Infinite Sequences and Series

Infinite Sequences and Series

In this chapter we shall study the theory of infinite sequences and series, and investigate their convergence.

sequence infinite}{a 1.

:Notation

,8,6,4,2 2.

Cos4, Cos3, Cos2, Cos1, Cos0, 1.

Examples

sequence infinite,a,a,a,a,a,a

Sequences Infinite 5.1

1nn

543210

oscillate. tosaid is }{a then ,-or nor to

limit, finite a toconvergenot does alim If 4.

divergent. be tosaid is }{a ,alim If

Calim if Cnumber finite a toconverges }{a 3.

n allfor aa if decreasing monotone is and

n, allfor aa if increasing monotone is}{a 2.

n. allfor k |a|such that

0kconstant a exists thereif bounded is}{a 1.

:Remark

1nn

nn

1nnnn

nn

1nn

1nn

1nn1nn

n

1nn

nn

n

n

1nnn

nn

n

22

nn

nn

nn2

2

n

alim ,n

2a 4.

oscillate }{a ,(-1)a , (-1)lim 3.

0n2n

)n(-)2n(limalim , n-2na 2.

3

2alim ,

43n

n2na 1.

:Example

不存在

是有界集合。則，及有一個下界有一個上界且若

的一個下界。是有下界且則集合，使得存在且對於若

的一個上界。是有上界且則集合，使得存在且對於若

是所有實數的集合。令

A

pqARA 3.

ApA

pxp A,xRA 2.

AqA

qxq A,x RA 1.

:Definition

R

Examples ：

1. 若 A={-1,-2,-3,-4,…}, 則 A 是有上界的集合，且 -1,0,1,皆是 A 的一個上界，其實大於或等於 -1 的實數都是 A的上界。

2. 若 A={1,2,3,4,5,…}, 則 A 是有下界的集合，且 0,-1,-2, 皆是 A 的一個下界，其實小於或等於 1 的實數都是 A 的下界。

3. 若 A={-3,-2,1,0,1,2,3,4}, 則 A 有一個上界 4 及有一個下界 -3

故 A 是一個有界集合。

Definition:

1. 令 A 是有上界的集合，若是 A 的一個上界且小於或等於A 的其他上界，則稱為 A 的最小上界，記為 lub(A) 或 sup(A) 即 lub(A)=sup(A)=

2. 令 A 是有下界的集合，若 g 是 A 的一個下界且 g 大於或等於 A 的其他下界，則 g 稱為 A 的最大下界，記為 glb(A) 或 inf(A) 即 glb(A)=inf(A)=g.

注意 1. 若 A 是有上界的集合，則 sup(A) 存在。

2. 若 A 是有下界的集合，則 inf(A) 存在。

3. 若 A 是有界集合，則 sup(A) 及 inf(A) 存在。Example:

1. 若 A={x | x<0}, 則 lab(A)=sup(A)=0, 但 sup(A) A 。

2. 若 A={1/n | n=1,2,3,…}, 則 lub(A)=1, glb(A)=0, 但是。

Aglb(A) A,lub(A)

但不收斂。是有界

如數列此定理的逆敘述不成立

注意有界。故數列則對於所有

令

故對於

或即使得

對於則存在一整數收斂到若

,

(-1),

:

a k,|a| n,

,|1-c|,1c1,|a|,1,|a|1,|a|maxk

|1-c||,1c|max|a|N,n

c1a1-c1c-a1-1,c-a

N,nN,c,a :Proof

bounded. is sequence convergentEvery

5.1.1 Theorem

1nn

1nnn

N21

n

nnn

1nn

其證明與上面的相似。

收斂。則列是有界且單調遞減的數注意

故得或對於得或對於得即對於

則是遞增數列因為使得存在故對任意

因此對於所有令存在故其上界

必定有因此的數列是一個有界且單調遞增令

a,a :

calim

|c-a|N,n

ca-c N,n

ca-cN,n

caa-c

,ac,a-cN,0,

c,an,),asup(c,)asup(,

a,a :Proof

converges. sequence decresing)or ncreasingmonotone(i boumdedEvery

5.1.2 Theorem

1nn1nn

nn

n

N

N

1NN

1nnN

n1nn1nn

1nn1nn

此數列有界。利用數學歸納法可證明

對於考慮數列

這裡

則是有下限且遞減的數列若

這裡

則是有上限且遞增的數列若

推論

n1n11nn

n1n

1nnnn

1nn

n1n

1nnnn

1nn

a2a1,n,2a,}{a

:Example

.ainf)}inf({ad d,alim

,}{a 2.

.asup)}sup({ac c,alim

,}{a 1.

:

1.831c c2)-c (

c2c

c2c solve To

alim2

a2limalim

thatnote wec, find To

. converges calim 5.1.2 Theoremby

,increasing monotone and bounded is a Sequence

. 22222a2a then 2,a if 2,2a

22

2

nn

nn

1nn

nn

1nn

n1nn1

. }{a of esubsequenc a also is ,4000

1,

400

1,

40

1,

4

1

. }{a of esubsequenc a is ,2

1,

2

1,

2

1,

2

1}{b

,5

1,-

4

1,

3

1,-

2

1,

1

1-

n

(-1)}{a 2.

.61,2,3,4,5, of esubsequenc a is ,,2,22,2 1.

:Examples

kkkfor ab if

, }{a of esubsequenc a called is }{b 1.

:Definition

1nn

1nn4321nn

1n

n

1nn

432

321kn

1nn1nn

n

.n allfor k,|a| ,,a,a,a :pf

esubsequenc convergent a has sequence boundedEvery

5.1.4 Theorem

. c toconverges }{a of esubsequencevery

ifonly and if c toconverges }{a sequenceA

5.1.3 Theorem

n321

1nn

1nn

k a a a ak - 4213

R

nn

1nn

nn

1nn

1nn

a inf limby denoted is and }{a oflimit lower the

called is E of boundlower greatest The .a sup limby

denoted is and }{a oflimit upper thecalled is E of

boundupper least The limit. ialsubsequent its all ofset

thebe Elet and sequence bounded a be }{aLet

:Definition

,

密密麻麻的點。一定會有某些地方是有限空間插入無窮多點

-1ainf lim 1a sup lim

[-1,1]E cosn,a ,}{a 4.

{-1,0,1}E -1,ainf lim 1a sup lim

}1,0,-1,1,0,-{1,0,-1,1,}{a 3.

1ainf lim 3a sup lim

{1,3}E odd isn if

n

13

even isn if n

11

a ,}{a 2.

1ainf lim 1a sup lim

1E ,n

11a ,}{a 1.

Example

nn

nn

n1nn

nn

nn

1nn

nn

nn

n1nn

nn

nn

n1nn

。夠大只要很近很近就應該要靠的與當然既會密密麻麻靠近夠大時當

靠近會密密麻麻的往意指

)mn,(

aac,)n(

c a,a,a c toconverges }{a

:Proof

. Nnm, allfor |a-a|such that Ninteger an is there0

eachfor ifonly and if converges }{a sequence The

criterion)Cauchy (The 5.1.6 Theorem

CriterionCauchy The 5.1.1

cainf limasup lim

ifonly and if c toconverges }{a sequence The

5.1.5 Theorem

mn

3211nn

mn

1nn

nn

nn

1nn

sequenceCauchy n

1sin 3.

sequenceCauchy n

(-1)1 2.

sequenceCauchy n

1 1.

:Example

. Nmn, allfor |a-a|

such that Ninteger an exist there0each

for if sequenceCauchy a called is }{a sequenceA 1.

:Definition

1n

1n

n

1n

nm

1nn

. casay then wecSlim If 4.

.divergent be tosaid is

a series then thelimit, finite a toconvergenot does S If 3.

. converges S sequence theif convergent be tosaid is series The 2.

. a series theof sum partialnth thecalled is aS 1.

:Definition

n!

e 3.

n

1cos(-1) 2.

3

1

2

1

1

1 1.

:Example

series infinite-aaaa

Series Infinite 2.5

1iin

n

1ii1nn

1nn

1ii

n

1iin

1n

n

1n

n

1nn321

. Nmn, allfor S-Sa

converges Sconverges a

sum partialnth aSa

:Pf

. Nnm, allfor asuch that Ninteger an is there

0given afor ifonly and if converges a

5.2.1 Theorem

mn

n

1mii

1nn1i

i

n

1iin

1ii

n

1mii

1ii

. 0Slim -Slim )S-(Slim alimlimit Take

. S-Sa and

c,Slim c,Slim have We

. a of sum partialnth thebe Slet

c,a that suppose :pf

. 0alim then converges a If

Properties

1-nn

nn

1-nnn

nn

1-nnn

1-nn

nn

1iin

1ii

nn

1ii

diverges. n

1 Hence Slim have We

2

k

2

1

12

1SS

2

3

8

4S

8

1

5

1SS

2

2

2

1

2

1

4

2S

4

1

3

1SS

2

1

2

1

1

1S since

0.n

1lim and ,

n

1 2.

diverges. 1n

n hence

0,11n

nlim Since ,

1n

n 1.

:Example

1n2k

k1-k22

448

224

2

n1n

1n

n1n

k

1-kk

.convergent absolutelynot are n

cos ,n

(-1) 2.

.convergent absolutely are ,n!

(-1) ,

n!

1- ,

2

1- 1.

:Example

.convergent is |a| if convergent absolutely is a series The

:Definition

n

2-

n

4 ,

n

1

n

2 ,

n

3 1.

:Example

ba)b (a 2. acca 1.

:convergent also are series following then the

constant, a is c if and series, convergent twoare b and a If

5.2.2 Theorem

1n1n

n

1n

n

1n

n

1n

1nn

1nn

1n42

1n1n

1nn

1nnn

1nn

1nn

1nn

1nn

1nn

.convergent are 3

1- ,

n!

(-1) ,

n

(-1) 1.

:Example

converges. a Hence

mn, as 0|a|converge |a|

criterion.Cauchy theapplyingby

,|a| and a seriesConsider

:Pf

.convergent is series convergent absolutelyEvery

5.2.3 Theorem

n

1n1n

n

1n2

n

1ii

n

1mii

1ii

n

1mii

1ii

holds are (ii) (i) Hence

. ba and criterion Cauchy Apply the :Pf

too.divergent is b then divergent, is a If (ii)

. a does so then converges, b If (i)

integer. fixed a is N where,Nnfor

basuch that termspositive of series twobe b and aLet

Test) Comparison .4(TheTheorem5.2

terms.positive of series theis 1,2,n 0,a ,a

Terms Positive of Seriesfor eConvergenc of Tests 5.2.1

n

1mii

n

1mii

1nn

1nn

1nn

1nn

00

nnnn

n1n

n

divergent. is n

1 hence ,

2

mlimSlim Since

2

m

2

1

12

1

4

1

3

1

2

1

1

1S

2

3

8

1

5

1

4

1

3

1

2

1

1

1S

2

2

4

2

2

1

4

1

3

1

2

1

1

1S

2

1

2

1

1

1S

i

1S,

n

1 1. :Examples

1.kfor convergent is n

1 2. series. harmonic thecalled is

n

1 1.

:Remark

1nn2n

m1-m2

2

4

2

n

1in

1n

1nk

1n

m

m

3

.convergent is , 1k , n

1 proves This

.convergent is }{S hence ,increasing and bounded is }{S

1-2

2

21-1

1

)2

1()

2

1()

2

1()

2

1(1)

2

1(

8

1

4

1

2

11

))2

1()

2

1()

2

1(()

8

1

8

1()

4

1

4

1()

2

1

2

1(1

)1)-(2

1

)(2

1()

15

1

8

1()

7

1

4

1()

3

1

2

1(1

i

1

i

1S

have we2nsuch that minteger exist theren,each For

1,2,...n ,i

1S 1,k ,

n

1 2.

1nk

nn

1-k

1-k

1-k

1-m1-k

31-k

21-k

11-k

1-k1-m1-k1-k1-k

k1-m

k1-m

k1-mkkkkkk

kmk1-mkkkkkk

1-2

1ik

n

1ikn

m

n

1ikn

1nk

m

.convergent is n

1~

12nn

1 3.

divergent. is n

1~

nn

43n 2.

.convergent is n

1~

nn

2 1.

:Examples

b ba ba that means b

alim

:Pf

divergent.both or convergentboth either are series twothen the

,constant somefor b

alim If terms.positive of series twobe b and aLet

5.2.5 Theorem

1n233

2

22

1nn

1n

1nnnn

n

n

n

n

n

n1n

n1n

n

llll

ll

n

.a

asup lim1

a

ainf lim if ,conclusion No 3.

1.a

ainf lim if diverges, a 2.

1.a

asup lim if converges, a 1.

:hold following theThen terms,positive of series a be aLet

Test) o5.2.6(Rati Theorem

.convergent is n

1~

1n

1~

1n

1sin 2.

1).x

sinxlim (Sincedivergent is

n

1~

n

1sin 1.

:Examples

. somefor b

alim that means b~a

:Remark

n

1n

nn

1n

n

n

1n

n1n

n

n

1n

n1nn

1nn

1n2

1n2

1n2

0x1n1n

n

n

nnn

ll

. n as 12n

2

21

12n1

a

a

. n as 02

1-2n

1-2n1

21

a

a

12n

1a ,

2

1a i.e

2

1

5

1

2

1

3

1

2

1

1

1aLet 2.

converges. a1ra

alim

)rrr(1aararraaaThen

. n asr a

a Suppose aaaaa

. 1r ,r-1

1rrr1r 1.

:Examples

n

n2n

12n

n

n

1-2n

2n

12nn2n4321n

n

nn

1n

n

3211

31

211

1nn

n

1n4321

1nn

32

0n

n

0cinf limcsup limclim

2

1c

4.Consider

-1cinf lim 1csup lim exist t doesn' clim

,9

1,-1

8sin,

7cos,

6

1,-1

5sin,

4cos,

3

1,-1

2sin,

1cosc

Consider 3.

a

asup lim 0

a

ainf limBut

exist.t doesn' a

alim

2n m if12n

2

1-2nm if2

1-2n

2n m ifa

a

1-2nm ifa

a

a

a Since

nn

nn

nn

nn

nn

nn

nn

n

m

1m

mm

1m

m

m

1m

m

n

n

2n

12n

1-2n

2n

m

1m

432432n

n

1n

nn

1n

n

n

1n

nn

1n

n

m as

2m1-2m

2m

1-2m

2m

m as

2m2m

12m

2m

12m

n

1n

54321n

n

n

n

n

1nn

nn

1n

nn

1n

n

1nn

1nn

1nn

n

1n

n

1n1nn

2

1

2

1

2

1

2

1

2

1

3

1

2

1

3

1aBut

.conclusion no a

asup lim1

a

ainf lim 5.2.6 TheoremBy

. 0a

ainf lim ,

a

asup lim have We

)2

3(

3

1

31

21

a

a

0)3

2(

3

1

21

31

a

a

a

a Since

3

1

2

1

3

1

2

1

3

1aConsider

odd isn if3

1

even isn if2

1

aLet 3.

divergent. is a hence 1,e

1)(nlim

en!

e1)!(nlim

a

alim Since

e

n!a 2.

.convergent is 1)!(n

1 hence 1,0

1)!(n1

2)!(n1lim

a

alim Since

1)!(n

1aConsider 1.

:Example

n

2n

nn

kn

n

n

1n

nn

kn

n

1nn

n

kn

n

1nn

n

kn

n

1nn

a

alim

a

alim

2kFor 2.

test.ratio theis 5.2.7 Theorem ,a

alim

a

alim

1kFor 1.

:Remark

diverges. a then 1,a

alim If 2.

converges. a then 1,a

alim If 1.

integer. positive fixed a bek and termspositive of series a be aLet

5.2.7 Theorem

.convergent is a 5.2.7, Theoremby , 5

1

a

alim But

.divergence a

alim get We

,5

1,

5

1,

5

1

a

a ),

5

1(

3

2,

2

3),

5

1(

3

2,

2

3),

5

1(

3

2,

2

3

a

a Since

5

13

5

12

5

13

5

12

5

13

5

1232aConsider 2.

exist.not does a

alim

odd. isn if 31

31

even. isn if 21

21

a

a Since ,aConsider

odd. isn if 3

1

even. isn if 2

1

aLet 1.

:Example

1nn

n

2n

n

n

1n

n

n

2n

n

1n

3322

1nn

n

2n

n

n

2n

n

2n

n

2n

1nn

n

n

n

converges. a2 ifonly and if converges a then

n, offunction decreasing monotone a is a where terms,positive of series a be aLet

Test)on Condensati schy'5.2.10(Cau Theorem

alimainf limasup lim then exist, alim If 1.

Properties

converges. a hence 1,2

1alimsup Since .aConsider

odd. isn if 3

1

even. isn if 2

1

a 1.

Example

1. if ,conclusion No (3)

1. if dinverges, a (2)

1. if converges, a (1)

Then asup limLet terms.positive of series a be aLet

t)Cauchy tesor test 5.2.8(Root Theorem

1n2

n

1nn

n1n

n

nn

nn

nn

nn

1nn

n

1

nn1n

n

n

n

n

1nn

1nn

n

1

nn1n

n

n

converges. tlima2 ifonly and if converges Slima hence

,tSt2

1 Since

t2

1)a2a22a(a

2

1

a28a4a2aaa2

1

)a(a)aaa(a)a(aaa2

1

aaaaS

hand,other In the

ta28a4a2aa

)aa(a)a(a)a(aa

a)aaa(a)a(aaS

get we2n2such that 0,minteger exist theren,each For

.1,2,mn, ,a2 t,aSLet

:Pf

mm

0n2

nn

n1n

n

mn1-m

1-m2

1-m4

221

2

2-m168421

21287654321

2321n

m2

m8421

1-212274321

27654321n

m1-m

m

1i2

im

n

1iin

n

1-m

1-m

1-m2-m

1-m

m

1mmm

m

i

1.k if converges, n

1 Hence

)2

1(

2

1

)(2

12a2

. n

1aConsider 2.

divergent. is n

1 Hence

12

12a2

. n

1aConsider 1.

:Example

1nk

n

0n1-k

0n1)-n(k

0nkn

n

0n2

n

1nk

1nn

1n

0n0nn

n

0n2

n

1n1nn

n

n

divergent. is a hence divergent, is b Since

1b

b

a

a ,

b

b

a

a

0b

1-

a

a

b

1 0

b

1-

a

a

b

1

case 0For

:pf

0 if diverges and 0 if converges aThen

. )b

1-

a

a

b

1(limLet

divergent. is b that Suppose

terms.positive of series twobe b and aLet

Test) s(Kummer' 5.2.11 Theorem

1nn

1nn

n

1n

n

1n

1n

n

1n

n

1n1n

n

n1n1n

n

n

1nn

1n1n

n

nn

1nn

1nn

1nn

divergent is 2

n as -1)(n-2

2n

b

1 -

a

a

b

1

n

1b ,2a 2.

.convergent is 2

1 test,sKummer'By

n as 1n-2n1)(n-2

2n

b

1 -

a

a

b

1

n

1bLet a

2

1 1.

:Examples

1n

n

1n

n

1n1n

n

n

1n 1nn

1n

n

1nn

n

n

1n

1n1n

n

n

1n 1nn

1n 1nnn

converges.

n

1 test sRaabe'by 1,2 Since

n

1o

n

21

n

1

n

21

n

1)(n

1n1

n1

a

a

n

1aConsider :Example

0.1- if diverges a and 01- if converges a test sKummer'by diverges, n

1 Since

n as 1-1n1

1-

a

a

n1

1

n as 0-n-a

an

n as 0n1

n-1-

aa

n

1o

n1

a

a

:proof


n

1o

n1

a

a that and termspositive of series a is a that Suppose

Test) s(Raabe' 5.2.12 Theorem

1n2

22

2

2

2

1n

n

1n2

1nn

1nn

1nn

1n

1n

n

1n

n

1n

n

1n

n

1nn

1n

n

1nn

diverges. n

1 test sGauss'by ,

n

10

n

11

n

11

a

a Since

n

1a

:Example

diverges. a test sKummer'by , diverges nlogn

1 Since

-11

)x-(11-lim

x

x)-log(1lim

1)(n1

1))(n1-log(1lim 0

n

1(nlogn)Olim and n as -1

1n

n1)log(n

because trueis This

0-1n

1(nlogn)O

1n

n1)log(nlim)

b

1-

a

a

b

1(lim

calculate and nlogn

1bput , 1consider thereforeusLet

1 if diverges and 1 if converges a test sRaabe'By

n

1o

n

1O Since

:proof


0 ,n

1O

n1

a

a that suppose , termspositive of series a be aLet

Test) s(Gauss' 5.2.13 Theorem

1n211

1n

n

1n1nn

1nn

1n

0x0xn1n

1n1n1n

n

nn

n

1nn

1

1nn

11n

n

1nn

series. galternatin are n!

(-1) ,

13n

(-1)

:Example

).series( galternatinan called is 1nfor 0a where,a(-1) 2.

.absolutelynot but converges a if series, convergentlly conditiona a called is a 1.

:Definition

13n

(-1) ,cosna 1.

:Example

. 1nfor negativeor positive bemay a where,a

Terms Negative and Positive of Series 2.2.5

1n

1-n

1n

n

n1n

n1-n

1nn

1nn

1n

n

1n1nn

n1n

n

交錯級數

converges.a(-1) criterion, sCauchy'By

mn, as 0a

odd is m-n if )a-(a--)a-(a-a

even is m-n if a-)a-(a--)a-(a-a

a(-1)a-aS-S Hence

n 0,a and decreasing monotone is a Since

a(-1)a-a(-1)S-S

have wen,mFor

. a(-1) of sum partialnth thebe SLet

:Pf

.convergent is series then the, 0alim and decreasing

monotone is a sequence thesuch that series galternatinan be a(-1)Let

5.2.14 Theorem

1nn

1-n

1m

n1-n3m2m1m

n1-n2-n3m2m1m

n1-m-n

2m1mmn

n1nn

n1-m-n

2m1mm

mn

n

1ii

1-in

nn

1nn1n

n1-n

.convergentlly conditiona is lnnn

(-1) hence

divergent, is lnnn

1but

,convergent is lnnn

(-1) since ,

lnnn

(-1) 4.

. duu

1dx

xlnx

1 since ,convergentlly conditiona is

nlnn

(-1) 3.

.convergentlly conditiona is n

(-1) 2.

.convergent are n

(-1) ,

n!

(-1) ,

n!

(-1) ,

n

(-1) 1.

:Example

2n

n

2n

2n

n

2n

n

ln2

lnxu

22n

n

1n

n

1n

n

2n

n

1n

n

1n

n

. f(x)dx ifonly and if a Hence

a-adxf(x)a-a

a-adxf(x)a

1a1adxf(x)dxf(x)dxf(x)1a1a1a Since

:pf

converges, f(x)dx ifonly and if converges athen ,1,2,3,n ,af(n) If

0. todecreasing monotone and continuous be Rx 0f(x)Let

:Properties

11i

i

n

n

1ii

n

11

n

1ii

n

n

1ii

n

1

n

2ii

1-n1

n

1-n

2

1

n

1n32

11n

nn

1a

y

x3 2 1

)(xfy

497253

54321i

i

2

1

2

1

2

1

2

1

2

1

2

1

1

1 :Rearrange

2

1

2

1

2

1

2

1

1

1a

Series ofent Rearrangem 3.2.5

. 1,2,3,n ,absuch that

JJ:ffunction onto and one-to-one a exists thereif

,a ofent rearrangem a called is b series The series.given a be aLet 2.

integers. positive ofset thedenotesJ 1.

:Definition

f(n)n

1nn

1nn

1nn

converges. |b| hence converges, |a| Since . |a| |b|such that Mexist e ther

b of sum partialnth each For .a ofent rearrangem a be bLet

:pf

sum. same thehas and convergent

absolutely remainsit ofent rearrangemany then ,convergent absolutely is a If

5.2.15 Theorem

. a ofent Rearrangem

-5

1

12

1-

10

1-

3

1

8

1-

6

1-

1

1

4

1-

2

1-b ,

6

1-

5

1

4

1-

3

1

2

1-

1

1a 2.

. a of seriesent rearrangem a is b

4

1

11

1

9

1

7

1

2

1

5

1

3

1

1

1b ,

4

1

3

1

2

1

1

1

n

1a 1.

:Example

1ii

1ii

M

1ii

N

1ii

1n 1ii

1nnn

1nn

1nn

1nn

1nn

1nn

1nn

1nn

1n1nn

converges. n

(-1)but diverges, b hence

, Slim have we1-4n

nlim Since

1-4n

nS

1-4n

1

12n

1S

)1-4n

1

12n

1()

2n

1

1-2n

1()

4

1

3

1()

2

11(

)2n

1

1-4n

1

3-4n

1()

4

1

7

1

5

1()

2

1

3

11(SThen

. n

(-1) of sum partialnth thedenote S and b of sum partial3nth thedenote SLet

6

1

11

1

9

1

4

1

7

1

5

1

2

1

3

11b

seriesent rearrangem theand n

(-1)Consider

:Example

1n

1-n

1nn

3nnn

2n2n

3n

1n

1-n

n1n

n3n

1nn

1n

1-n

. babababa

0,1,2,n ,baC

where,C series theis b and a series ofproduct sCauchy' The 1.

:Definition

Series oftion Multiplica 5.2.4

book text See :proof

-or todiverge or tonumber given any to

converge toas so rearranged be alwayscan series convergentlly conditionaA

5.2.16 Theorem

,, 2.

, 1.

:Remark

0n2-n21-n1n0

k-n

n

0kkn

0nn

0nn

0nn

此新級數不一定收斂。若重新排列出新級數條件收斂的級數斂值。級數一定會有相同的收不管如何重新排列的新絕對收斂的級數

1)2

13

2

13(11)

2

13(11)3333)(1

2

1

2

1

2

1

2

1(1 .2

3

13

2

510CCCCC

43213

1

2

1

1

1n

1n

1C

0n

14)-(n

5

13)-(n

4

12)-(n

3

11)-(n

2

1n

1

1C

3

13

3

1130

4

11

3

12

2

13

1

1babababaC

2

50

3

11

2

12

1

1bababaC

102

11

1

1babaC

0010

1baC

C seriesproduct sCauchy' The nb ,1n

1a 1.

:Examples

22432

432

43210

0n0n0nn

n

031221303

0211202

01101

000

0nn

0n0nn

0n0nn

. 3

4

41

-1

1

2

1 Cget We

odd. isn if 0

even. isn if 2

1(-1)

2

1

2

(-1)

2

(-1)

2

1baC

Cproduct sCauchy' The

3

2

2

(-1)b 2,

2

1a 1.

:Example

st. toabsolutely convergesC then ,convergent absolutely are b and a If (ii)

st.C and converges C then ,absolutely converges b and a of oneleast at If (i)

then,b t,as that Suppose , b and a ofproduct sCauchy' be CLet

5.2.17 Theorem

0n2n

0nn

n

n

0i

i-n

n

0in

i-n

i-n

i-nn

0ii

n

0ii-nin

0nn

0nn

n

0nn

0nn

0nn

0nn

0nn

0nn

0nn

0nn

0nn

0nn

0nn

0nn

0nn

0nn

0nn

D.x ,(x)f(x)S where

Don f(x) touniformly converge tosaid is (x)f then D,on xt independen is N If

Nn ,S-(x)Ssuch that Ninteger an exists there0,given aFor 3.

series. theof sum thebe tosaid is s(x) then s(x), toDin every x for converges (x)f If 2.

D.on f(x) toconverge tosaid is (x)f sequence then thef(x),(x)flim

Din every x for such that Don defined f(x)function a exists thereIf 1.

R.Dset aon defned functions of sequence a be (x)fLet

:Definition

Function of Series and Sequences 3.5

n

1iin

1nn

nn

1nn

1nnnn

1nn

. 0lim thusN,n ,f(x)-(x)fsup that followsIt

. D xallfor f(x)-(x)f , Nnfor such that

on only depends that Nexist thereD,on uniformly f(x)(x)f since 0,given aFor

. n as 0 that show To, Don uniformly f(x)(x)f that Suppose )(

. Don xuniformly f(x)(x)f i.e

Dx Nn |0-|f(x)-(x)f Hence . Nn |0-|such that

on xt independen Nexist there0lim and Dx ,f(x)-(x)f Since

. Don uniformly f(x)(x)f that show To 0,lim that Suppose )(

:proof

n as 0 ifonly and if

Don f(x) touniformly converges sequence Then the ,f(x)-(x)fsup as Define

. f(x) toconverges and RDon defined functions of sequence a be (x)fLet

5.3.1 Theorem

nn

nDx

n

n

n

nn

n

nnnn

nn

nn

nnn

n

nDx

nn

1nn

. [0,1]on uniformly 2x (x)f ,1n

12nx(x)f 5.

. Don uniformly 1(x)f Hence

. n as 0 n

2

n

x2 sup1-

n

x2cos sup1-(x)f sup Since

. [0,1]D ,n

x2cos(x)fLet 4.

. [0,1]on uniformly 0(x)f Hence

. n as 0 n

0-)n

xsin( sup And

nn

x0-)

n

xsin( Since

. [0,1] x),n

xsin((x)fLet 3.

)[0,xx cosx-1 2.

)[0,x 0x-sinx 1.

:Examples

n1n

1nn

n

DxDxn

Dxn

1n1nn

n

[0,1]xn

n

. Don uniformly converges (x)f Hence

. mn, as 0|M||(x)f|(x)f

criterion sCauchy' By the

:proof

. Don uniformly converges (x)f then converges, M if

then D, xallfor

1,2,n ,M|(x)f|such that constants of M

sequence a exists thereIf R.Don defined functions of series a be (x)fLet

Test)-M sss'(Weierstra 5.3.2 Theorem

1ii

m

1nii

m

1nii

m

1nii

1nn

1nn

nn1nn

1nn

. Ron f(x) touniformly converges (x)f

2x1

2x02xf(x)

2xn

2-1

2x0n

12x

(x)f 2.

. [0,1]on uniformly converges nxn

x(x)f Hence

converges. n

1 and [0,1], x,

n

1

0n

x

nxn

x Since

[0,1] x,nxn

x(x)f 1.

:Exampes

. Don S(x) touniformly converge tosaid is (x)fThen

. Don S(x) touniformly converges (x)S If Dx ,(x)f(x)SLet

:Definition

n

n

1n23

2

1nn

1n333

2

23

2

23

2

n

n

1ii

1nn

n

1iin

. Nmn, as ,4

|-|get we(1)in x xaslimit theBy taking

(3)|x-x| 4

|-(x)f| n,each for have we,(x)flim From

(2)Dx N,n as ,4

|f(x)-(x)f|or

(1)Nmn, as ,4

|(x)f-(x)f| have Wef. toconvergentuniformly (x)f From

:proof

. (x)flim limf(x)lim(x)flim limi.e f(x),limlimThen D. ofpoint limit a

is x where,(x)flim If D.set aon f(x) toconvergentuniformly be (x)fLet

5.3.3 Theorem

Series and Sequences Convergent Uniformlyof Properties 1.3.5

mn0

0nnnnxx

n

mn1nn

nnxxxx

nxxnxx

nn

0nnxx1nn

0

0000

0

(x)flimlimlimf(x)lim Hence

x-x as 24

|-f(x)|

get we(2)by

x-x and Nn as2

|(x)f-f(x)|

|-||-(x)f||(x)f-f(x)|

|--(x)f(x)f-f(x)||-f(x)| Since

Nn as 4

|-|Then ,limLet

sequence convergent a is criterion Cauchy By

nxxn

nn

0xx

00

0

n

0nnnn

0nnnn0

0n0nn

1nn

00

1ii

xx

n

1ii

xxn

n

1ii

xxn

nxxn

nnxx

n

1ii

nxx1n

nxx

n

1iin

1nn

xx1n

nxx

nxx

1nn

00nn

nxxn

nnx x

00nn

0nnxx

0

nn

1nn

(x)flim(x)flimlim(x)flimlim

(x)Slimlim(x)Slimlim(x)flimlim(x)f lim

1,2,3,n ,(x)f(x)SLet

:proof

. (x)flim (x)f lim

thenexist, (x)flim n,each for If D.on uniformly convergent be (x)fLet

5.3.2Corollary

)f(x)(xflim(x)flimlim(x)flimlim

5.3.3 TheoremBy

)f(x)(xflim and )(xf(x)flim Since D,each xFor

:proof

Dx f(x)(x)flimLet D.on continuous is f(x)Then D.set aon

f(x) touniformly converges that functions continuous of sequence a be (x)fLet

5.3.1Corollary

000

0000

00

0

00

0

(x)flimlim100lim(x)flimlimget We

0nx1

nxlim(x)flimBut

11lim(x)flimlim have We

1x

n1

xlim

nx1

nxlim (x)flim

(0,1) x,nx1

nx(x)f 1.

:Examples

5.3.2Corollary toSimilarly

:proof

. Don continuous is s(x)Then D.set aon

s(x) touniformly converges that functions continuous of series a be (x)fLet

5.3.3Corollary

nn0xn

n0xn

0xn

0x

0xn

n0x

nnn

n

1n1nn

1nn

5!

1

4!

1

3!

1

2!

1

1!

1

1

1e

n!

1

n!

xlim

n!

xlimelime

5.3.2Corollary By , 0,1on e touniformly converges n!

1 And

. [0,1]x n!

1

n!

x Since .

n!

xe theorem,sTaylor'By

n!

x(x)fConsider 3.

22

1

x)(1

xlim

x)(1

xlim

5.3.2Corollary By

,22

1on uniformly converges

x)(1

xget weconverges.

)21

(1

2 And .,2

2

1x

)21

(1

2

x)(1

x Since .

x)(1

xlim Find .

x)(1

x(x)fLet 2.

0n0n

n

1x0n

n

1x

x

1x

1

x

0n

n

0n

nx

0n

n

0nn

1n1-n

1n1-n1x

1n1-n1x

1n1-n

1n 1-n

1-n1-n

1n1-n1x

1n1-n

1nn

. b][a,on s(x)function some

touniformly converges )(f)x-(x)(xf(x)fget We

. b][a,on uniformly converges )(f and converges )(xf Since

. )(f)x-(x)(xf-(x)f

such that x),(x exists there theorem,mean value by the ,xFor x (1)

:proof

. (x)f(x)s (2)

. b][a,on s(x)function some touniformly converges (x)f (1)

thenb],[a,on uniformly

converges (x)f that and b][a,point x oneat least at converges (x)f that Suppose

. 1nfor b][a,on abledifferenti is (x)f wherefunctions, of series a be (x)fLet

5.3.4 Theorem

1nnn0

1n0n

1nn

1n 1nnn0n

nn00nn

0n0

1nn

1nn

1nn0

1nn

n1n

n

.

2

10,x x),--ln(1

n

x Hence

. 0c 0,n

0s(0) From . cx)--ln(1s(x)get We

. n

xs(x) ,

x-1

1(x)s have We.

2

10,x,

x-1

1x handother In the

. x(x)f(x)s And

. 2

10,on s(x) touniformly converges

n

x 5.3.4 TheoremBy .

2

10,on uniformly converges

x(x)f And converges. n2

1)

2

1(f have We.

n

x(x)fConsider

:Example

(x)f

5.3.2)Corollary (By h

f(x)-h)(xflim

h

f(x)-h)(xflim

e)convergencuniformly ( h

f(x)-h)(xflim

h

(x)f-h)(xflim

h

s(x)-h)s(xlim(x)s

b)(a,each xFor (x)fs(x) (2)

1n

n

1n

n

1n

n

1n

1-n

1n 1n

1-nn

1n

n

1n 1n

1-nn

1n 1nnn

1n 1n

n

n

1nn

1n

n

0h1n

n

0h

1nn

0h

1nn

1nn

0h0h

1nn

. convergent )isc)-(xa(or xa|Rxeconvergenc ofregion The 4.

).( econvergenc of interval thecalled is ))c,-(c(or

),(- interval theand series, theof )( econvergenc of radius thebe

tosaid is then ),c-x(or x ifdivergent is and )c-x(or x if

convergent is )c)-(xa(or xasuch that 0number aexist thereIf 3.

c).-(xin seriespower a called is c)-(xa form The 2.

constant. some are a wherein x, seriespower a called is xa form The 1.

:Definition

function. seriespower and c)-(xa ,xa

SeriesPower 5.4

0n

nn

0n

nn

0n

nn

0n

nn

0n

nn

n0n

nn

0n

nn

0n

nn

收斂區間收斂半徑

.absolutely converges xa implies This converges. From

. 1 ,kx

xk

x

xxa xa Now .1,2,3,n ,k xa

such that 0kconstant a have Weconverges. xa and |x||x|such that

),(-exist x thereSince converges. xa that show To ),(-each xFor

:pf

),(- xallfor absolutely converges

xaThen 0. that Suppose .xa of econvergenc of radius thebe Let

5.4.1 Theorem

e.convergenc of interval theis (-1,1) ,x of

econvergenc of radius theis 1 1,|x| ifdivergent is and 1|x| if convergent is x

:Example

0n

nn

0n

n

n

0

n

0

n0n

nn

n0n

0n

n0n0

00n

nn

0n

nn

0n

nn

0n

n

0n

n

. p

1|x| if diverges xa and converges xa implies

p

1|x| Hence

. |x|pxa

xalim Now .convergent absolutely is xa1

xa

xalim

xa test toratio Apply the

:proof

0p,

p0,

p0,p

1

is xa of econvergenc of radius theThen,

. pa

alim that Suppose series.power a be xaLet

5.4.2 Theorem

0n

nn

0n

nn

nn

1n1n

n0n

nnn

n

1n1n

n

0n

nn

0n

nn

n

1n

n0n

nn

. 2

1|x|on absolutely converges

n

x2 2.

. 1 is x of econvergenc of radius The 1.|x|on convergent absolutely is x Hence

1|x|1x

x Since x 1.

:Example

.2.Theorem5.4 toSimilarly xa to5.2.8 Theoremin root test Apply the

:proof

0q,

q0,

q0,q

1

then

q|a|limsup that Suppose series.power a be xaLet

5.4.3 Theorem

1n

nn

0n

n

0n

n

n

1n

0n

n

0n

nn

n

1

nn0n

nn

. r][-r, x,1,2,3,k ,x

k)!-(n

n!a

dx

S(x)d

and r r],[-r,on orders all of derivative has S(x) then ,xaS(x) If 2.

r wherer],[-r,on uniformly converges xa 1.

following thehave Then we 0).( econvergenc of radius a with seriespower a be xaLet

5.4.4 Theorem

. (-1,1) is econvergenc ofregion The 1.|x|on absolutely converges nx

|x|nx

1)x(nlim nx 5.

. 0 converges of radius The 0.at xonly converges xn!

. |x|1)(nlimxn!

x1)!(nlim Since xn! 4.

. R,on x absolutely converges series The

. 01n

|x|lim

n!x

1)!(nxlim Since

n!

x 3.

kn

k-nnk

k

0n

nn

0n

nn

0n

nn

0n

n

n

1n

n0n

n

0n

n

nn

1n

n0n

n

nn

1n

n0n

n

. r][-r, x,xk)!-(n

n!a(x)S Similarly,

. nxaxaxa(x)S

r][-r,for x 5.3.4 TheoremBy

. r][-r,on convergentuniformly is xnat assert thacan then We1.nlim Since

asup limansup limnasup lim

. xna and xa Compare (2)

. r][-r,on uniformly converges xa

5.3.2), (Theoremtest -M s WeiertrasBy the .convergent is ra Since

. 0nfor ,ra|xa| then r,|x| If (1)

:Pf

kn

k-nn

(k)

1n

1-nn

1n

nn

0n

nn

1n

1-nn

n

n

1

nn

n

1

nn

n

1

nn

1n

1-nn

0n

nn

0n

nn

0n

nn

nn

nn

n

1

n

1

x

0n

n(k)

2n

2-n

0k

k

1n

1-n

1n

1-n

0n

n

0n

n

0n

n

3432

32n

2-n2

1n

1-n

0n

n

n

1

nn

n

1

nn

1n

nn

n

n

443322

0n

nn

en!

xS(x) Hence 1,2,k S(x),(x)Sget We

S(x)n!

1)x-n(n(x)S And

S(x)k!

x

1)!-(n

x

n!

nx(x)S have WeR x,

n!

xS(x)Consider

R xallfor converges n!

x Since

n!

x 3.

1 x,-1x-1

25x64x53x42x32

1 x,-1x-1

2 1)x-n(n 1 x,-1

x-1

1 nx 1x ,

x-1

1x 2.

3asup lim

even isn ifx3

odd isn ifx2|x|axa

0n if1

even isn if3

odd isn if2

a

x3x2x32x1xa 1.

:Examples

. [-1,1]on convergentuniformly is 1)(n

x , 5.4.5 TheoremBy

.convergent absolutely is 1)(n

1 Since 1.xon converges

1)(n

x 1.

:Examples

],[-on convergentuniformly is xa

Test,-M rassBy Weierst . ],[-x ,axa

:proof

. ],[-on convergentuniformly is xa then

,convergent absolutely is a If 0 ,xon converge xaLet

5.4.5 Theorem

0n2

n

0n2

n

0n2

n

nn

nn

nn

0n

nn

nn

0n

nn

. 3

11,-on convergentuniformly is

x-1

x

n2n

2 Hence

converges.

31

1

31

n2n

2 and

11

1-

n2n

2 Since . ,1

3

1-on convergent is series The

above. thetosimilarly x-1

x

n2n

2 3.

. ,13

1- is econvergenc ofregion theand ,1

3

1-on convergentuniformly

. x1

x

n2n

2 Hence converges.

r1

1

n2n

2 ,

31

1

31

n2n

2 Since

1x31-2x13

22

3x1

12

12

1x1

1-12

1-

21

x1

x2

1-2

1x1

x2

1Z1Z2Zn2n

2lim

. Zn2n

2 asrewritten

x1

x

n2n

2 2.

1n

n

2

n

1n

n

2

n

1n

n

2

n

1n

n

2

n

1n

n

2

n

1n

n

2

n

1n

n

2

n

n

1

n2

n

n

1n

n2

n

1n

n

2

n

norm. a is ||||check 3,21(-2),3,1,maxA ,

213

2-1A

m,1,j n,,1,2,i ,amaxA 2.

457)2(13(-2)1A

213

2-1A , aA normEuclidean The 1.

:Examples

matrix.k m a is D where,DAAD (4)

matrix. mn a is B where,BABA (3)

scalar. a is c where,AccA (2)

. 0A ifonly and if 0A and 0,A (1)

:properties following A with the offunction

valued-real a is ,Aby denoted A, of norm The m.norder ofmatrix a beA Let

5.5.1 Definition

1,2,3,k 4k

k

1

2k!

1

A

Matrices of Series and Sequence 5.5

ij

2222

2

2

1n

1i

m

1j

2ij2

k-

k

.n 1,j m,1,ifor aalim if aAmatrix n m the toconverge to

said is A sequence The 1.kfor n m orders of matrices be aALet

5.5.2 Definition

norm. a is ||||check ,458

11

03

12

A Hence

. 45845-,8458maxA)A(e have We

4582

1801601916-

0-23

3-140

10

01-

23

314det solve toisA A of eigenvalue The

23

314

11

03

12

101

132AA

101

132A ,

11

03

12

A

A.A of eigenvaluelargest theis A)A(e where,A)A(eA:norm spectral The 3.

ijijkk

ij

1kkijkk

s

21

S

S

max

2

max21

maxs

divergent. is a series theof oneleast at ifdivergent is A series The 2.

n.1,j m,1,i,Sa and ,n 1,2,j m,1,2,i allfor converges a if

only and if SSmatrix n m the toconverge tosaid is A matrices. of series

infinitean called is AThen n.m orders of matrices of sequence a be ALet 1.

:Definition

. k as 0A-A ifonly and ifA toconverges A 1.

:Remark

k as 12k since matrix,any toconvergenot does

k

1sin

k!

1

12kk

1

A 2.

.A 21

10A ,

2k

1cos

1k

k

2

1

A 1.

:Example

1kijk

1kk

ij1k

ijk1k

ijk

nmij1k

k

1kk1kk

k1kk

k

k

1k

k

1kk

32A

A

1kk

1k1k24

1kk

1kk

1-1k1k

1k-

1k1k1-k

1k 1k-

1-k

1kk

A3!

1A

2!

1AIe

e Define matrix.n n beA Let

:Definition

diverges A Hence . diverges k

1 Since

k

1

2k

1

coskk

1

A 2.

o1-e

ee2

toconverges A

o1-e

ee2

0e-1

1

e21-1

1

0e

1)!-(k

1

2

1

0e1)!-(k

1

2

1A 1.

:Example

converges. k!

3 ,

k!

30

0ee 3.

e0

0ee 2.

e0

2ee

e01)!-(k

12e

k!

10

k!

2k

k!

1

2!

1

1!

110

4!

8

3!

6

2!

420

2!

1

1!

11

10

61

3!

1

10

41

2!

1

10

21

10

01

10

21

10

21

2!

1

10

21

10

01e 1.

:Examples

0k

k

0k

k30

01

10

01

1k

0k

0k0k

10

21

1-

0k

k A)-(I toconverges AThen

1Ah matrix wit symmetricn n a beA Let

5.5.1Corollary

A

A Hence

AA handother In the A have We

vector somefor A Since

:pf

. AThen A. of eigenvalueany be andmatrix n n symmetric a beA Let

5.5.3 Theorem

005

1

2

1

2

1A ,

005

1

2

1A 2.

2

30

02

3

10

02

1

A 00

00

3

10

02

1

limAlim

13

1

2

1A ,

3

10

02

1

A 1.

:Example

0Alim Hence

0AlimAlim have we1,A And

1,2,k ,AA Since

:proof

norm.matrix any is |||| where1,A if 0,AlimThen matrix.n nan beA Let

5.5.1 Theorem

0k

k

0k

k

0k

k

0k

k0k

k

0k

kk

k

k

k

k

22

2

k

k

k

k

k

k

kk

k

k

. 1,2,n ,aC where,Clim Find (v)

. lnalim Find (iv)

. blim Find (iii)

.convergent is b that Show (ii)

. alim Find (i)

. 1,2,i ,an

1b and

n

1cosaLet 2.

. blim Find a,a,amaxb Define

a)n

1

2sin( SequenceConsider 1.

5 Exercise

n

1n

1iinn

n

nn

nn

1nn

nn

n

1iin

1n1nn

nn

n21n

1nn1n

3

n-

3

n (f)

n1

n1- (e)

2

ncos

2

nsin (d)

3

nnsin (c) cosn

n

11 (b) cosn (a)

bygiven is a if ainf lim and asup lim Find 4.

. an

1C here w

, cClim that Show 0,Calim that Suppose (b)

. blim find and convergent is b that Show

. 1,2,n ,a,amaxb Define (a)

terms.positive of sequence bounded a is a that Suppose 3.

n

n

nnn

nn

n

1n

1iin

nn

nn

nn1nn

n1n

1nn

. n as ca

that show (a)

. 1,2,i 0, wherelim and calim that Suppose 7.

bounded. also islimit ialsubsequent its

all of Eset then thesequence, bounded a is a if that Show 6.

odd isn if3

even isn if2a (d)

1][2na (c) n)(1

n(-1)a (b)

n!

na (a)

by given is a if a

ainf lim and

a

asup lim Find 5.

n

1ii

n

1iii

i

n

1ii

nn

n

1nn

n

n

n

nn

n

n

n

n

nn

1n

nn

1n

n

. alim Find ,a3

)a3(1a and 3aLet 10.

limit. its find and converges a that Show

. 1,2,n ,)a(2a and 1aLet 9.

limit. its find and converges a that Show . 1,2n 0,b

b3a

)a(3baa and 1a wherea sequence thehave that weSuppose 8.

. n as converges n

ayet converge,not

does that a sequencce a of mplecounterexa a givingby hold

alwaysnot does (a)in case special theof converse that theShow (b)

nn

n

n1n1

1nn

21n1n1

1nn

2n

2nn

1n11nn

n

1ii

1nn

divergent. is a1

a that show ,

n

1aLet 13.

sequence. bounded anot is a ifeven trueis (a) that Show (b)

diverges. a1

a that show then sequence, bounded a is a If (a)

terms.positive of seriesdivergent a be aLet 12.

series. theof sum partialth -n theis S wheresequence, bounded a is

S ifonly and if converges a then ,1,2,i 0,a if that Show 11.

1n n

n

1n1nn

1nn

1n n

n1nn

1nn

n

1nn1n

nn

constant. fixed a isk where, n

1lim Find (b)

. 1n

1lim 1,

n

1lim that Show (a)

. n

1 ,

n

1 seriesConsider 15.

converges. S

a that deduce then

series, theof sum partialnth theis S where, 2,3,n ,S

1-

S

1

S

a

thatShow terms.positive of seriesdivergent a be aLet 14.

n1

kn

n1

2n

n1

n

1n2

1n

1n2n

n

nn1-n

2n

n

1nn

1n2

n

1n

2

n

1n

n2n

1nn

nn

n2

n

qpnn-3

n

nn

n

nn

nn

nn

nn1n

1nn

1)n(n

cosnx (d) x1)(n (c)

xn

10 (b) x

3

2n (a)

uniformly. converges series following theofeach for which x of values theDetermine 17.

1-na (j) n-n1a (i)

0qp ,n-n

1a (h) ena (g)

n

1nsina (f) n-n2na (e)

n

4(-1)a (d)

12n

1

2n642

1)-(2n531a (c)

)elog(1

n)log(1a (b) 1-na (a)

where,a series theof econvergencfor Test 16.

2

n as zero togonot doesproduct thisof nth term that theShow :Hint

divergent. isproduct that thisShow 0,1,n ,1n

(-1)a e wher

itself, with a ofproduct sCauchy' heConsider t 19.

. 12

11 exceeds b series that the whereas,

12

10 than less is a series the

of sum that theShow negative. oneby followed are termspositive two Where

. 6

1-

11

1

9

1

4

1-

7

1

5

1

2

1-

3

11by given

a ofent rearrangemcertain a be bLet , n

(-1)aConsider 18.

n

n

1nn

1nn

1nn

1nn

1nn

1n

1-n

1nn

. [0,1]on uniform convergentnot does h that show (b)

. [0,1]on uniform convergent are g and f that show (a)

. (x)(x)gf(x)h and

b; ,b

asay xration is x if

n

1b

irration 0, xifn

1

(x)g ),n

1x(1(x)fLet 22.

. (x)for expansion series a give , so If

)?,[1on abledifferenti (x) is (b)

. 0 where),[1on uniformly converges n

1 that show (a)

. (x)n

1 series heConsider t 21.

. )[0,on uniform is econvergenc not theor whether Determine (b)

. (x)flim Find (a)

. 0 x,nx1

nx(x)f where, (x)ffunction theof sequence heConsider t 20.

n

nn

nnn

nn

1nx

1nx

nn

2n1nn

Infinite Sequences and Series

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