Gio trnh GII TCH 3
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PHNG TRNH VI PHN V L THUYT CHUI BI 1. CHNG I. L THUYT CHUI
1. i cng v chui s nh ngha iu kin cn chui hi t
Cc tnh cht c bn
t vn : 1 1 1 11 22 4 8 2n
+ + + + + + =
C phi l c cng mi cc s hng ca v tri th thnh v phi? 1 + ( 1)+1 + ( 1) + .... = ? 1. Chui s: nh ngha: Vi mi s t nhin n, cho tng ng vi mt s thc an, ta c dy s k hiu l { }na . nh ngha:
Cho dy s {an}, ta gi tng v hn 1 2 3a a a+ + + l chui s, k hiu l 1
nn
a
=
,
an l s hng tng qut. Sn = a1 + a2 + a3 + ... + an l tng ring th n. Nu lim n
nS S
= th ta bo chui hi t,
c tng S v vit: 1
nn
a S
=
= .
Khi dy {Sn} phn k th ta bo chui 1
nn
a
=
phn k.
V d 1. Xt s hi t v tnh 0
n
n
q
=
1
2 11 , 11
nn
nqS q q q q
q
+
= + + + + = + + + + = +
Do Sn c th ln bao nhiu tu , nn c lim nn
S
=
Chui cho phn k
V d 4. Chui nghch o bnh phng: 21
1
n n
=
( )2 2 21 1 1 1 1 1 1 1 11 1 1
2.2 3.3 . 1.2 2.3 12 3nS
n n n nn= + + + + = + + + + < + + + +
1 1 1 1 1 1 1 1 11 2 21 2 2 3 3 4 1n n n
= + + + + + =
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1 1n
n
n
=+ phn k
V d 6. ( ) ( ) ( )1
1 1 1 1 1n
n
=
= + + + +
C ( ) chn l.
1lim 11
n
n
n
n
=
Khng tn ti ( )lim 1 nn
( )1
1 n
n
=
phn k.
V d 7. Tm tng (nu c) ca chui s sau ( )223 5 2 14 36 1
n
n n
++ + + +
+ (S: 1)
V d 8. 1
11
n
n
n
n
=
+ (PK)
Tnh cht. Gi s lim , limn nn n
a a b b
= =
( )lim n nn
a b a b
+ = + ( )lim .n n
na b a b
=
lim , 0.nn n
a a bb b
=
2. Chui s dng nh ngha Cc nh l so snh
Cc tiu chun hi t
1. nh ngha: 1
, 0n nn
a a
=
>
Nhn xt. 1
nn
a
=
hi t khi v ch khi Sn b chn.
Trong bi ny ta gi thit ch xt cc chui s dng 2. Cc nh l so snh. nh l 1. Cho hai chui s dng, n na b , n tu hoc t mt lc no tr i
1n
n
b
=
hi t 1
nn
a
=
hi t
1n
n
a
=
phn k 1
nn
b
=
phn k
PGS. TS. Nguyn Xun Tho Email: [email protected] Chng minh.
1 2 1 20
n n
n n
a a a b b bS T
+ + + < + + +
<
Rt ra cc khng nh.
V d 1. 1
13 1nn
=+
Chui dng 3 1 3
1 13 1 3
n n
n n
+ >
, n tu , chn m sao cho 2mn < , c
( ) ( )
( ) ( ) ( )
2 1 1
1
1 2 11 1 1
1
1 1 1 1 1 112 3 4 7 2 2 1
2 4 2 1 1 11 12 4 22 2 2
1 1 1, 0 1
1 1 2
mn p p p p p pm m
m
p p p p mm p p
m
p
S S
aa
a a
= + + + + + + + + +
+ + + + = + + + +
= < < = <
Dy Sn b chn trn 1
1p
n n
=
hi t.
KL: Chui hi t vi p > 1 v phn k vi 0 < p 1.
V d 6. 3
1
1
3n n
= +
Chui dng
3 3 / 23
1 133 1
nan n
n
= =
+ +
; 3 / 21
nbn
=
lim 1nn n
a
b=
1n
n
b
=
hi t
31
1
3n n
= + hi t
PGS. TS. Nguyn Xun Tho Email: [email protected] V d 7
a1) ( )2ln 1 2 1
n
n n
=
+ + (PK) a2) ( )2sin 1 1
n
n n
=
+ (PK)
b1) 21
sin2n
nn
=
pi
(PK); b2) ( )11
1 2 1nn
n
=
(HT)
c1) 5
1
cos
1n
n n
n
=
+
+ (HT) c2) 3
1
sin1n
n n
n
=
+
+ (PK)
d1) ( )2
2 1n
n n
=
+ (PK) d3) ( )12
1nn
n e
=
(PK)
d3)3 7 3
1
1sin
2 3n
n
n n
=
+
+ + (HT)
e) Xt s hi t 1)
=
4 51
ln
n
n
n (HT) 2)
+
11
arcsin lnnn
(PK)
3) pi
=
+
2 3
1ln 1 arctan
2nn
n (HT)
3) Cc tiu chun hi t a) Tiu chun DAlembert
1lim nn n
a la
+
=
Khi 1l < 1
nn
a
=
hi t
Khi 1l > 1
nn
a
=
phn k.
Chng minh l < 1: T 1lim n
n n
a la
+
= , chn > 0 b l + < 1 1n
n
a
a+
< l + , n n0.
Mt khc c 00
0
11
1 2. .
nn nn n
n n n
aa aa a
a a a+
= ( ) 00
n nnl a
+ 0, n
Do lim nn
a l
=
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l > 1: T 1lim nn n
a la
+
= , chn b l > 1 1 1n
n
a la
+ > > an + 1 > an
phn k Nhn xt. Khi l = 1 khng c kt lun g
V d 1. 1
1!
nn
=
1 0!n
an
= >
( ) ( )1 1 1 ! 1lim lim : lim lim 0 1
1 ! ! 1 ! 1n
n n n nn
a n
a n n n n+
= = = =
( )1
1 3 3 3:1 ! ! 1
n nn
n
a
a n n n
++
= =
+ +
1lim 0 1nn n
a
a+
= <
Chui cho hi t
V d 3. Xt s hi t, phn k ca chui ( )( )1.3.5 2 11 1.3 1.3.5
2 2.5 2.5.8 2.5.8 3 1n
n
+ + + +
( )( )
1.3.5 2 1 02.5.8 3 1n
na
n
= >
( ) ( )( )( )
( )( )
1
1
1.3.5 2 1 2 1 1.3.5 2 1 2 1:
2.5.8 3 1 3 2 2.5.8 3 1 3 22lim 13
n
n
n
n n
n n na n
a n n n n
a
a
+
+
+ += =
+ +
= 1
nn
a
=
phn k
Nhn xt. Nu l = 1, khng c kt lun g
V d 5. 1
2 13 2
n
n
n
n
=
+
2 1 03 2n
na
n
= > +
2 13 2
nn
na
n
=
+
2lim 13
nn
na
= <
Chui cho hi t
V d 6. Xt s hi t, phn k 2
1
1 n
n
n
n
=
+ (PK)
V d 7.
a1) 2 ln2
21
3 14 cos
n n
n
n n
n n
=
+ +
+ (HT) a2)
=
+ +
+
3 ln2
21
2 13 sin
n n
n
n n
n n (HT)
PGS. TS. Nguyn Xun Tho Email: [email protected]
a3)( )
2
21
5
2 1
n n
nnn
n
n
= + (HT)
b1) ( )4
1
23
n n
n
n
n
+
=
+
+ (HT) b2) ( )4
1
32
n n
n
n
n
+
=
+ + (PK)
c) ( )
= +
2
21
5
3 1
n n
nnn
n
n (HT)
c) Tiu chun tch phn C mi lin h hay khng gia:
( ) lim ( )b
ba a
f x dx f x dx
+=
v 1 1
limk
n nkn n
a a
= =
=
1 2 11 1
( ) ( )n n
nf x dx a a a a f x dx + + + + , + =lim ( ) 0x f x
Nu f(x) l hm dng gim vi mi x 1, f(n) = an, khi
1n
n
a
=
v 1
( )f x dx
cng hi t hoc cng phn k.
V d 8. 2
1ln
nn n
=
1( )ln
f xx x
= dng, gim vi 2x v c +
=lim ( ) 0x
f x
( ) ( ) ( ) ( )( )22 2
ln( ) lim lim ln ln lim ln ln ln ln2ln
bb
b b n
d xf x dx x bx
= = = =
1
( )f x dx+
phn k
2
1ln
nn n
=
phn k
Tng qut c th xt ( )21
ln pn n n
=
hi t ch khi p > 1.
Hnh 14.4
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V d 9. Chng minh rng: 1 1 11 ln22 3 4
+ + =
[ ] [ ]
21 1 1 1 1 1 1 1 1 11 12 3 4 2 1 2 3 2 1 2 4 2
1 1 1 1 1 1 1 1 1 1 1 11 2 1 12 3 2 2 4 2 2 3 2 2 3
1 1ln2 (1) ln (1) , lim 1 ln2
n
n
Sn n n n
n n n n
n o n o nn
= + + + = + + + + + +
= + + + + + + + = + + + + + + + +
= + + + + = + + +
=
vi
ln2 (1) ln2o n+ khi Mt khc ta c
( )
2 1 2
2 1 2
1
1
12 1
lim lim ln2
1ln2
n n
n nn
n
n
S Sn
S S
n
+
+
+
=
= ++
= =
=
V d 10. Tng t nhn c 1 1 1 1 1 31 ln2.3 2 5 7 4 2
+ + + + =
V d 11. Xt s hi t phn k ca chui s sau
a) ( )21
1ln
2nn
n
=+
(HT); b) ( )
( )21ln 1
3n
n
n
=
+
+ (HT) c) 2
2
ln3n
n
n
=
(HT)
Happy new year 2011 !Happy new year 2011 !Happy new year 2011 !Happy new year 2011 !
PGS. TS. Nguyn Xun Tho [email protected]
HAPPY NEW YEAR 2011HAPPY NEW YEAR 2011HAPPY NEW YEAR 2011HAPPY NEW YEAR 2011
PHNG TRNH VI PHN V L THUYT CHUI BI 2
3. Chui s vi s hng c du bt k Chui vi s hng c du bt k Chui an du
Tnh cht ca chui hi t tuyt i
1. t vn . 2. Chui vi s hng c du bt k
nh ngha:
=
1
nn
a c gi l hi t tuyt i
=
1
nn
a hi t. Chui
=
1
nn
a c gi
l bn hi t
=
1
nn
a phn k v
=
1
nn
a hi t.
nh l.
=
1
nn
a hi t 1
nn
a
=
hi t.
V d 1. Xt s hi t tuyt i ca chui s sau
a) ( ) +
=
2
21
12
n n
nn
n ; b)
=
21sin
n
n
c) ( )( )pi=
+1sin 2 3
n
n
(HTT) d)
=
31
sin
n
n
n (HTT)
Hng dn.
a) ( ) +
=
2
21
12
n n
nn
n
+) Xt
=
1 2
nn
n
+) +
=
PGS. TS. Nguyn Xun Tho [email protected] Nhn xt.
1/ Nu
=
1
nn
a phn k theo tiu chun DAlembert hoc Cauchy
=
1
nn
a phn k
2/
=
1
nn
a phn k
=
1
nn
a phn k (ng hay sai?)
3. Chui an du
nh ngha. ( )
=
> 11
1 , 0n n nn
a a c gi l chui an du
Ch . ( )
=
>1
1 , 0n n nn
a a cng c gi l chui an du.
nh l Leibnitz
Dy { }na gim, > 0na , lim 0nn
a
= ( )
=
11
1 n nn
a hi t v c ( ) 1 11
1 n nn
a a
=
Chng minh: +) = 2n m : C ( ) ( ) ( )
= + + + 2 1 2 3 4 2 1 2m m mS a a a a a a { }2mS tng ( ) ( ) ( )
=
PGS. TS. Nguyn Xun Tho [email protected]
i) ( )
=
+ 2
11
2 1n
n
n
n (PK)
k) ( )
=
+
+ 1112
nn
n
n
n (PK)
l) ( )
=
+ 1 2
1
11 lnn
n
n
n (HTT)
m) ( )
=
11
ln1 n
n
n
n (Bn HT)
o) ( ) ( )3 7 3
1
1 sin 2,
2 3n
n n
n n
=
+
+ + (HTT)
p) ( )
=
1
1ln
n
nn n
(Bn HT)
Hng dn.
b) +) ( )
=
1
1
1 n
n n l chui an du
+)
1n
gim v c
=
1lim 0n n
+) Hi t theo Leibnitz
+)
=
1
1
n n phn k bn hi t
d) +) ( )
=
1
1
16 5
n
n
n
n l chui an du
+)
=
1lim6 5 6n
n
n
=
1 6 5n
n
n phn k
+) ( )
1lim 16 5
n
n
n
n
+) ( )
=
1
16 5
n
n
n
n phn k.
4. Tnh cht ca chui hi t tuyt i
a)
=
=1
nn
a S chui s nhn c t chui ny bng cch i th t cc s hng
v nhm tu cc s hng cng hi t tuyt i v c tng S
b) Cho
=
=1
nn
a S ,
=
1
nn
a phn k c th thay i th t cc s hng ca n
chui thu c hi t v c tng l mt s bt k cho trc hoc tr nn phn k.
nh ngha. Cho
= =
1 1
,n nn n
a b , khi ta nh ngha php nhn chui:
= = =
=
1 1 1n n n
n n n
a b c , 11
n
n k n kk
c a b + =
=
c)
=
= 11
nn
a S ,
=
= 21
nn
b S
= =
= 1 2
1 1n n
n n
a b S S
V d 3.a) Xt s hi t ca tch cc chui s sau: 1
1
n n n
=
v 11
12nn
=
.
PGS. TS. Nguyn Xun Tho [email protected]
b) Xt s hi t ca chui s ( )
= =
+
+ 1 2
1 1
1 21 tan .ln1
nk
n k
n kn kk k
Hng dn.
a) +)
=
1
1
n n n hi t tuyt i
+)
=
11
12nn
hi t tuyt i
+)
= =
1
1 1
1 1.
2nn nn n hi t
HAVE A GOOD UNDERSTANDING!HAVE A GOOD UNDERSTANDING!HAVE A GOOD UNDERSTANDING!HAVE A GOOD UNDERSTANDING!
PGS. TS. Nguyn Xun Tho [email protected]
PHNG TRNH VI PHN V L THUYT CHUI BI 3
4. Chui hm s t vn . 1. Chui hm s hi t nh ngha: Cho dy hm s ( ){ }nu x xc nh trn X , ta nh ngha chui hm s
( ) ( ) ( )
=
+ + 1 21
nn
u x u x u x (1)
( )
=
1
nn
u x hi t ti 0x chui s ( )
=
01
nn
u x hi t
( )
=
1
nn
u x phn k ti 0x chui s ( )
=
01
nn
u x phn k
Tp cc im hi t ca (1) gi l tp hi t ca n. Tng ca chui hm s l hm s xc nh trong tp hi t ca n. V d 1. Tm tp hi t ca cc chui hm s sau
a)
=
11
n
n
x b)
=+
2 21
cos
n
nx
n x c)
=
1
1x
n n ( 1x > ) d)
=
1 !
n
n
x
n ( )
e) ( )( )
=
+
+
2
21
sin 2 43 1n
n x
n ( ) f) ( )
=
1 cos1
1 n n xn
e ( 2 22 2
k x kpi pipi pi + < < + )
g) ( )( )+
=
1
1
15 3
n
nnn n x
( 135
x > )
Hng dn.
a)
=
11
n
n
x
+) Xt chui s
=
101
n
n
x (2)
+) (2) hi t vi
PGS. TS. Nguyn Xun Tho [email protected] V d 2. Tm tp hi t ca cc chui hm s sau
a) 1) ( ) ( ) +
=
+
1 2 3
21
13 2 3
n n
nn
x
n ( 3 3x < )
2) ( )
=+ +
1
11 1 nn n x
( 0 2x x> )
3) ( )
=+ +
31
11 2 nn n x
( 1 3x x> )
b) 1) ( )
=
+
3
221
4 3
1
n
n
n x
xn ( 3 ;1
5
)
2) ( )
=
+
22
1 111
n n
n
x
xn ([ )0 ; + )
c) ( )( )
=
+
+ +
2
0
11 2
n
n
x x
n n (0 1x )
2. Chui hm s hi t u
nh ngha. ( )
=
1
nn
u x hi t u n ( )S x trn tp X > 0 b tu
( ) 0 n : ( ) > 0n n , ta c ( ) ( ) 0 b tu
( ) 0 n : ( ) > > 0p q n , ta c ( ) ( ) < ,p qS x S x x X .
Tiu chun Weierstrass. Nu c ( ) , ,n nu x a n x X v
=
1
nn
a hi t
( )
=
1
nn
u x hi t tuyt i v u trn X .
V d 3. Xt s hi t u ca chui hm ( )
=
+
1
2 21
1 n
n x n
+) ( )
+
1
2 2 21 1
,
n
xx n n
+)
=
21
1
n n hi t
+) Chui cho hi t tuyt i v u trn V d 4. Xt s hi t u ca chui hm
a)
=
+
2 21
sin,
n
nxx
n x (HT) b) [ ]
=
31
, 2 ; 22
n
nn
xx
n n (HT)
c)
=
1
cos,
3
nn
nxx (HT) d) ( ) ( )
=
21
11 , 1; 1
nn
n
xx
n (HT)
e)
=
+
5 21
,
1
n
nxx
n x (HT) f)
=
>1
, 0!
n
n
xx
n (HTK)
Hng dn.
PGS. TS. Nguyn Xun Tho [email protected]
b) +) 4 / 331
, 22
n
n
xx
n n n +)
=
4 / 31
1
n n hi t
+) Chui cho hi t u v hi t tuyt i trn [ ]2 ; 2 . V d 5. Xt s hi t u ca chui hm
a) 1)
=
+
1
21 0
sin ,1
n
n
xdxnx x
x (HT) 2)
=
+
1
21 0
cos ,1
n
n
xdxnx x
x (HT)
b) 1) [ ]
=
+ +
+ 11 2 1
, 1; 123
n
nn
n xx
x (HT)
2) [ ]
=
+ +
+ + 2
1
1 2 1, 1; 1
2 2
n n
n
n xx
n x (HT)
c) Chng minh rng chui hm
=
1
2x
nx
n
e hi t u vi 0x
d) 1) Chng minh rng chui ( )
=
+ + 2
0
11
n
n x n hi t u trn
2) Chng minh rng chui ( )
=
+ + 2
0
12
n
n x n hi t u trn
3. Tnh cht ca chui hm s hi t u
nh l 1. Chui ( )
=
1
nn
u x hi t u v ( )S x trn X , ( )nu x lin tc trn X , vi
n ( )S x lin tc trn X .
nh l 2. ( )
=
1
nn
u x hi t u n ( )S x trn [ ];a b , ( )nu x lin tc trn [ ];a b , n
( ) ( ) ( )
= =
= =
1 1
b b b
n nn na a a
S x dx u x dx u x dx
nh l 3. ( ) ( )
=
=1
nn
u x S x trn ( );a b , cc hm ( )nu x kh vi lin tc trn ( );a b ,
( )
=
1
nn
u x hi t u trn ( );a b ( )S x kh vi trn ( );a b v c
( ) ( ) ( )
= =
= =
1 1n n
n n
S x u x u x
PGS. TS. Nguyn Xun Tho [email protected] V d 6. Xt tnh kh vi ca cc hm sau
a) ( ) ( )
=
=
+11 n
n
xf xn x
; b) ( )
=
= 21arctan
n
xf xn
( ( )2
4 21
,
n
nf x xn x
=
= +
)
Hng dn. a) +) x n l chui an du hi t theo Leibnitz +) ( ) ( ) = + 2n
nu x
n x lin tc
=
1
, nn
x n u hi t u theo Dirichlet
+) ( ) ( ) ( )
=
= +
21
1 ,n
n
nf x x nn x
V d 7 a) Tm min hi t v tnh tng 1) ( ) ( )
+
=
+3 2
0
113 1
nn
n
x
n ((0 ; 2] ,
21 1 2 3( 1) ln arctan3 3 3 6 33 3
x xS xx x
pi = + +
+ )
2) ( ) ( )+
=
+
+3 2
0
113 1
nn
n
x
n (( 2 ; 0) ,
21 2 1 2 1( 1) ln arctan3 3 3 6 31
x xS xx x
pi+ + = + + +
+ + )
b) Tm min hi t v tnh tng 1) ( ) ( )
=
+1
1
1 1n
n
n
xn
; 2) ( ) ( )( )
=
+ 11
1 1 1n n
n
n x ((0 ; 2) , 2
21xS
x
= )
Hng dn. b1) Hi t vi +
PGS. TS. Nguyn Xun Tho [email protected] PHNG TRNH VI PHN V L THUYT CHUI
BI 4 5 Chui lu tha
nh ngha Cc tnh cht Khai trin thnh chui lu tha t vn 1. nh ngha. 20 1 2 nna a x a x a x+ + + + + (1)
K hiu l 0
nn
n
a x
=
, na l cc s thc, x l bin s.
Ta bo chui lu tha hi t (phn k) ti 0x chui s 00
nn
n
a x
=
hi t (phn k),
chui 0
nn
n
a x
=
hi t trn khong ( );a b chui s 00
nn
n
a x
=
hi t, 0x tu ( ; )a b .
V d 1. 20
1nn
x x x
=
= + + +
bit hi t khi 1x < , c 0
11
n
n
xx
=
=
Phn k khi 1x
nh l 1 (Abel). 0
nn
n
a x
=
hi t ti 0 0x hi t tuyt i ti 0:x x x<
Chng minh. +) 01
nn
n
a x
=
hi t 0lim 0nnn
a x
= 0 0,n
na x M n N
+) 0 00 0
n nn n
n nx x
a x a x Mx x
=
+) 0
1xx
< 01
n
n
xMx
=
hi t (nh l so snh 1) 0
nn
n
a x
=
hi t tuyt i
Nhn xt. T nh l Abel suy ra: Nu 0
nn
n
a x
=
phn k ti 0x phn k ti 0:x x x>
nh l 2. Nu 1lim nn n
a
a+
= (hoc lim n n
na
= ) th bn knh hi t R ca chui lu
tha 1
nn
n
a x
=
c xc nh bi 1
, 0
0,, 0
R
< <
= = + =