toán cc-chương 8 _ chuỗi số - chuỗi lũy thừa _.image.marked
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chuỗi sốTRANSCRIPT
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Chng 7: chui s chui lu tha
7.1. chui s
7.1.1. nh ngha chui s.
nh ngha 7.1. Dy s l mt tp hp gm v hn cc s thc c sp
xp theo mt quy lut no .
Thng thng, ngi ta k hiu dy s bi:
un = u1, u2, u3, ..., un, ...
trong uk c gi l s hng th k ca dy s (k = ,1 ), un c gi l
s hng tng qut ca dy s.
V d 7.1.
(i) Tp hp: 1, 3, 5, 7, 9 khng phi l dy s v n ch c 5 s hng.
(ii) Tp hp: 1, 3, 5, 7, 9,... l dy s v n c v hn s thc v c sp
xp theo quy lut s ng sau bng s ng ngay trc n cng vi 2. Dy
s ny c vit gn nh sau: 2n 1.
nh ngha 7.2. Cho dy s un. Tng tt c cc s hng ca dy s trn
(k hiu l nn
u
1
) c gi l mt chui s. Vy:
nn
u
1
= u1 u2 u3... un...
7.1.2. Cc loi chui s.
Chui s dng l mt chui m tt c cc s hng ca n u dng.
Chui s m l mt chui m tt c cc s hng ca n u m.
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Chui s an du l mt chui m hai s hng bt k ng cnh
nhau th c du ngc nhau.
Mt chui s khng phi l chui s dng, khng phi l chui s
m, khng phi l chui s an du th c gi l chui s bt k.
V d 7.2.
(i) nn
1
2 , n
n
1
3 1 l cc chui s dng.
(ii) nn
2 11
1 , n
n
1
1 2 l cc chui s m.
(iii) nn
1
1 , nn
n
1 21
1 l cc chui s an du.
(iiii) nn
cosn
2 11
1 , n
sinn
1
l cc chui s bt k.
7.1.3. S hi t ca chui s.
Cho chui s nn
u
1
= u1 u2 u3... un... Ta thnh lp dy cc tng
ring nh sau: S1 = u1, S2 = u1 u2, S3 = u1 u2 u3,..., Sn = u1 u2 u3...
un,...
nh ngha 7.3. Nu chui s nn
u
1
c nnlim S
tn ti, hu hn. Th chui
nn
u
1
c gi l hi t. Khi ,
nn
u
1
= nnlim S
=I,
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I c gi l tng ca chui s. Trong trng hp ngc li, chui nn
u
1
c gi l phn k.
Nhn xt 7.1. Chui nn
u
1
phn k khi nnlim S
khng tn ti hoc tn ti
nhng l s v hn.
V d 7.3. (i) Chui nn
1
1
2 hi t v c tng = 1 v:
Sn = n n... 1 1 1 1
12 4 2 2
n nn nlim S lim
11 1
2.
(ii) Chui n
n
1
phn k v:
Sn = n n
... n
1
1 22
nn n
n nlim S lim
1
2.
(iii) Chui nn
1
1 phn k v:
Sn = n khi n k...
khi n k
1 2 11 1 1 1
0 2 (k nguyn, dng).
nnlim S
khng tn ti.
Tnh cht 7.1.
(i) Nu nn
u
1
v nn
v
1
l cc chui hi t th n nn
u v
1
cng hi t.
(ii) Nu nhn tt c cc s hng ca mt chui s vi mt s khc khng th
khng lm thay i s hi t hay phn k ca chui s .
(iii) Nu thm vo hoc bt i mt s hu hn cc s hng ca chui s th
khng lm thay i s hi t hay phn k ca chui s .
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nh l 7.1 (Tiu chun Cauchy mt chui s hi t). iu kin cn v
chui s nn
u
1
hi t l: ( >0),(N > 0: m,n > N) Sm Sn< .
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H qu 7.1.1. iu kin cn chui s nn
u
1
hi t l nnlim u
0 .
Chng minh. Theo nh l 7.1, v chui s nn
u
1
hi t nn:
( >0),(N > 0: m = n1,n > N) Sm Sn< .
( >0),(N > 0: n1 > N) un< .
nnlim u
0 . (pcm)
Nhn xt 7.2. Nu mt chui s khng tho mn iu kin nnlim u
0
(ngha l gii hn trn khng tn ti hoc tn ti nhng l s khc 0) th
chui phn k. Tuy nhin, h qu 7.1.1 ch l iu kin cn nn mt
chui s tho mn iu kin nnlim u
0 , th cha kt lun c chui s
hi t hay phn k.
V d 7.4.
(i) Chui n
nn
1
1
2 1 phn k v: n
n n
nlim u lim
n
1 1
02 1 2
.
(ii) Chui nn
1
1 phn k v: nnlim u
khng tn ti.
Ch 7.1. Chng ta cng nhn cc kt qu sau:
(i) Chui s nn
q
1
(q l hng s) c gi l chui s nhn. Chui ny
hi t khi q< 1 v phn k khi q 1.
(ii) Chui s sn n
1
1 (s l hng s) c gi l chui Dirichlet. Chui
ny hi t khi s > 1 v phn k khi s 1.
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Nhn xt 7.3. i vi chui s Dirichlet v chui s nhn chng ta ch cn
nhn vo s hoc q l c th kt lun c chui hi t hay phn k.
Chng hn:
(i) nn
1
2 l chui s phn k v n l chui s nhn c q = 2 > 1.
(ii) nn
1
1
3l chui s hi t v n l chui s nhn c q =
1
3 < 1.
(iii) n n
3
1
1l chui s phn k v n l chui s Dirichlet c s =
1
3 < 1.
(iiii) n n
2
1
1l chui s hi t v n l chui s Dirichlet c s = 2 > 1.
7.2. S hi t ca chui s dng
Nhc li: Chui s nn
u
1
l chui s dng nu un > 0 (n =1, 2, 3,...).
Trong phn ny chng ta a ra cc iu kin mt chui s
dng hi t.
7.2.1. Du hiu so snh 1.
nh l 7.2. Cho hai chui s dng nn
u
1
, nn
v
1
; tn ti s c > 0 v tn ti
s N nguyn dng sao cho:
un c.vn (n > N).
Khi , (i) Nu chui s nn
v
1
hi t th chui s nn
u
1
hi t.
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(ii) Nu chui s nn
u
1
phn k th chui s nn
v
1
phn k.
(Kt qu trn vn ng cho chui s khng m).
Nhn xt 7.4. p dng c du hiu so snh 1 xt s hi t ca mt
chui s ta phi tin hnh qua cc bc nh sau:
Bc 1: Kim tra tnh dng ca chui s cho.
Bc 2: a ra chui s th hai nn
v
1
tho mn cc iu kin: l
chui s dng ; bit hi t hay phn k ri; so snh c vi chui s
cho.
a ra chui s th hai ta phi da vo chui s cho v chui
s nhn (hoc chui s Dirichlet).
V d 7.5. Xt s hi t ca cc chui: a) n n
2
1
1
3 1, b)
n
nn
1
5
3 2.
Gii. a) n n
2
1
1
3 1. Ta c un = n 2
1
3 1 > 0 (n =1, 2, 3,...) vy chui
cho l chui dng.
Chui s nn
v
1
= n n
2
1
1 l chui dng v l chui Dirichlet hi t v
s = 2 > 1. Mt khc:
un = n 21
3 1 <
n
2
1 1
33vn (n =1, 2, 3,...).
Theo du hiu so snh 1, chui cho hi t.
b)n
nn
1
5
3 2. Ta c un=
n
n 5
3 2>0 (n=1,2,...) vy chui cho l chui dng.
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Chui s nn
v
1
= n
n
1
5
3 l chui dng v l chui nhn phn k v
q > 1. Mt khc: un = nn
n
5 5
3 2 3 vn (n =1, 2, 3,...).
Theo du hiu so snh 1, chui cho phn k.
Nhn xt 7.5. Nu trong v d 7.5, phn a) mu s 3n2 1 c thay bi
3n2 k hoc phn b) mu s 3n 2 c thay bi 3n k (vi k l hng s
dng). Th chng ta khng th p dng du hiu so snh 1 c. khc
phc, sau y chng ta a ra du hiu so snh 2.
7.2.2. Du hiu so snh 2.
nh l 7.3. Cho hai chui s dng nn
u
1
, nn
v
1
c nn n
ulim
v= k 0, hu
hn. Th hai chui s trn cng hi t hoc cng phn k.
Chng minh.
T gi thit ca nh l ta suy ra k > 0. Theo nh ngha gii hn ta c:
n
n n
ulim
v= k ( > 0), (N > 0: n > N) k < n
n
uv
< k (7.1)
V (7.1) ng vi mi > 0 nn cng ng vi 0 =k2
> 0. Ngha l tn
ti N0 > 0 sao cho vi mi n > N0 th:
k2
vn = (k k2
)vn < un < (k k2
)vn = k32
vn . (7.2)
(i) Nu nn
u
1
hi t nn N
u
0
hi t. T (7.2) ta c: k2
vn < un ( n > N0). p
dng du hiu so snh 1 ta c nn N
v
0
hi t nn
v
1
hi t.
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(ii) Nu nn
u
1
phn k nn N
u
0
phn k. T (7.2) ta c: un < k32
vn ( n
>N0). p dng du hiu so snh 1 ta c nn N
v
0
phn k nn
v
1
phn k.
(iii) Nu nn
v
1
hi t nn N
v
0
hi t. T (7.2) ta c: un < k32
vn ( n >N0).
p dng du hiu so snh 1 ta c nn N
u
0
hi t nn
u
1
hi t.
(iiii) Nu nn
v
1
phn k nn N
v
0
phn k. T (7.2) ta c: k2
vn < un (n >
N0). p dng du hiu so snh 1 ta c nn N
u
0
phn k nn
u
1
phn k.
T (i) n (iiii) (pcm)
Nhn xt 7.6. p dng c du hiu so snh 2 xt s hi t ca chui
s chng ta cng phi chi qua cc bc nh trong nhn xt 7.4.
V d 7.6. Xt s hi t ca cc chui: a) n n
2
1
1
3 2, b)
n
nn
1
5
3 2.
Gii. a) n n
2
1
1
3 2. Ta c un = n 2
1
3 2> 0 (n =1, 2, 3,...) vy chui
cho l chui dng.
Chui s nn
v
1
= n n
2
1
1 l chui dng v l chui Dirichlet hi t v
s = 2 > 1. Mt khc:
n
n nn
u nlim lim
v n
2
23 2=
1
3 0, hu hn.
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Theo du hiu so snh 2, chui cho hi t.
b) n
nn
1
5
3 2.
Ta c un = n
n 5
3 2> 0 (n =1, 2, 3,...) vy chui cho l chui dng.
Chui s nn
v
1
= n
n
1
5
3 l chui dng v l chui nhn phn k v
q > 1. Mt khc:
n nn
n nn nn
ulim lim
v
5 3
3 2 5= 1 0, hu hn.
Theo du hiu so snh 2, chui cho phn k.
Ch 7.2. Trong du hiu so snh 2, ta mi chng minh c cho trng
hp nn n
ulim
v = k 0, hu hn. Nu k = 0 hoc k = th du hiu so snh 2
cn ng na khng ? Sau y chng ta xt c th cho tng trng hp.
(i) Nu k = 0 nn n
ulim
v= k
( > 0), (N > 0: n > N) < nn
uv
< (7.3)
V (7.3) ng vi mi > 0 nn cng ng vi 0 = 2 > 0. Ngha l tn
ti N0 > 0 sao cho vi mi n > N0 th:
2vn < un < 2vn. (7.4)
Nu nn
v
1
hi t nn N
v
0
hi t. T (7.4) ta c: un < 2vn ( n >N0).
p dng du hiu so snh 1 ta c nn N
u
0
hi t nn
u
1
hi t.
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Nu nn
u
1
phn k nn N
u
0
phn k. T (7.4) ta c: un < 2vn ( n
>N0). p dng du hiu so snh 1 ta c nn N
v
0
phn k nn
v
1
phn k.
Nh vy, nu k = 0. Th du hiu so snh 2 khng ng na m ch
c th kt lun nh sau:
Nu k = 0 th t chui nn
v
1
hi t nn
u
1
hi t.
Nu k = 0 th t chui nn
u
1
phn k nn
v
1
phn k.
(ii) Nu k = nn n
ulim
v=
(M > 0), (N > 0: n > N) nn
uv
> M (7.5)
V (7.5) ng vi mi M > 0 nn cng ng vi M0 = 20 > 0. Ngha l
tn ti N0 > 0 sao cho vi mi n > N0 th:
un > 20vn. (7.6)
Nu nn
v
1
phn k nn N
v
0
phn k. T (7.6) ta c: un> 20vn ( n
>N0). p dng du hiu so snh 1 ta c nn N
u
0
phn k nn
u
1
phn k.
Nu nn
u
1
hi t nn N
u
0
hi t. T (7.6) ta c: un< 20vn ( n > N0).
p dng du hiu so snh 1 ta c nn N
v
0
hi t nn
v
1
hi t.
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Nh vy, nu k = . Th du hiu so snh 2 khng ng na m ch
c th kt lun nh sau:
Nu nn
u
1
hi t nn
v
1
hi t.
Nu nn
v
1
phn k nn
u
1
phn k.
Nhn xt 7.7. Mun p dng cc du hiu so snh1 v du hiu so snh 2
xt s hi t ca chui s dng. Chng ta phi a ra c chui
dng th hai bit hi t hay phn k ri v so snh c vi chui
cho. Tuy nhin, vic a ra chui th hai tho mn cc iu kin nh trn
khng phi trng hp no cng thun li. khc phc sau y chng ta
a ra hai iu kin khc (thun li hn) xt s hi t ca chui s
dng.
7.2.3. Du hiu DAlembert.
nh l 7.5. Cho chui s dng nn
u
1
c nn n
ulim
u
1 = k. Khi :
(i) Nu k < 1. Th chui nn
u
1
hi t.
(ii) Nu k > 1. Th chui nn
u
1
phn k.
(iii) Nu k = 1. Th cha kt lun c v s hi t hay phn k ca
chui nn
u
1
.
V d 7.7. Xt s hi t ca cc chui:
a) n
n n
2
1
2
3 2, b) n
n
n
1
5 13
3, c)
n
nn
2
1 3 2.
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Gii. a) n
n n
2
1
2
3 2. Ta c un =
nn 23 2
> 0 (n =1, 2, 3,...) vy chui
cho l chui dng ; un+1 = n
n n
1
2
2
3 6 1.
k =
n
nnn nn
nulim lim
u n n
1 2
12
2 3 22
2 3 6 1> 1.
Theo du hiu DAlembert, chui cho phn k.
b) nn
n
1
5 13
3. Ta c un = n
n 5 133
> 0 (n =3, 4, 5,...), u1 = 8
3 < 0 vy
chui cho khng phi l chui dng. Nhng chui nn
n
3
5 13
3 l chui
dng ; un+1 = nn
1
5 8
3. k =
nn
nn nn
nulim lim
u n
1
1
3 5 8 1
33 5 13 < 1. Theo
du hiu DAlembert, chui nn
n
3
5 13
3 hi t. Theo tnh cht v s hi t
ca chui s th chui cho hi t.
c) n
nn
2
1 3 2. Ta c un =
nn 23 2
> 0 (n =1, 2, 3,...), vy chui cho l
chui dng ; un+1 = n
n n
21
3 6 1. k =
n
n nn
n nulim lim
u n n n
2
12
3 2 11
3 6 1.
Theo du hiu DAlembert, cha kt lun c s hi t hay phn k
ca chui cho. Mt khc, chui nn
v
1
= n n
1
1 l chui s dng v l
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chui Dirichlet phn k. Ta li c:
n
n nn
u nlim lim
v n
2
2
10
33 2, hu
hn. Theo du hiu so snh 2, chui cho phn k.
7.2.4. Du hiu Cauchy.
nh l 7.6. Cho chui s dng nn
u
1
c n nnlim u
= k. Khi :
(i) Nu k < 1. Th chui nn
u
1
hi t.
(ii) Nu k > 1. Th chui nn
u
1
phn k.
(iii) Nu k = 1. Th cha kt lun c v s hi t ca chui nn
u
1
.
V d 7.8. Xt s hi t ca cc chui: a) n
n n
2
1
11 , b)
n
n
nn
1 3 1
.
Gii. a) n
n n
2
1
11 . Ta c un =
n
n
2
11 > 0 (n =1, 2, 3,...), vy chui
cho l chui dng k = n
nn
n nlim u lim e
n
11 > 1.
Theo du hiu Cauchy, chui cho phn k.
b) n
n
nn
1 3 1
. Ta c un = n
nn
3 1
> 0 (n =1, 2, 3,...), vy chui cho
l chui dng k = n nn n
nlim u lim
n
1
3 1 3 < 1.
Theo du hiu Cauchy, chui cho hi t.
Nhn xt 7.8.
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(i) Cc du hiu so snh 1, so snh 2, du hiu DAlembert v du hiu
Cauchy p dng c xt s hi t ca chui s m bng cch nhn tt
c cc s hng ca chui s m vi (1).
(ii) Trn y chng ta trnh by cc iu kin mt chui s dng
hi t hay phn k, theo th t: du hiu so snh 1, so snh 2, du hiu
DAlembert v du hiu Cauchy. Tuy nhin, khi p dng chng ta nn i
theo chu trnh ngc li. Tc l, s dng cc du hiu DAlembert v
Cauchy trc, nu khng c (k =1) th s dng du hiu so snh 2, cng
khng c th s dng du hiu so snh 1 nu khng c na th s
dng iu kin cn chui s hi t. V d sau y khng nh iu .
V d 7.9. Xt s hi t ca chui n
nn n e
2
1
1 11 .
Gii. D dng kim tra c chui trn khng p dng c cc du hiu
DAlembert v Cauchy (v trong c hai trng hp ta lun tm c k
=1). Vic s dng cc du hiu so snh 1 v so snh 2 xt s hi t ca
chui s ny gp rt nhiu kh khn. Mt khc, nnlim u e
12 0 v:
tx t
x tln t tx limlim x x ln lim t tx tn
n xlim u lim e e e e
e x
2 00
1 11 11 1 12 1 21 11 .
Nn chui cho phn k.
7.3. S hi t ca chui s an du
7.3.1. nh ngha chui s an du.
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nh ngha 7.4. Chui s an du l mt chui s c mt trong cc dng
sau: n nn
u
1
1 hoc n nn
u
11
1 , trong un > 0 (n = 1, 2, 3,...).
V d 7.9.
(i) Cc chui s sau y l cc chui s an du: n
n n
1
1,
n
n n
1
21
1
2.
(ii) Chui s n
n n
2
1
1
2khng phi l chui an du v u1 = 1< 0, u2 =
1
2> 0.
Nhn xt 7.9.
(i) S hng tng qut ca n nn
u
1
1 l (1)nun ch khng phi l un .
(ii) S hng tng qut ca n nn
u
11
1 l (1)n+1un ch khng phi l un .
7.3.2. S hi t ca chui s an du.
nh l 7.7 (nh l Leibnitz). Nu chui s an du n nn
u
11
1 tho
mn cc iu kin sau:
(i) u1 u2 u3 ... un ...;
(ii) nnlim u
= 0.
Th chui an du trn hi t v c tng u1.
Chng minh.
Vi mi k= 1,2,3,... ta c: ukuk+1 0 (v u1 u2 u3 ... un ...) nn
S2k = u1 u2 u3 u4 ... u2k u2k
= (u1 u2) (u3 u4) ... (u2k-1 u2k). (7.7)
= u1 (u2 u3) ... (u2k-2 u2k-1) u2k u1. (7.8)
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17
T (7.7) v (7.8) suy ra S2k l dy khng gim v b chn trn bi
u1 khi k . Theo tiu chun tn ti gii hn th hai th tn ti
kklim S
2 = I u1. (7.9)
Mt khc, v nnlim u
= 0 nn kklim u
2 1 = 0. M S2k+1= S2k u2k+1. Do
:
kklim S
2 1 = kklim S
2 kklim u
2 1 = I 0 = I u1. (7.10)
T (7.9) v (7.10) suy ra nnlim S
= I u1. (pcm)
V d 7.10. Xt s hi t ca chui s n
n n
1
1
1
3.
Gii. Chui s cho l chui s an du vi un = n 1
3 v ta c:
u1 = 1
4 > u2 =
1
5 > u3 =
1
6 > ....; n
n nlim u lim
n
1
3 = 0.
Theo nh l Leibniz, chui s cho hi t v c tng 1
4.
Ch 7.3.
(i) nh l Leibnitz pht biu cho chui an du dng n nn
u
11
1 v l
iu kin chui n nn
u
11
1 hi t. i vi chui an du dng
n nn
u
1
1 chng ta ch p dng c nh l Leibnitz sau khi nhn tt c
cc s hng ca n nn
u
1
1 vi (1), do ch kt lun uc s hi t ca
n nn
u
1
1 m khng kt lun c n nn
u
1
1 c tng u1.
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(ii) i vi chui an du dng n nn
u
11
1 m cc gi thit ca nh l
Leibnitz ch ng khi n > N (vi N l s nguyn dng no ). Th vn kt
lun c s hi t ca chui an du nhng khng kt lun c chui
c tng u1.
V d 7.11. Xt s hi t ca cc chui s (i) n
n n
1
1, (ii)
n
n
n
n
2
1
1
16.
Gii. (i) Chui s cho l chui s an du vi un = n1
. Nhn tt c cc
s hng ca chui s cho vi (1) ta c chui s mi: n
n n
1
1
1 v ta
c:
u1 = 1 > u2 = 1
2 > u3 =
1
3 > ....; n
n nlim u lim
n
1 = 0.
Theo nh l Leibnitz, chui s n
n n
1
1
1 hi t v c tng 1. Theo
tnh cht v s hi t ca chui s suy ra chui s cho hi t.
(ii) Chui s cho l chui s an du vi un = n
n 2 16. Nhn tt c cc s
hng ca chui s cho vi (1) ta c chui s mi: n
n
n
n
1
21
1
16 v ta
c:
nn n
nlim u lim
n
20
16, un+1 un =
n n
n n
2
2 2
160
1 16 16 (n 4).
-
19
Xy dng chui n nn
v
11
1 nh sau: vn = n
u khi n , , , ;u khi n .
4 1 2 3 4
4 Th
ba chui n
n
n
n
2
1
1
16,
n
n
n
n
1
21
1
16 v n n
n
v
11
1 cng hi t hoc cng phn
k (tnh cht v s hi t ca chui s), m n nn
v
11
1 l chui an du
tho mn nh l Leibnitz nn chui n
n
n
n
2
1
1
16 hi t.
Nhn xt 7.10. (i) Nu chui s an du c iu kin (ii) ca nh l
Leibnitz khng tho mn (tc l, nnlim u
khng tn ti hoc tn ti nhng
bng k 0). Th chui s an du phn k v khng tho mn iu kin
cn mt chui s hi t.
(ii) Nu chui s an du c iu kin (i) ca nh l Leibnitz khng
tho mn v iu kin (ii) ca nh l tho mn. Th ta cha th kt lun
g v s hi t hay phn k ca chui s an du , v nh l Leibnitz ch
l mt iu kin m khng phi l iu kin cn chui s an du
hi t. V d sau y s khng nh iu .
V d 7.12. (a) Cho chui s n
n n
1
1
1. y l chui s an du hi t. Ta
c chui s mi: 1 8 31 1
3002 3
... l chui hi t tho mn iu
kin (ii) nhng khng tho mn iu kin (i) ca nh l Leibnitz.
(b) Chui: n n
... ...
2 3 4 2 1 2
1 1 1 1 1 1
2 3 2 3 32 l chui s an du hi
t (tng ca hai chui hi t) v l chui tho mn iu kin (ii) nhng
-
20
khng tho mn iu kin (i) ca nh l Leibnitz.
(c) Xt chui s c cho bi: u2k-1= k
k
2 21
2 1, u2k=
k
k
2 11
2 (k = 1,2,3,
...). Chui c th nh sau: 1 k k
... ...kk
2 2 2 11 11 1 1 1
2 4 23 5 2 1
= k
k kk k
1
2 2 1
2 2 1.
y l chui s an du phn k tho mn iu kin (ii) nhng
khng tho mn iu kin (i) ca nh l Leibnitz.
7.4. Chui s bt k
7.4.1. Chui s tr tuyt i.
nh ngha 7.5. Cho nn
u
1
l chui s bt k th chui nn
u
1
c gi l
chui s tr tuyt i ca chui s nn
u
1
.
Nu chui nn
u
1
hi t cn chui nn
u
1
phn k th chui nn
u
1
c
gi l chui bn hi t.
Nu c hai chui nn
u
1
v nn
u
1
u hi t th chui nn
u
1
c gi l
chui hi t tuyt i.
V d 7.13. D dng kim tra c cc kt qu sau:
-
21
Chui s n
n n
1
1 l chui bn hi t; Chui s
n
n n
2
1
1 l chui
hi t tuyt i.
7.4.2. Tnh cht.
nh l 7.8. Nu chui nn
u
1
hi t th chui nn
u
1
cng hi t.
Nhn xt 7.11.
kim tra mt chui s c hi t tuyt i hay khng ta ch cn
kim tra s hi t ca chui tr tuyt i ca n.
7.5. Chui lu tha
7.5.1. Chui hm.
nh ngha 7.6. Cho fn(x) l dy cc hm s xc nh trn min X . Th
tng tt c cc hm s ca dy hm s trn c gi l mt chui hm s
trn min X, k hiu l nn
f x
1
. Vy: n nn
f x f x f x ... f x ...
1 21
V d 7.14. n
x x xln ln x ln ... ln ...
n n
1 2
l mt chui hm trn (0;).
n
sinnx sin x sin x ... sinnx ...
1
2 l mt chui hm trn
(;).
nh ngha 7.7. Cho chui nn
f x
1
cc hm xc nh trn X, x0 X. Th
nn
f x
0
1
l mt chui s.Nu chui s ny hi t th chui hm
-
22
nn
f x
1
c gi l hi t ti x0 v im x0 c gi l im t ca chui
hm nn
f x
1
. Tp hp tt c cc im t ca mt chui hm c gi l
min hi t ca chui hm .
V d 7.15. Chui nn
x
1
c min hi t l (1;1). Chui n
sinnx
1
hi t ti
cc im k vi mi k = 1, 2, 3,...
7.5.2. Chui lu tha.
nh ngha 7.8. Chui lu tha l mt chui hm s c mt trong cc
dng sau:
nn
n
a x
1
(1), nnn
a x x
01
(2),
trong x0, an (n =1, 2, 3,...) l cc s thc.
V d 7.16. nn
x
1
l chui lu tha vi an = 1 ( n =1, 2, 3,...).
n
n
xn
1
2
3 1l chui lu tha vi an = n
1
3 1 (n =1, 2, 3,...).
Nhn xt 7.12.
(i) Chui lu tha dng (1) lun hi t ti x = 0, chui lu tha dng
(2) lun hi t ti x = x0.
(ii) Chui lu tha dng (2) lun a c v dng (1) bng cch t
y = x x0. V vy xt s hi t ca chui lu tha chng ta ch cn xt s
hi t ca chui lu tha dng (1).
-
23
7.6. S hi t ca chui lu tha
7.6.1. nh l Abel.
nh l 7.9. Nu chui lu tha nnn
a x
1
hi t ti x0 0 th chui lu tha
hi t tuyt i ti mi x mx 0 ,N0> 0: n > N0 n
na x0 M.
Ti x = 0 chui nnn
a x
1
hi t v c tng bng 0.
Vi x 0, ta c: anxn =
n nn
nx x
a x Mx x
00 0
( n > N0).
M n
n
xx
1 0
l dng v l chui nhn hi t khi x x0. Vy theo du
hiu so snh 1 ca chui s dng, vi mi x m x x0 thn
nn
n N
a x
0 1
hi
t. Theo tnh cht v s hi t ca chui s th chui n
nn
n
a x
1
hi t. Do
, chui nnn
a x
1
hi t tuyt i khi x x0.(pcm)
H qu 7.9.1. Nu chui lu tha nnn
a x
1
phn k ti x1 0 th chui lu
tha phn k ti mi x mx> x1.
-
24
Chng minh.
Gi s tn ti x0 sao cho: x0> x1v chui n
nn
a x
1
hi t ti x0. Th
x1 0, theo nh l Abel chui n
nn
a x
1
hi t tuyt i khi x x0. Do
chui hi t ti x1. Tri vi gi thit, suy ra (pcm)
Nhn xt 7.13. i vi chui lu tha m bng cch no chng ta tm
c hai im x0, x1 0 (x0 < x1); ti x0 chui hi t, ti x1 chui phn k.
Th kt lun c chui hi t trn (x0; x0), phn k trn (;x1)
(x1;), cn trn [x1; x0] v [x0;x1] phi xt ring.
Vn t ra l vi mt chui lu liu c hay khng mt s a > 0 sao
cho chui hi t khi x < a v phn k khi x > a?
Ngi ta chng minh c vi mi chui lu tha lun tm c
mt s r 0 sao cho chui hi t khi x < r v phn k khi x > r.
nh ngha 7.8. Nu tn ti s r 0 sao cho chui lu tha nnn
a x
1
hi t
khi x < r v phn k khi x > r (khi r > 0). Th r c gi l bn knh hi
t ca chui lu tha nnn
a x
1
.
7.6.2. Phng php tm bn knh hi t ca chui lu tha.
Da vo cc du hiu D Alembert v Cauchy v s hi t ca chui
s dng ngi ta chng minh c nh l sau:
nh l 7.10. Nu chui lu tha nnn
a x
1
c nn
n
alim k
a
1 hoc
nnn
lim a k
. Th bn knh hi t r ca chui lu tha c tnh theo
-
25
cng thc:
r =
khi k ;
khi k ;
khi k .k
0
0
10
V d 7.16. Tm min hi t ca cc chui sau:
a) n
n
xn
1 3 1, b)
n
n
x
n
2
1
1, c)
nn
n
nx
n
2
1
12
2 1.
Gii. a) n
n
xn
1 3 1. y l chui lu tha vi an = n
1
3 1> 0 ( n = 1, 2,
3,...) v: k = nn n
n
a nlim lim
a n
1 3 2 1
3 1. Vy bn knh hi t ca chui
cho l r = 1. Hay chui cho hi t khi x < 1 v phn k khi x > 1.
Ti x = 1, ta c chui s: n n
1
1
3 1. y l chui s dng phn k (so
snh vi chui n n
1
1). Vy chui lu tha cho phn k ti x = 1.
Ti x = 1, ta c chui s: n
n n
1
1
3 1. y l chui s an du hi t
(s dng nh l Leibnitz). Vy chui lu tha cho hi t ti x = 1.
Vy min hi t ca chui lu tha cho l: [1; 1).
b) n
n
x
n
2
1
1. y l chui lu tha vi an = n2
1> 0 ( n = 1, 2, 3,...) v:
k = n
n nn
nalim lim
a n
2
12
11 .
-
26
Vy bn knh hi t ca chui cho l r = 1. Hay chui cho hi t
khi x 1 < 1 v phn k khi x 1 > 1.
Ti x 1 = 1, ta c chui s: n n
2
1
1. y l chui s dng v l chui
Dirichlet hi t. Vy chui lu tha cho hi t ti x 1 = 1.
Ti x 1 = 1, ta c chui s: n
n n
2
1
1. C
n
n n
2
1
1=
n n
2
1
1 hi t
nn chui n
n n
2
1
1 hi t. Vy min hi t ca chui lu tha cho l:
x 1[1; 1] x [2; 0].
c) n
n
n
nx
n
2
1
12
2 1. y l chui lu tha vi an =
nnn
1
2 1> 0 ( n =
1, 2, 3,...) v: k = n nn n
nlim a lim
n
1 1
2 1 2.
Vy bn knh hi t ca chui cho l r = 2. Hay chui cho hi t
khi (x 2)2 < 2 v phn k khi (x 2)2 > 2.
Ti (x 2)2 = 2, ta c chui s: n
n
nn
1
1
2 1. y l chui s dng v l
chui hi t (s dng du hiu Cauchy). Vy chui lu tha cho hi t
ti (x 2)2 = 2.
Ti (x2)2=2 v nghim. Vy min hi t ca chui lu tha cho l:
0 (x 2)2 2 2 x 2 2 2 2 x 2 2 .
V d 7.17. Tm min hi t ca cc chui sau:
a) nn
n! x
1
, b) n
n
xn !
1
1, c) n
n
nx
n
3
1
2 132
2.
-
27
Gii. a) nn
n! x
1
l chui lu tha vi an = n! > 0 ( n = 1, 2, 3,...) v:
k = n
n n nn
n !alim lim lim n
a n!
1
11 .
Vy bn knh hi t ca chui cho l r = 0 hay chui lu tha
cho hi t ti duy nht im x = 0.
b) n
n
xn !
1
1 l chui lu tha vi an = n!
1 > 0 ( n = 1, 2, 3,...) v:
k =
n
n n nn
a n!lim lim lim
a n ! n
1 1 0
1 1.
Vy bn knh hi t ca chui cho l r = hay chui lu tha cho
hi t ti mi im.
c) nn
nx
n
3
1
2 132
2 l chui lu tha vi an =
nn
32 13
2 > 0 ( n = 7, 8, 9,...)
v: k =
n
n nn
n nalim lim
a n n
3
13
2 2 131
1 2 2 11.
Vy bn knh hi t ca chui cho l r = 1. Hay chui cho hi t
khi x 2 < 1 v phn k khi x 2 > 1.
Ti x 2 = 1, ta c chui s: n
nn
31
2 13
2. y khng phi l chui s
dng, nhng n
nn
37
2 13
2 l chui s dng hi t (so snh vi
n n
2
7
1). Nn
n
nn
31
2 13
2 l chui s hi t . Vy chui lu tha cho hi t ti x 2 =
1.
-
28
Ti x 2 = 1, ta c chui s: nn
nn
3
1
2 131
2. y khng phi l
chui s an du v s hng th 6 bng 1
218 v s hng th 7 bng
1
345,
nhng chui nn
nn
3
7
2 131
2 l chui s an du hi t (theo Leibniz). Vy
chui nn
nn
3
1
2 131
2 hi t hay chui lu tha cho hi t ti x 2 = 1.
Vy min hi t ca chui lu tha cho l: [3; 1].
Nhn xt 7.14. Cho chui lu tha nnn
a x
7
.
Ti x = 0 chui hi t v c tng bng 0.
Ti x 0. Nu tn ti s N0 nguyn, dng sao cho an 0 ( n > N0) v
tn ti gii hn:
n
n nnn n
nn
a x alim x lim
aa x
11 1 = k(x), (7.11)
hoc n nn n nn nlim a x x lim a
= k(x). (7.12)
Th ta c th s dng du hiu DAlembert hoc du hiu Cauchy i
vi chui s dng tm min hi t ca lu tha cho nh sau:
Vi x 0. V an 0 (n>N0) nn chui n
nn N
a x
0 1
l chui s dng.
Nu gii hn (7.11) [hoc (7.12)] tn ti th theo du hiu DAlembert
(hoc du hiu Cauchy) chui nnn N
a x
0 1
hi t khi k(x) < 1 v phn k
khi k(x) > 1. T c th suy ra c s hi t ca chui lu tha cho.
-
29
V d 7.18. Tm min hi t ca chui: nnn
nx
1
2 1
3.
Gii. Chui cho l chui lu tha vi an = nn 2 13
> 0 ( n = 1, 2, 3, ...)
chui hi t ti x = 0 v c tng bng 0.
Ti mi x 0 ta c: nnn
nx
1
2 1
3 l chui s dng v
k(x) =
nnn n
n nn n nnn
xna x alim x lim x lim
aa x n
11 1
1
2 1 3
33 2 1.
Theo du hiu DAlembert chui nnn
nx
1
2 1
3 hi t khi
x 1
3, phn
k khi x 1
3. Do , chui lu tha cho hi t tuyt i khi x (3; 3).
Vi mi x > 3 chui lu tha cho l chui s dng c k(x) > 1.
Theo du hiu DAlembert chui lu tha cho phn k.
Vi mi x < 3 chui lu tha cho l chui s an du v vit c
di dng: nnn
nx
1
2 1
3 = n nn
n
nx
1
2 11
3 = n n
n
u x
1
1 , vi
un(x) = n n
n n
n nx
2 1 2 13
3 3 = 2n 1 nn nlim u x lim n 2 1 .
Vy iu kin (ii) ca nh l Leibnitz khng tho mn. Hay chui lu
tha cho phn k khi x < 3.
Vi x = 3 ta c chui s n
n
1
2 1 nn nlim u lim n 2 1
chui phn k, hay chui lu tha phn k ti x = 3.
-
30
Vi x = 3 ta c chui s nn
n
1
1 2 1 .
nnn nlim u lim n 1 2 1 khng tn ti.
chui phn k, hay chui lu tha phn k ti x = 3.
MHT ca chui lu tha cho l (3;3).
7.6.3. o hm v tch phn tng s hng ca chui.
nh l 7. 11. Nu chui lu tha nnn
a x
1
hi t trn min X v vi mi x
X ta c nnn
a x
1
= S(x). Th ta c th o hm v tch phn tng s hng
ca chui. C th: n nn nn n
a x na x S x
1
1 1
( x X);
n
n nn
n n
a xa x dx S x dx
n
1
1 1 1.
V d 7.18. Tnh cc tng sau:
(i) S1(x) = 1 2x 3x2 4x3 ....
(ii) S2(x) = x2 2x3 3x4 4x5 ....
(iii) S3(x) = x 1
2x2
1
3x3
1
4x4 ...
Gii. Ta c nn
x
1
l chui lu tha hi t v nn
xx
x
1 1
(khi 1 < x < 1).
Theo nh l 7.11, o hm hai v ca ng thc trn ta c:
S1(x) = 1 2x 3x2 4x3 .... =
x 21
1 (x (1;1)).
Nhn c hai v ca ng thc trn vi x2 ta c:
-
31
S2(x) = x2 2x3 3x4 4x5 .... =
x
x
2
21
(x (1;1)).
Mt khc, n nn
x x x x x .... x ...
2 3 40
1 l chui lu
tha hi t v nn
xx
0
1
1 (x (1;1)). Theo nh l 7.11, tch phn
hai v ca ng thc trn ta c:
S3(x) = x 1
2x2
1
3x3
1
4x4 ... =
dxln x C
x
11 (x (1;1)).
7.6.4. Khai trin hm s thnh chui lu tha.
nh l 7. 12. Nu hm f(x) c o hm mi cp ti x0 v o hm cc cp
ca n lin tc trong mt ln cn no ca x0 th n khai trin c
thnh chui lu tha ti x0 . Khi ,
f(x) = kkk
f x x xk !
0 00
1.
V d 7.19. Khai trin cc hm s sau thnh chui lu tha cho tng
trng hp c th:
(i) f(x) = sin x theo lu tha ca x.
(ii) f(x) = cos x theo lu tha ca x.
(iii) f(x) = ln x theo lu tha ca x 1.
Gii. (i) f(x) = sin x l hm c o hm mi cp ti x = 0 v
f(n)(x) = sin x n 2 . Nn f(x) = sin x khai trin c thnh chui lu tha ti x = 0 v
f(x) = sin x = nnsinx x
x ... x ...! ! n !
3 52
3 5
-
32
(ii) f(x) = cos x l hm c o hm mi cp ti x = 0 v
f(n)(x) = cos x n 2 . Nn f(x) = cos x khai trin c thnh chui lu tha ti x = 0 v
f(x) = cos x = 1 nncosx x
... x ...! ! n !
2 42
2 4
(iii) f(x) = ln x l hm c o hm mi cp ti x = 1 v
f (x) = xx
11
, f (1) =1, f(n)(x) = (1)n1(n1)! xn, f(n)(1) = (1)n1(n1)!.
Nn f(x) = ln x khai trin c thnh chui lu tha ti x = 1 v
f(x) = ln x = (x1) (x1)2 x x ....! !
3 41 1
1 12 3
Cu hi n tp chng 7
Cu 1: nh ngha chui s; nh ngha tng ca chui s.
Cu 2: nh ngha s hi t ca chui s. Nu cc tnh cht v s hi t
ca chui s; chng minh tnh cht th 3 v s hi t ca chui s.
Cu 3: Cho hai chui s n nn n
u A , v B
1 1
. Khi c th ni g v s hi
t ca chui n nn
u v
1
nu:
a) C hai chui (A) v (B) u hi t?
b) Mt trong hai chui hi t cn chui kia phn k?
c) C hai chui (A) v (B) u phn k?
Cu 4: 1. Pht biu iu kin cn v Cauchy v s hi t ca chui s,
t suy ra iu kin cn mt chui s hi t.
-
33
2. Cho chui s nn
u
1
c nnlim u a
0 hoc khng tn ti nnlim u
th c kt
lun g v s hi t hay phn k ca chui s cho?
Cu 5: nh ngha chui s dng. Pht biu cc du hiu: So snh 1, So
snh 2, Dalambe v Cauchy v s hi t ca chui s dng. Chng minh
du hiu so snh 2; mi du hiu l diu kin cn hay iu kin hay
iu kin cn v mt chui s dng hi t?
Cu 6: Cc du hiu: So snh 1, So snh 2, Dalambe v Cauchy v s hi
t ca chui s dng c p dng xt s hi t ca cc chui s sau y
c khng? ti sao?
a) Chui s m.
b) Chui s khng m.
Cu 7: Cho chui s dng nn
a
1
hi t v nn n
blim
a 1. Th c th khng
nh chui nn
b
1
cng hi t hay khng? ti sao?
Cu 8: Cho hai chui s n nn n
u A , v B
1 1
l cc chui s hi t, ng
thi n n nu t v n , ,... 1 2 . Chng minh rng chui s nn
t
1
(C) cng hi
t.
Nu cc chui (A) v (B) cng phn k th c th ni g v s hi t
ca chui (C)?
Cu 9: Cho nn
a
1
l chui s khng i du v tho mn iu kin:
nnlim na k .
0
-
34
Th chui cho phn k.
Cu 10: Pht biu v chng minh nh l Lepnit v s hi t ca chui s
an du. nh l Lepnit c ng cho chui an du n nn
u
1
1 khng? ti
sao? Nu c t nht mt trong hai gi thit ca nh l Lepnit v s hi t
ca chui s an du khng c tho mn th nh l cn ng na
khng? Ly v d minh ho.
Cu 11: Cho chui s bt k nn
u
1
. Nu mi lin h gia s hi t ca
chui s nn
u
1
v s hi t ca chui s nn
u
1
. Ly v d minh ho.
Cu 12: Nu cc nh ngha v: chui hm; chui lu tha; im t v
min hi t ca chui lu tha.
Cu 13: Pht biu v chng minh nh l Abel v s hi t ca chui lu
tha. T hy chng t rng nu chui lu tha nnn
a x
1
phn k ti im
x0 0 th n phn k ti mi x m x > x0. mi
Cu 14: nh ngha bn knh hi t ca chui lu tha. Nu phng php
tm bn knh hi t ca chui lu tha.
Cu 15: Nu iu kin c th o hm v tch phn tng s hng ca
chui lu tha trn min hi t ca n. p dng kt qu trn vo vic tnh
tng.