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Infinite Sequences and Series In this chapter we shall study the theory of infinite sequences and series, and investigate their convergence. Examples ：. 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 。 - PowerPoint PPT Presentation
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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
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n, allfor aa if increasing monotone is}{a 2.
n. allfor k |a|such that
0kconstant a exists thereif bounded is}{a 1.
:Remark
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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
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n
N
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1NN
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n1nn1nn
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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
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13
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11
a ,}{a 2.
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00
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b ba ba that means b
alim
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.convergent is n
1~
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n
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. somefor b
alim that means b~a
:Remark
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n
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ll
. n as 12n
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12n1
a
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. n as 02
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. 1r ,r-1
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0n
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4.Consider
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,9
1,-1
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6
1,-1
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Consider 3.
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asup lim 0
a
ainf limBut
exist.t doesn' a
alim
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a
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a Since
nn
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n
m
1m
mm
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m
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432432n
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m as
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1aBut
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asup lim1
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a
asup lim have We
)2
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1
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21
a
a
0)3
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1
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a
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a Since
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1
2
1
3
1
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1
3
1aConsider
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1
even isn if2
1
aLet 3.
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1)(nlim
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alim Since
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:Example
n
2n
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n
n
1n
nn
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n
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n
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a
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a
alim
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:Remark
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alim If 2.
converges. a then 1,a
alim If 1.
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1
a
alim But
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,5
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1,
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31
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a
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aLet 1.
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n
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3322
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n
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n
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n
converges. a2 ifonly and if converges a then
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Test)on Condensati schy'5.2.10(Cau Theorem
alimainf limasup lim then exist, alim If 1.
Properties
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1
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1
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t)Cauchy tesor test 5.2.8(Root Theorem
1n2
n
1nn
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n
nn
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a28a4a2aaa2
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im
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1-m
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i
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1(
2
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. n
1aConsider 2.
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12
12a2
. n
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:Example
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n
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n
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n
0n2
n
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n
n
divergent. is a hence divergent, is b Since
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b
a
a ,
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a
a
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b
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n
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n
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n
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n
n
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n
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n
n
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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
1 if diverges and 1 if converges aThen
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
1 if diverges and 1 if converges aThen
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
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en!
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where,a series theof econvergencfor Test 16.
2
n as zero togonot doesproduct thisof nth term that theShow :Hint
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. [0,1]on uniform convergentnot does h that show (b)
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