 1. . 9n + lS _ i _ m - u w

1-0

U+1 0

Sn+ 4n-o

ence by D'Alemrt's ratio test the given eries converges

34/52

3. UNT I

(iii For the sries (i ), w hav

1 . 3 . 5 . (4 - -2U = (4n 6) 0 (4n - 4) '

1 3 5 (4n )(4n 5)(4n 3 ) ?Un+I = ·-.2 4 6 · (4n - 6)(4n 4)(4n - 2) 4n

Thrre

m _ = lim (41 - 2){4n - 4) 4n

_ = im16n 8n }

no Un I no (4 5)(4n 3) (4 4) x2 no 16n - 32n + 5 . =

Hnc by D'Almbt atio st the sis convrs if�> 1 or x 1 fx = I , the test ils But it = 1 , then

x

( 7 3 5 · · · n 41 3 5 (4n 7) n� = = .2 · 4 6 · · (4n - 6)(4n 4)

(6) (

4)2 4 6 n 4 - 4 n n

I 3 5 · · · n uTake

vn= 2 6

Thn_ A

(ite and non-zo. Hnc he seris convgs• 4 n Vnrr = 1

Th, th sris convrgs whn x � 1 and divrges whn > I .

3.24 Cauchy's Root Tes

IfL is a seris of positiv trms, heni th ses

n

convrges f im n! < l L n_

ii th sris '" n

diverges if im n�> I.n-

Whn im (n = 1 th rt est ls to giv any informaton rgardng convrgnce ofn-

th sisLun)·

35/52

InnJte Srie and Diferet Cculs 3.35

EXPLE 3.35

Examine the convergence of the series

2 32 43-x + x +-x3 + 12 23 24

Soluton: The th tem of the gven seies is

herere,

( l)1 + xI

� 1. n

+ 1 x _.

1m u, - m · 1 m 1 xn-o n-o · ,.0 •

(. 08)

Hence by Cauchy's root test the given sees converges ifx < 1 nd diverges fx > 1.When x = l we hve

ITake v =

.hen,

im u, = Jm ( 1 + !"= e (ite nd non-zero). v1 1-

Bt L v, L � is divergent So Lu diverges r x Hence, the given seriesconverges ifx < and diverges ifx � 1

EXPLE 336xamine the convegence of the lowing series

36/52

3. UNIT l

Terereim � = im {n + lr im -31 ( �)= e < I . n - no 3 3

Hene by Cauy's rt test te given series is onvergent Ignoring te st te e t te o te series isTerefe

+ )U = n + 2 :.

I)I +

. l .

+ m = m 2 x = m = 2x = x. I

+ O 1 + Hene, te series onverges i x I x = then ') n l)I - l + -(1 + )n ·1 n = + 1 ;

)"

[ 1 ;)]d s m = � 0 Sice l 0, e sees diverges fr x l . hus then-

gven series nveges r x 0 L 1 +J1

Soluon: i Negleting te st tem te t tem o the gven ss sx

un = .


37/52

38/52

3.38 UNIT Ill

nd sI. 1 I I l1m u = m

( ) =

-< ·-o l el

Hence, by Caucy's t test, the gven series conveges.

EXERCISE 3.2

Comprson Test

Examine the convergence of the owing sees:01 . L ( + I! n=Io I

2 .

l

o

n

A. Divergen

Ans vegento

I3· + I)" As. Converges rp

< � and dverges rp � �

4. L I)i - I

1 2 1 23 2l + 23 + .

- l J - I J4 l6 3 - I 43 - 1 53 1 + . .? ' . . V

2

(

: I ). . .1 2 3

B. t v+1 2J+1 34+· ·

1 19· � 2 . 33 . 4 + . .

As vegent

As Divergent

As. Convergen

Ans. Divergent

As. vege

Ans. Convergen

39/52

nfnle Srles nd Dierenta CaJcuu 3.39

IO (+ l)(2)· ( 1 )(+ 2) Ans. Convergent

1 L s!Wot: take U = in n = !, then lim Un = I butL ! diverge Ans. Divergent V

D'Alember' Ratio TeExamine he llowing seres r convergene

2 ! 3 ! 4 l12. •+ + + 4+L (

3x+5)13 ( + I !

14 I (�

I S f n· I ( I (+ 2

Ans ConvergenAns. Convergent r all value ofx.

Ans Convergent.

Ans.

Convergent r x < l, divergen x $ 1As. Convergent

x x l? 2+ 3J 40 . Ans Cnvergent rx � I and dvergent r x > I

183 6 9 . ·3

_

L 4 7 · 1 · (3n+ 3219 L 1

2

Ans. Convergent.Ans. Convergent

Ans Convege r x < 1 and diverges ix 1

s.Converges r x $ I and diverge x

Ans Convergent

40/52

3.4 U IT Il

Cauchy's Rt esExamie the llowing series r covergece

( + It x

AnConverges x < I and dvergs ifx � l .

L. 1+24 L(1 r g

A Convgn

5 L 5n-" An Convergent2 4 + 26. x - x2 +-x3 + ·]2 2 34 I

A. Coverges r x < and diverges ifx � 1 .

27 L [( :T'- :f As. Convergent.

2s I' A Converges r x < I, dvrges r x � I .

33 ALTERNATING SERIES AND SERIES OF POSTIVEAND NEGATIVE ERMS

331 Series of Aternating Terms (lebn es)A sees in whch stive ad negative tems ocur aeately is called aAlteaig Se.

Regarding covergece behavior ofa ateatig seies, we have the llowng theo-rem, know as Leb1tz 's Rule.

Theorem3.1 0:

(Lebtz e)If1, osve a onoo1cay eceases othe m ero then e ateig sei u1 - : th "" · ts coverge

EXPLE 339

Show that the series I k

+;

k

-�k

+ s covergent r all positve vaues ofkSouon: For the given series the th term is u, =�· Then u � Unt ad lim . = 0.

I n

r

Therere by eibnit's rule the given seies converges I parcular, he ereI I I

- + - + . onverge

41/52

ie S ad Dferetil Calclus 3.41

EXPLE 340Eamine he convergece of the erie

I I I I

"- 3 . 4 5 6 - 7·8

+

Soluo: Fo he given sere, we haveI I

Un = =2n-

12n2n (2 �) ·

There Jim U = 0. uher,n

I, I

.8n 2

Un Un+ = = > 0 2n - 1)2n) 2n 1)(2n 2 2n 1)2n}2n 1)2n 1

r al n hu {} i mootonicall decreang to zero herere b Leibniz' rulethe given ere covergent

EXPLE 3.41Eamine r convergece of the ere

(I) I I 1 h + , w ere x ot a negatve teger.x+I x+2 x 3ii)

I II - 3+5 -7

.

(iii)lo 2 g3 g4

+

I 2 3 4 5v

6

T+1 -2

26

Souo (i) Ifx > - I, the te alteating om he giig. Ifx < 1 (excepegatve teger he tem a lmatel alerg thr g Snce he removal of itenumber ofte oe not aect he covegece of he ere, we may aume the sries to

be aleaing n thi cae alo The nt e f the seie i U ad o Jim Un

Fuher,

x+ n -o

I I IUn Ln+I

= > 0

x n x n 1 x+ n) x+ n I and so } i mootonicaly decreaing quence Hence by eibnt et, the given

Isee coverge.

42/52

3.42 UNIT Ill

(ii Clealy, {Un} is monotonic dcrasing and

Jim Un Jim-- = 0n-o n2 - I

Hnce, th givn altating ss is convrgnt(iii The th t of th givn altating series is

hen,

log{ 1 ) 1)

1

lim Un lim Jog � (0 orm) = lm 2

)

('Hospita len (1 + 1 ) 0 n ·

m 1 = 0n 2 2)o examine th monooncity w mk us of a coollay o Man Valu Thoreccording to whic "a fncon is monotoc decreasig if drivative is ngatv"o, t us ak

hn

) log

>0.

(�) - 2 log) = 2 log < O i l - 2 log < O4

that is, if log >�· that is, if > 65. B > I and so th condition is tsThus + l )> 2) r U � l

which shows that Un > u,1. Henc, by Leibnitzs test, th givn sies is convergent(iv Th th tm of h gvn sris is

1 1 lLn - 1 - s - oS + 1 S 5

hus the tes ar monotonically dcrsing d Un tends to a nit imit e bycooa to ibnis test, the given sees oscilats nitly.


43/52

EXPLE 3.42

nOnt S ad DUTroJ Clcus 3.43

Exam h cvgc of h srs

I I I I

- + . .

log2 log3 log4 g 5

Soluon: For hs ss, h nh m s

Th,

ad

IU

=

log(n + 1 ) "

1 . l ' 11m U m 0 lg l )

1 IL UI = log 1

g(n + 2) ) O

Hc by Lbzs s h gv ss covrgs

EXPLE 343

Eam h covrgc of h sris

l l I I

_

_ . 1 + v 1 - 1 +

. . ·.

S

o

lu

o

n

:

h gv srs s a alatg ss . W obsrv ha

irU V + -lr

n � 2)

ad tha u- 0

as n- .

Bu h sris s o covrg. I c h Lbzs s s o aplcab bcaus h rms do o dcras mooocally Furh,

Thr,

c _ t) [f rv U n - -y 1 n n 1 n 1

�

u

� 1r _�

L

n - 1 L n

-

l

Th rs srs o h g covgs by btz's s whl h scod ss dvgsr u, whos ms a h dc of h rms o hs wo sriesvrgs o o

44/52

3.4 UNIT m

3.32 Absolute Convergence of a Series

A ries Lun

cotaiing th sitive ad egative tes is said to basolutely co-vee, i L un is converget

hus, the series which becomes covergent when all its negative tesamade si-tive is caled absolutely conveget series

Fo eample the sees

1 1 I1 - + 2 2 23

b l h I I I .

1s a so u y converget ecaus e ss I 2 2 3 · 1s convergen. the res Lun covege and he series L I Un I diverg, the the sere L"n is

id to b codjtional convergtor example the ses

1 1 1 1 1! - -- 2 3 4 5 6

coverges by Leibtz's tes but he sesI 1 l I

I 2 3 4 5 6 . .

d th

. I 1 I I I1s 1verget. Hece e ses 1 - + 3 - 4 S - 6 · 1s cod1t1ona convergen.

Theem 31 1 : An bely cvn its crg

Proof: Sppose lu, converges Thrfre, fr a given e > 0, here ess a sitveinteger n0 such tat

Therefre,

I Un+ l I Un+ + + Um I < fr m n>

Un+I Un+ + � I Un+I Un+ + Un I . + U r m > 1.

ence, b Cauch's pnciple o convergece, the series :n is convegent


45/52

Ifte Series ad DlrnaJ Caclu 3.45

Rema 32 The convese of Theorem 320 is not e. For exaple, the seiesI - + - + � + · is convergent by Leibnitz's test, bu he series L I Un I -I

I l I l 1 d

2 + 3 4 + 5 + 6 ·· · 1s 1vergent

Thor 1 : ln n oltely convrt e th er b itvetc· on y i on\'rn d r fed b nive te nly s nn.

EXPLE 3.44Show th te exponential seies

xl x + + · 2! 3 n!s absote convegent all vls ofx.Soluon: For the given sries we have

u�,

= I

�e)

- n

;

1

and so

_! I = 0 > I .n-x

Hence the series is absoltely convergent by extended D'Ales test

EXE 345

Exaine he logritic sees

absote onvegence

x n+I x + - +( ) 2 nluon It is an alteating series r wich

I: l n I - r+I =

46/52

3.46 UNIT Il

Theere

lm U I _ lno U+l - �and so, by extended DAlembet's to es he gven seres converges absolutely l�

> 1 that s, f I x I < 1.When x = , the seres becomes 1 - + �- + ., whch converges by LebnsB h

l l dtest ut t e sees 1 +2+3 4+ · · 1verges

Wen x 1 , te seies becomes ( �+ j+ + ) and s hence dvergentWhen x< 1, the tens are al negatve. Removng a common negatve sgn the e

ecomes stve Snce

I. U 11m = l the nth ten does not tend to zero. For x y, thenn

( lo n l = 1 logx - lg n l x -

- o,. gn 0sce - s n-o

n

Therere o as n - o. The seies therere oscllates nnteyn

E>PLE 3.46

snO s n6Show tht te sees L. and ;are absolutely convergent p > 1Soluton: nce

'

I .'

cosn8'

si nfnd sce

L.

converges r p > 1, 1t lowsthat

and

convege

r p > .

47/52

Innite Sris and DlfernlaJ Cuus 347

EXPLE 347

Examne absolute convergence ofthe hyrgeomerc series

1 a

·

P � + a(a l )P(P 1) x + . . . l y l .

2(y 1)

Soluon: We have

lim � lim l (n I)(ny) . ! = 1n- Un+I n a + n){P n) J I '

Therere by D' Alembers raio test the hyrgeometrc seres is absolutely convergen

if I

> 1 hat s if I x < 1

When x = 1, we have

( 1 +!) ( 1 + )Un n n l - a - P 1 b ' . I ·-= P= I 0 2 oma eansoUn+1 ( l �) l + ; n n

Therere b Gausss tes the series s converget f l + y - - p > l or ify> a + p anddivergent f l y - a - < or if $ a+ p

When x > l im �

1 1 Therere the series diverges whaever p and ybe n- Un+ Xmay

EXPLE 3.48

Dscuss covegence of the Binomil Series

1 nu+ mm l)x . m(m

-

1). .

. m - 1 + I ) x . !

outon: We hve

I. Un · n 1 i 1 I lm - m - m n- Un - n-o m - n + I x no m _ l ! J ·n nHence the series is absolutely convergent if x

< l . Fuher,

I � i I Un+I I m Un n = m x n n Un

48/52

3.48 UNIT Il

Therere Lim Un = lim I i� 0 i IX > 1 n-

Hence the series cannot conerge when!i > 1 .

Whenx 1, the tes alteaelypositve and negative aer a cetain stage and the seres oscilates innitely. Wenx = I , the series is

m(m ) n m(m I) · · (m - n - I - m 2

(1n!

+ · ·

Whateer m may b, the terms are of the same s aer a ceain value of. We haven m 1

O

-= 1 + 2 .+1 n n

Hence, by Gauss's test he sees is convergent im + I > l , tha is ifm > 0and divergentif m < 0 If m = 0, the seies reduce to single tem 1. Ifx = l the series ism(n - I m(m - 1 ) · · (m - n - l )

l + m + 2

The tes are alteatey positive nd egatie aer a ceain value of n Fro abovewe have

'. = 1 +� + 0 - .Un+ IHence, by Gauss's test the sees conveges absolutely f m + 1 > that s if m > 0

EXPLE 3.49

Examine the lowing series convegence/absolute convergence

L(

l )n l

sin 1

i 13i(ii)

x x3 x4 5x + - -

2 3 4 5 x3 x4

x + · V v .

Solution:

(i The 1h term of the seies isU =(-1r+'s n d I 1 - ' sin n I . n a so tin 3 � 3 •

49/52

Innie Sre and Dfereta Caculus 3.49

aking vn a �3 we have L, =L:3, wich converges. herere by comparisontes lun conveges. Hence, te given ries is absoltely convergent(ii) We have u and

Ths u_X n I _ (n 1Un+ - ; . � .

lim I_ I = 1 !) I ! I = I ! I·1-0 Un+ n X XHence, by extended D'Alembes ratio tet te give eies converges absolutey if

I I>

I, tat i i x < I Ix = I, te ies comesl I I I 1l 2 + 3 - 4 + 5 - 6 · wich is convergent by Leibitz's tet Ix = - ten the series come 1 � -I 1 I I ) h t .5- = +2 + 3 , w C S vergent Hence e gven sees convergesr -1 < x � I and converge abolutely r -1 < x l, tatis i x < 1, tha i i -1 < x < I. If = , then the serie becomes x I1 + .· J '

wose th t i u 0 as o Aso U 2 u• Hence y eintz est2u1 converge. lfx = 1 then the sees becoe

1 1 1 - - - · · J


50/52

3.S UNIT llJ

o ( 1 I I ) h h d"- I + · w 1c 1s 1vergen. J 'Hence he gven sere converges r - I

51/52

InniteSe ad Dferea Caculm 3.51

EXERCISE 33

Altematfng Seres

1 . Exmine the convergence of

(i) - x - 4 . l l+2 l+3 l+

Ans Convergent

Hnt: Terms decr monotonclly t �= - 0 a n - o. Hence,l + x

by Lebnitz's test, the given alteating sees converges

(i• r

. ,2 + I

Hi: Un - � (nite and so the sees oscilltes niely

lute Convergence/Condtonal Convergence

Show tht the sees

1 1 1 1 1 1

(t) 2 -2 . 2+3 . 23 - 4 . 24 + .

( .. ) �

( l) n 3 · 6 (3n) 11 L

-2 . -

3

-

.

)

3

{iii) (- l n+

nd

ivL, 0 < x < 1 bsoluey convegent+

An. Convergent

An. Convegent


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3.5 2 UNT Ill

3 Sow at e series

(i) (-•rJ

(ii2 •r1 - og ·and

. c 1rl1�

og{ + l

ae conditionally convergent

4. Sow that e series

converges absolutely.

cosx cos 2 cos 3x--+· 1 23 3

I

cosIU

I

IHint:

-

$3 andL

3 converges 1

34 DIFFERENTIAL CALCULUS

34.1 Differential Coefficents of Elementaryand Trigonometric

Let be a fnction o x then is dierentiaJ coecient : a nction o x whc weassme at i possesses a derivative i it is er dierentiated. Te derivative : scalled st dierential coecient o frst order derivative o wrt x Te derivative o:ie!(:) is called second order derivative o .X,whic'e denote az readadee two over dee x uae] The derivative f ie !(�) is caled thid oderdeivative o wrt x, wich is denoted as hs i is dieentiated times sc-

e i el t te it i lled th de de i ti e t d it i de te Z

Sequence and Its Convergence

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