Bai_tap_giai_tich_Tap1-DoanChi-Goc

8/9/2019 Bai_tap_giai_tich_Tap1-DoanChi-Goc

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Mc lc

L i ni u iii

Cc k hiu v k hi nim vii

Bi tp

1 S thc 3

1.1 Cn trn ng v cn di ng ca tp cc s thc. Linphn s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.2 Mt s bt ng thc s cp . . . . . . . . . . . . . . . . . . 11

2 Dy s thc 19

2.1 Dy n iu . . . . . . . . . . . . . . . . . . . . . . . . . . 23

2.2 Gii hn. Tnh cht ca dy hi t . . . . . . . . . . . . . . 30

2.3 nh l Toeplitz, nh l Stolz v ng dng . . . . . . . . . 37

2.4 im gii hn. Gii hn trn v gii hn di . . . . . . . . 42

2.5 Cc bi ton hn hp . . . . . . . . . . . . . . . . . . . . . . 48

3 Chui s thc 63

3.1 Tng ca chui . . . . . . . . . . . . . . . . . . . . . . . . . 67

3.2 Chui dng . . . . . . . . . . . . . . . . . . . . . . . . . . 75

3.3 Du hiu tch phn . . . . . . . . . . . . . . . . . . . . . . . 90

3.4 Hi t tuyt i. nh l Leibniz . . . . . . . . . . . . . . . 93

3.5 Tiu chun Dirichlet v tiu chun A bel . . . . . . . . . . . . 99

i


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ii Mc lc

3.6 Tch Cauchy ca cc chui v hn . . . . . . . . . . . . . . 102

3.7 Sp xp li chui. Chui kp . . . . . . . . . . . . . . . . . . 104

3.8 Tch v hn . . . . . . . . . . . . . . . . . . . . . . . . . . . 111

Li gii

1 S thc 121

1.1 Cn trn ng v cn di ng ca tp cc s thc. Linphn s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121

1.2 Mt s bt ng thc s cp . . . . . . . . . . . . . . . . . . 131

2 Dy s thc 145

2.1 Dy n iu . . . . . . . . . . . . . . . . . . . . . . . . . . 145

2.2 Gii hn. Tnh cht ca dy hi t . . . . . . . . . . . . . . 156

2.3 nh l Toeplitz, nh l Stolz v ng dng . . . . . . . . . . 173

2.4 im gii hn. Gii hn trn v gii hn di . . . . . . . . 181

2.5 Cc bi ton hn hp . . . . . . . . . . . . . . . . . . . . . . 199

3 Chui s thc 231

3.1 Tng ca chui . . . . . . . . . . . . . . . . . . . . . . . . . 231

3.2 Chui dng . . . . . . . . . . . . . . . . . . . . . . . . . . 253

3.3 Du hiu tch phn . . . . . . . . . . . . . . . . . . . . . . . 285

3.4 Hi t tuyt i. nh l Leibniz . . . . . . . . . . . . . . . 291

3.5 Tiu chun Dirichlet v tiu chun Abel . . . . . . . . . . . . 304

3.6 Tch Cauchy ca cc chui v hn . . . . . . . . . . . . . . 313

3.7 Sp xp li chui. Chui kp . . . . . . . . . . . . . . . . . . 321

3.8 Tch v hn . . . . . . . . . . . . . . . . . . . . . . . . . . . 338

T i liu tham k ho 354


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Li ni u

Bn ang c trong tay tp I ca mt trong nhng sch bi tp gii tch

(theo chng ti) hay nht th gii .Trc y, hu ht nhng ngi lm ton ca Vit Nam thng s dng

hai cun sch ni ting sau (bng ting Nga v c dch ra ting Vit):

1. "Bi tp gii tch ton hc"ca Demidovich (B. P. Demidovich;1969, Sbornik Zadach i Uprazhnenii po Matematicheskomu Analizu,Izdatelp1stvo "Nauka", Moskva)

v

2. "Gii tch ton hc, cc v d vbi tp"ca Ljaszko, Bojachuk,

Gai, Golovach (I. I. Lyashko, A . K. Boyachuk, YA . G. Gai, G. P. Golobach; 1975, Matematicheski Analiz v Primerakh i Zadachakh,Tom 1, 2, Izdatelp1stvo Vishaya Shkola).

ging dy hoc hc gii tch.

Cn ch rng, cun th nht ch c bi tp v p s. Cun th haicho li gii chi tit i vi phn ln bi tp ca cun th nht v mt sbi ton khc.

Ln ny chng ti chn cun sch (bng ting Ba Lan v c dchra ting Anh):

3. "Bi tp gii tch. Tp I: S thc, Dy s vChui s"(W. J.Kaczkor, M. T. Nowak, Zadania z Analizy Matematycznej, Czesc Pier-wsza, Liczby Rzeczywiste, Ciagi i Szeregi Liczbowe, WydawnictwoUniversytetu Marii Curie - Sklodowskiej, Lublin, 1996),

4. "Bi tp gii tch. Tp II: Lin tc vVi phn "(W. J. Kaczkor, M.T. Nowak, Zadania z Analizy Matematycznej, Czesc Druga, Funkcje

iii


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iv Li ni u

Jednej Zmiennej{Rachunek Rozniczowy, Wydawnictwo Universytetu

Marii Curie - Sklodowskiej, Lublin, 1998).

bin dch nhm cung cp thm mt ti liu tt gip bn c hc v dygii tch. Khi bin dch, chng ti tham kho bn ting Anh:

3*. W. J. Kaczkor, M. T. Nowak, Problems in Mathematical Analy-sis I, Real Numbers, Sequences and Series, A MS, 2000.

4*. W. J. Kaczkor, M. T. Nowak, Problems in Mathematical Analy-sis II, Continuity and Differentiation, AMS, 2001.

Sch ny c cc u im sau:

Cc bi tp c xp xp t d cho ti kh v c nhiu bi tp hay. Li gii kh y v chi tit. Kt hp c nhng tng hay gia ton hc s cp v ton hc

hin i. Nhiu bi tp c ly t cc tp ch ni ting nh, Ameri-can Mathematical Monthly (ting Anh), Mathematics Today (tingNga), Delta (t ing Balan). V th, sch ny c th dng lm ti liu

cho cc hc sinh ph thng cc lp chuyn cng nh cho cc sinhvin i hc ngnh ton.

Cc kin thc c bn gii cc bi tp trong sch ny c th tm trong

5. Nguyn Duy Tin, Bi Ging Gii Tch, Tp I, NX B i Hc QucGia H Ni, 2000.

6. W. Rudin, Principles of Mathematical Analysis, McGraw -HilBook Company, New York, 1964.

Tuy vy, trc mi chng chng ti trnh by tm tt l thuyt gipbn c nh li cc kin thc c bn cn thit khi gii bi tp trong chngtng ng.

Tp I v II ca sch ch bn n hm s mt bin s (tr phn khnggian metric trong tp II). Kaczkor, Nowak chc s cn vit Bi Tp GiiTch cho hm nhiu bin v php tnh tch phn.

Chng ti ang bin dch tp II, sp ti s xut bn.


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Li ni u v

Chng ti rt bit n :

- Gio s Phm X un Y m (Php) gi cho chng ti bn gc tingAnh tp I ca sch ny,

- Gio s Nguyn Hu Vit Hng (Vit Nam) gi cho chng ti bngc ting Anh tp II ca sch ny,

- Gio s Spencer Shaw (M) gi cho chng ti bn gc ting Anhcun sch ni ting ca W. Rudin (ni trn), xut bn ln th ba, 1976,

- TS Dng Tt Thng c v v to iu kin chng ti bin dchcun sch ny.

Chng ti chn thnh cm n tp th sinh vin Ton - L K5 H o

To C Nhn Khoa Hc Ti Nng, Trng HKHTN, HQGHN, ck bn tho v sa nhiu li ch bn ca bn nh my u tin.

Chng ti hy vng rng cun sch ny s c ng o bn c nnhn v gp nhiu kin qu bu v phn bin dch v trnh by. Rt mongnhn c s ch gio ca qu v bn c, nhng kin gp xin gi v:Chi on cn b, Khoa Ton C Tin hc, trng i hc Khoahc T nhin, i hc Quc gia HNi, 334 Nguyn Tri, ThanhXun, HNi.

X in chn thnh cm n.

H Ni, X un 2002.Nhm bin dch

on Chi


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Cc k hiu v khi nim

R - tp cc s thc R+ - tp cc s thc dng Z - tp cc s nguyn N - tp cc s nguyn dng hay cc s t nhin Q - tp cc s hu t (a, b) - khong m c hai u mt l a v b

[a, b] - on (khong ng) c hai u mt l a v b

[x] - phn nguyn ca s thc x Vi x R, hm du ca x l

sgn x =

1 vi x > 0,

1 vi x < 0,0 vi x = 0.

Vi x

N,

n! = 1 2 3 ... n,(2n)!! = 2 4 6 ... (2n 2) (2n),

(2n 1)!! = 1 3 5 ... (2n 3) (2n 1).

K hiu nk

= n!

k!(nk)! , n, k N, n k, l h s ca khai trin nhthc Newton.

vii


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viii Cc k hi u v khi ni m

Nu A

R khc rng v b chn trn th ta k hiu supA l cn

trn ng ca n, nu n khng b chn trn th ta quy c rngsupA = +.

Nu A R khc rng v b chn di th ta k hiu infA l cndi ng ca n, nu n khng b chn di th ta quy c rnginfA = .

Dy {an} cc s thc c gi l n iu tng (tng ng n iugim) nu an+1 an (tng ng nu an+1 an) vi mi n N. Lpcc dy n iu cha cc dy tng v gim.

S thc c c gi l im gii hn ca dy {an} nu tn ti mt dycon {ank} ca {an} hi t vc. Cho S l tp cc im t ca dy {an}. Cn di ng v cn trn

ng ca dy , k hiu ln lt l limn

an v limn

an c xc nh

nh sau

limn

an =

+ nu {an} khng b chn trn, nu {an} b chn trn v S = ,supS nu {an} b chn trn v S = ,

limn

an = nu {an} khng b chn di,+ nu {an} b chn di v S = ,

infS nu {an} b chn di v S = ,

Tch v hn

n=1

an hi t nu tn ti n0 N sao cho an = 0 vin n0 v dy {an0an0+1 ... an0+n} hi t khi n ti mt giihn P0 = 0. S P = an0an0+1 ... an0+n P0 c gi l gi tr catch v hn.

Trong phn ln cc sch ton nc ta t trc n nay, cc hm

tang v ctang cng nh cc hm ngc ca chng c k hiul tg x, cotg x, arctg x, arccotg x theo cch k hiu ca cc sch cngun gc t Php v Nga, tuy nhin trong cc sch ton ca Mv phn ln cc nc chu u, chng c k hiu tng t ltan x, cot x, arctan x, arccot x. Trong cun sch ny chng ti ss dng nhng k hiu ny bn c lm quen vi nhng k hiu c chun ho trn th gii.


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Bi tp


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Ch-ng 1

S thc

Tm tt l thuyt

Cho A l tp con khng rng ca tp cc s thc R = (, ).S thc x R c gi l mt cn trn ca A nu

a x, x A.

Tp A c gi l b chn trn nu A c t nht mt cn trn.S thc x R c gi l mt cn di ca A nu

a x, a A.

Tp A c gi l b chn di nu A c t nht mt cn di.

Tp A c gi l b chn nu A va b chn trn v va b chn di.R rng A b chn khi v ch khi tn ti x > 0 sao cho

|a| x, a A.

Cho A l tp con khng rng ca tp cc s thc R = (, ).S thc x R c gi l gi tr ln nht ca A nu

x A, a x, a A.Khi , ta vit

x = max{a : a A} = max aaA

.

3


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4 Chng 1. S thc

S thc x

R c gi l gi tr b nht ca A nu

x A, a x, a A.

Khi , ta vitx = min{a : a A} = min a

aA.

Cho A l tp con khng rng ca tp cc s thc R = (, ). GisA b chn trn.

S thc x R c gi l cn trn ng ca A, nu x l mt cntrn ca A v l cn trn b nht trong tp cc cn trn ca A. Tc l,

a x, a A,

> o, a A, a > x .Khi , ta vit

x = sup{a : a A} = sup aaA

.

Cho A l tp con khng rng ca tp cc s thc R = (, ). GisA b chn di.

S thc x R c gi l cn di ng ca A, nu x l mt cndi ca A v l cn trn ln nht trong tp cc cn di ca A. Tc l,

a x, a A,

> o, a A, a < x + .Khi , ta vit

x = inf{a : a A} = infaaA

.

Tin v cn trn ng ni rng nu A l tp con khng rng,

b chn trn ca tp cc s thc, th A c cn trn ng (duy nht).Tin trn tng ng vi: nu A l tp con khng rng, b chn

di ca tp cc s thc, th A c cn di ng (duy nht).

T suy ra rng A l tp con khng rng, b chn ca tp cc s thc,thA c cn trn ng, v c cn di ng.

Nu tp A khng b chn trn, th ta qui c sup A = +; Nu tpA khng b chn di, th ta qui c infA = ;


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Tm tt l thuy t 5

Cho hai s nguyn a, b. Ta ni rng b chia ht cho a hoc a chia b,

nu tn ti s nguyn c, sao cho b = a.c. Trong trng hp ta ni a lc ca b (hoc b l bi ca a) v vit a|b.

Cho hai s nguyn a1, a2. S nguyn m c gi l c chung caa1, a2 nu m|a1, m|a2. S nguyn m c gi l bi chung ca a1, a2nu a1|m, a2|m.

c chung m 0 ca a1, a2 c tnh cht l chia ht cho bt k cchung no ca a1, a2) c gi l c chung ln nht ca a1, a2 vuc k hiu l (a1, a2).

Bi chung m 0 ca a1, a2 c tnh cht l c ca bt k bi chungno ca a1, a2 c gi l bi chung nh nht ca a1, a2 v uc khiu l [a1, a2].

Nu (a, b) = 1 th ta ni a, b nguyn t cng nhau.

S nguyn dng p N c gi l s nguyn t, nu p ch c haic (tm thng) l 1 v p.

Ga sm l s nguyn dng. Hai s nguyn a, b c gi l ng dtheo modulo m, nu m|(a b). Trong trng hp ta vit

a = b (modm).

Ta gi r l s hu t (hay phn s), nu tn ti p, q Z sao chor = p/q. Phn s ny l ti gin nu (p,q) = 1. S v t l s thc nhng khng phi l s v t. Tp hp cc s

hu t tr mt trong tp cc s thc, tc l, gia hai s thc khcnhau bt k (a < b) tn ti t nht mt s hu t (r: a < r < b).

Phn nguyn ca s thc x, c k hiu l [x], l s nguyn (duynht) sao cho x 1 < [x] x. Phn l ca s thc x, c k hiu l{x}, l s thc xc nh theo cng thc {x} = x [x].

Cc hm s s cp ax, loga x, sin x, cos x, arcsin x, arccos x c nhngha theo cch thng thng. Tuy nhin, cn ch rng, ti liu ny dng

cc k hiu tiu chun quc t sau

tan x = sin x/ cos x, cot x = cos x/ sin x,

cosh x =ex + ex

2, sinh x =

ex ex2

,

tanh x = sinh x/ cosh x, coth x = cosh x/ sinh x.

Tng t ta c cc k hiu v hm ngc arctan x, arccot x.


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6 Chng 1. S thc

1.1 Cn trn ng v cn d-i ng ca tp ccs thc. Lin phn s

1.1.1. Chng minh rng

sup{x Q : x > 0, x2 < 2} =

2.

1.1.2. Cho A R khc rng. nh ngha A = {x : x A}. Chngminh rng

sup(A) = infA,inf(A) = supA.

1.1.3. Cho A, B R l khng rng. nh nghaA+B = {z = x + y : x A, y B} ,AB = {z = x y : x A, y B} .

Chng minh rng

sup(A +B) = supA+ supB,

sup(A

B) = supA

infB.

Thit lp nhng cng thc tng t cho inf(A+B) v inf(AB).1.1.4. Cho cc tp khng rng A v B nhng s thc dng, nh ngha

A B = {z = x y : x A, y B} ,1

A=

z =

1

x: x A

.

Chng minh rngsup(A B) = supA supB,

v nuinfA

> 0th

sup 1A

=

1

infA,

khi infA = 0 thsup

1A

= +. Hn na nu A v B l cc tp s thc b

chn th

sup(A B)= max {supA supB, supA infB, infA supB, infA infB} .


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1.1. Cn tr n ng v cn di ng. Li n phn s 7

1.1.5. Cho A v B l nhng tp con khc rng cc s thc. Chng minh rng

sup(A B) = max {supA, supB}

v

inf(A B) = min {infA, infB} .

1.1.6. Tm cn trn ng v cn di ng ca A1,A2 xc nh bi

A1 = 2(1)n+1 + (1)n(n+1)

2 2 +3

n : n N ,A2 =

n 1n + 1

cos2n

3: n N

.

1.1.7. Tm cn trn ng v cn di ng ca cc tp A v B , trong A = {0, 2; 0, 22;0, 222; . . . } v B l tp cc phn s thp phn gia 0 v 1m ch gm cc ch s 0 v 1.

1.1.8. Tm cn di ng v cn trn ng ca tp cc s (n+1)2

2n, trong

n

N.

1.1.9. Tm cn trn ng v cn di ng ca tp cc s (n+m)2

2nm, trong

n, m N.

1.1.10. Xc nh cn trn ng v cn di ng ca cc tp sau:

A =m

n: m, n N, m < 2n

,(a)

B =

n [n] : n N .(b)1.1.11. Hy tm

sup

x R : x2 + x + 1 > 0 ,(a)inf

z = x + x1 : x > 0

,(b)

inf

z = 2x + 21x > 0

.(c)


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8 Chng 1. S thc

1.1.12. Tm cn trn ng v cn di ng ca nhng tp sau:

A =

m

n+

4n

m: m, n N

,(a)

B =

mn

4m2 + n2: m Z, n N

,(b)

C =

m

m + n: m, n N

,(c)

D =

m

|m| + n : m Z, n N

,(d)

E = mn1 + m + n : m, n N .(e)1.1.13. Cho n 3, n N. Xt tt c dy dng hu hn (a1, . . . , an), hy

tm cn trn ng v cn di ng ca tp cc s

nk=1

akak + ak+1 + ak+2

,

trong an+1 = a1, an+2 = a2.

1.1.14. Chng minh rng vi mi s v t v vi mi nN tn ti mt s

nguyn dng qn v mt s nguyn pn sao cho pnqn < 1nqn .

ng thi c th chn dy {pn} v {qn} sao cho pnqn < 1qn2 .

1.1.15. Cho l s v t. Chng minh rng A =

{m + n : m, n

Z

}l

tr mt trong R, tc l trong bt k khong m no u c t nht mt phn tca A.

1.1.16. Chng minh rng {cos n : n N} l tr mt trong on [1, 1].1.1.17. Cho x R \ Z v dy {xn} c xc nh bi

x = [x] +1

x1, x1 = [x1] +

1

x2, . . . , xn1 = [xn1] +

1

xn.


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1.1. Cn tr n ng v cn di ng. Li n phn s 9

khi

x = [x] +1

[x1] +1

[x2] +1

. . . +1

[xn1] +1

xn

.

Chng minh rng x l s hu t khi v ch khi tn ti n N sao cho xn l mts nguyn.Ch . Ta gi biu din trn ca x l mt lin phn s hu hn. Biu thc

a0 + 1

a1 +1

a2 +1

. . . +1

an1 +1

an

c vit gn thnh

a0 +1||a1 +

1||a2 + . . . +

1||an .

1.1.18. Cho cc s thc dng a1, a2, . . . , an, tp0 = a0, q0 = 1,p1 = a0a1 + 1, q1 = a1,pk = pk1ak +pk2, qk = qk1ak + qk2, vi k = 2, 3, . . . , n,

v nh ngha

R0 = a0, Rk = a0 +1||a1 +

1||a2 + . . . +

1||ak , k = 1, 2, . . . , n.

Rk c gi l phn t hi t thk n a0 +1||a1 +

1||a2 + . . . +

1||an

.

Chng minh rng

Rk =pkqk

vi k = 0, 1, . . . , n.

1.1.19. Chng minh rng nu pk, qk c nh ngha nh trong bi ton trnv a0, a1, . . . , an l cc s nguyn th

pk1qk qk1pk = (1)k vi k = 0, 1, . . . , n.S dng ng thc trn kt lun rng pk v qk l nguyn t cng nhau.


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10 Chng 1. S thc

1.1.20. Cho x l mt s v t, ta nh ngha dy

{xn

}nh sau:

x1 =1

x [x], x2 =1

x1 [x1] , . . . , xn =1

xn1 [xn1], . . . .

Ngoi ra, chng ta cho t a0 = [x], an = [xn], n = 1, 2, . . ., v

Rn = a0 +1||a1 +

1||a2 + . . . +

1||ak .

Chng minh rng lch gia sx v phn t hi t thn ca n c cho bicng thc

x Rn = (1)n

(qnxn+1 + qn

1)qn,

trong pn, qn l c nh ngha trong 1.1.18. T hy suy ra rng x nmgia hai phn t hi t lin tip ca n.

1.1.21. Chng minh rng tp {sin n : n N} l tr mt trong [1, 1].1.1.22. S dng kt qu trong bi 1.1.20 chng minh rng vi mi s v t x

tn ti dy

pnqn

cc s hu t, vi qn l, sao cho

x pn

qn

1, k = 1, . . . , n l cc s cng dnghoc cng m th

(1 + a1) (1 + a2) . . . (1 + an) 1 + a1 + a2 + . . . + an.

Ch . Nu a1 = a2 = . . . = an = a th ta c bt ng thc Bernoulli:(1 + a)n 1 + na, a > 1.

1.2.2. S dng php qui np, hy chng minh kt qu sau: Nu a1, a2, . . . , an

l cc s thc dng sao cho a1 a2 . . . an = 1 tha1 + a2 + . . . + an n.1.2.3. K hiu An, Gn v Hn ln lt l trung bnh cng, trung bnh nhn v

trung bnh iu ho ca n s thc dng a1, a2, . . . , an, tc l

An =a1 + a2 + . . . + an

n,

Gn = n

a1 a2 . . . an ,Hn =

n1a1

+ 1a2

+ . . . + 1an

.

Chng minh rng An Gn Hn.1.2.4. S dng kt qu Gn An trong bi ton trc kim tra bt ng thc

Bernoulli

(1 + x)n 1 + nx vi x > 0.

1.2.5. Cho n N, hy kim tra cc khng nh sau:1

n+

1

n + 1+

1

n + 1+ . . .

1

2n>

2

3,(a)

1

n + 1 +1

n + 2 +1

n + 3 + . . . +1

3n + 1 > 1,(b)1

2 0 v n

N ta c

xn

1 + x + x2 + x3 + . . . + x2n 1

2n + 1.

1.2.7. Cho {an} l mt cp s cng vi cc s hng dng. Chng minh rng

a1an n

a1a2 . . . an a1 + an

2.


n

n

n! n + 1

2, n N.

1.2.9. Cho ak, k = 1, 2, . . . , n, l cc s dng tho mn iu kinn

k=1

ak 1.

Chng minh rngn

k=1

1

ak n2.

1.2.10. Cho ak > 0, k = 1, 2, . . . , n (n > 1) v t s =n

k=1

ak. Hy kim

tra cc khng nh sau:

n nk=1

aks ak

1 n 1 1

n

nk=1

s akak

,(a)

nk=1

s

s ak n2

n 1 ,(b)

n

n

k=1

aks + ak

1 n + 1.(c)

1.2.11. Chng minh rng nu ak > 0, k = 1, . . . , n v a1 a2 . . . an = 1th

(1 + a1) (1 + a2) . . . (1 + an) 2n.1.2.12. Chng minh bt ng thc Cauchy (1):

nk=1

akbk

2

nk=1

a2k

nk=1

b2k.

(1)Cn gi l bt ng thc Buniakovskii- Cauchy - Schwarz


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1.2. Mt s bt ng thc s cp 13

1.2.13. Chng minh rng nk=1

ak

2+

n

k=1

bk

2 12 nk=1

a2k + b

2k

12 .

1.2.14. Chng minh rng nun

k=1

a2k =n

k=1

b2k = 1 th

nk=1

akbk

1.

1.2.15. Cho ak > 0, k = 1, 2, . . . , n, hy kim tra nhng khng nh sau

nk=1

ak

nk=1

1

ak n2,(a)

nk=1

ak

nk=1

1 akak

nn

k=1

(1 ak),(b)

(loga a1)2 + (loga a2)

2 + . . . + (loga an)2 1

n,(c)

vi iu kin a1

a2

. . .

an = a

= 1.

1.2.16. Cho > 0, chng minh rngn

k=1

akbk

1n

k=1

a2k +

4

nk=1

b2k.

1.2.17. Chng minh cc bt ng thc sau:

nk=1

|ak|

n

n

k=1

a2k

12

nn

k=1

|ak|.

1.2.18. Chng minh rngn

k=1

akbk

2

nk=1

ka2k

nk=1

b2kk

,(a) n

k=1

akk

2

nk=1

k3a2k

nk=1

1

k5.(b)


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14 Chng 1. S thc

1.2.19. Chng minh rngn

k=1

apk

2

nk=1

ap+qk

nk=1

apqk ,

vi mi p, q v mi b s dng a1, a2, . . . , an.

1.2.20. Tm gi tr nh nht ca tngn

k=1

a2k vi iu kinn

k=1

ak = 1.

1.2.21. Cho p1, p2, . . . , pn l c c s dng. Tm gi tr nh nht ca tngn

k=1pka2k vi iu kinn

k=1 ak = 1.1.2.22. Chng minh rng

nk=1

ak

2 (n 1)

n

k=1

a2k + 2a1a2

.

1.2.23. Chng minh cc bt ng thc sau:

n

k=1 (ak + bk)2

12

n

k=1 a2k

12

+ n

k=1 b2k

12

,(a)

nk=1

a2k

12

nk=1

b2k

12

n

k=1

|ak bk|.(b)

1.2.24. Cho p1, p2, . . . , pn l cc s dng. Tm gi tr nh nht ca

nk=1

a2k +

n

k=1

ak

2

vi iu kin

nk=1

pkak = 1.

1.2.25. Chng minh bt ng thc Chebyshev.

Nua1 a2 . . . an v b1 b2 . . . bn,

hoca1 a2 . . . an v b1 b2 . . . bn,


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thn

k=1

ak

nk=1

bk n

nk=1

akbk.

1.2.26. Gi sak 0, k = 1, 2, . . . , n v p N, chng minh rng1

n

nk=1

ak

p

1

n

nk=1

apk.

1.2.27. Chng minh bt ng thc

(a + b)2

(1 + c)a2

+ 1 + 1c b2vi s dng c v s thc a, b bt k.

1.2.28. Chng minh rnga2 + b2 a2 + c2 |b c|.

1.2.29. Cho cc s dng a, b, c, kim tra cc khng nh sau:

bc

a+

ac

b+

ab

c (a + b + c),(a)

1

a

+1

b

+1

c 1

bc+

1

ca+

1

ab,(b)

2

b + c+

2

a + c+

2

a + b 9

(a + b + c),(c)

b2 a2c + a

+c2 b2a + b

+a2 c2

b + c 0,(d)

1

8

(a b)2a

a + b

2

ab

1

8

(a b)2b

vi b a.(e)

1.2.30. Cho ak R, bk > 0, k = 1, 2, . . . , n, t

m = minakbk : k = 1, 2, . . . , nv

M = max

akbk

: k = 1, 2, . . . , n

.

Chng minh rng

m a1 + a2 + . . . + anb1 + b2 + . . . + bn

M


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16 Chng 1. S thc

1.2.31. Chng minh rng nu 0 < 1 < 2 < .. . < n 1 th

tan 1 0, bk > 0, k = 1, 2, . . . , n, t

M = max

akbk

: k = 1, 2, . . . , n

.

Chng minh rnga1 + a22 + . . . + a

nn

b1 + M b22 + . . . + Mn1bnn

M.

1.2.34. Chng minh rng nu x l mt s thc ln hn cc s a1, a2, . . . , an

th 1x a1 +

1x a2 + . . . +

1x an

n

x a1+a2+...+ann

.

1.2.35. t ck =nk

, k = 0, 1, 2, . . . , n. Chng minh bt ng thc

c1 +

c2 + . . . +

cn

n(2n 1).

1.2.36. Cho n 2, chng minh rngn

k=0 n

k

2n 2n

1

n1.

1.2.37. Cho ak > 0, k = 1, 2, . . . , n v k hiu An l trung bnh cng cachng. Chng minh rng

nk=1

Apk p

p 1n

k=1

Ap1k ak

vi mi s nguyn p > 1.


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1.2.38. Cho ak > 0, k = 1, 2, . . . , n, t a = a1 +a2 + . . . +an. Hy chng

minh rngn1k=1

akak+1 a2

4.

1.2.39. Chng minh rng vi mi hon v b1, b2, . . . , bn ca cc s dnga1, a2, . . . , an ta u c

a1b1

+a2b2

+ . . . +anbn

n.

1.2.40. Chng minh bt ng thc Weierstrass.

Nu 0 < ak < 1, k = 1, 2, . . . , n v a1 + a2 + . . . + an < 1 th

1 +n

k=1

ak 0 (an < 0) vi mi n;

- khng m (khng dng) nu an 0 (an 0) vi mi n;- n iu tng (gim) nu an+1 an (an+1 an) vi mi n;- tng (gim) ngt nu an+1 > an (an+1 < an) vi mi n;

- hi t ti a R (hoc c gii hn hu hn l a), nu vi mi s > 0cho trc b ty , tn ti n N sao cho

|an a| < , n n.

Trong trng hp nh th, ta ni dy {an} hi t, v gi a l gii hn cady {an} v vit

limn

an = a;

- phn k ra +, nu vi mi s > 0 cho trc ln ty , tn tin N sao cho

an > , n n.

19


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20 Chng 2. Dy s thc

Trong trng hp nh th, ta vit

limn

an = +;

- phn k ra , nu vi mi s > 0 cho trc ln ty , tn tin N sao cho

an < , n n.Trong trng hp nh th, ta vit

limn

an = ;

- dy Cauchy (hoc dy c bn), nu vi mi s > 0 cho trc b ty, tn ti n N sao cho

|am an| < , m, n n.

nh l hi t n iu ni rng dy s n iu (tng hoc gim)v b chn c gii hn hu hn.

Tiu chun Cauchy ni rng dy s hi t khi v ch khi n l dyCauchy.

Cc tnh cht c bn ca gii hn l

- Mt dy hi t th b chn.

- Bo ton cc php tnh s hc, tc l, nu

limn

an = a, limn

bn = b,

thlimn

(an bn) = a b, , R;limn

(anbn) = ab; limn

(an/bn) = a/b vi b = 0.

- Bo ton th t theo ngha sau: nu

limn

an = a, limn

bn = b, an bn; vi n n0 no ,

tha b.

- nh l kp: Cho ba dy s thc {an}, {bn}, {cn}. Nulimn

an = a, limn

bn = a, an cn bn, vi n n0 no


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Tm tt l thuy t 21

th limn

cn = a.

Cho {an} l dy s thc v {nk} l dy cac s t nhin tng ngt, tcl n1 < n2 < < ak < ak+1 < . Khi , ta gi {ank} l mt dycon ca dy {an}. S thc a c gi l gii hn ring hay limgii hn ca {an}, nu tn ti mt dy con {ank} hi t ti a, tc l,

limk

ank = a.

nh l Bolzano - Weierstrass khng nh rng, mi dy s thcb chn c t nht mt im gii hn.

Tp cc gii hn ring ca mt dy s thc b chn {an} c gi tr lnnht. Gi tr ny c gi l gii hn trn ca dy {an} v c k hiul

limn

an.

Tp cc gii hn ring ca mt dy s thc b chn {an} c gi tr bnht. Gi tr ny c gi l gii hn di ca dy {an} v c k hiul

limn

an.

Ni rng {an} l dy truy hi cp h nuan = f(an1,...,anh), n h,

trong f l hm s thc no .

Ni rng {an} l cp s cng nu n c dngan = a0 + nd,

(a0 l s hng u, d l cng sai).

Ni rng {an} l cp s nhn nu n c dngan = a0q

n,

(a0 l s hng u, q l cng bi).

Cc k hiu ca Landau. Cho hai dy {an} v {bn}. Ta ni rng- Dy {bn} chn dy {an}, nu tn ti hng s C > 0 v tn ti s

n0 N sao cho|an| C|bn|, n n0.


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22 Chng 2. Dy s thc

Trong trng hp ta vit

an = O(bn).

- Dy {an} khng ng k so vi {bn}, nu vi mi > 0 tn ti sn N sao cho

|an| |bn|, n n,tc l

limn

anbn

= 0.

Trong trng hp ta vit

an = (bn).

- Dy {an} tng ng vi {bn}, nu

an bn = (bn),

tc llimn

anbn

= 1.

Trong trng hp ta vitan bn.


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2.1. Dy n i u 23

2.1 Dy n iu

2.1.1. Chng minh rng:

(a) Nu {an} l dy n iu tng th limn

an = sup {an : n N},

(b) Nu {an} l dy n iu gim th limn

an = inf{an : n N} .

2.1.2. Gi sa1, a2,...,ap l nhng s dng c nh. Xt cc dy sau:

sn =

an1 + an2 + ... + a

np

pv

xn =

n

sn, n N

.

Chng minh rng {xn} l dy n iu tng.Gi . Trc tin xt tnh n iu ca dy

sn

sn1

, n 2.

2.1.3. Chng minh rng dy {an}, vi an = n2n , n > 1, l dy gim ngt vtm gii hn ca dy.

2.1.4. Cho {an} l dy b chn tho mn iu kin an+1 an 12n , n N.Chng minh rng dy {an} hi t.Gi . Xt dy an 12n1 .2.1.5. Chng minh s hi t ca cc dy sau:

an = 2

n +

11

+1

2+ ... +

1n

;(a)

bn = 2

n + 1 +

1

1+

12

+ ... +1n

.(b)

Gi . Trc tin thit lp bt ng thc:

2(n + 1 1) 2, xt dy

{an

}c xc nh theo cng thc truy hi

a1 = c2, an+1 = (an c)2, n 1.

Chng minh dy {an} tng ngt.2.1.8. Gi s dy {an} tho mn iu kin

0 < an < 1, an(1 an+1) > 14

vi n N.

Thit lp s hi t ca dy v tm gii hn ca n.

2.1.9. Thit lp s hi t v tm gii hn ca dy c xc nh theo biu thc

a1 = 0, an+1 =6 + an vi n 1.

2.1.10. Chng minh dy c cho bi

a1 = 0, a2 =1

2, an+1 =

1

3(1 + an + a

3n1) vi n > 1

hi t v xc nh gii hn ca n.

2.1.11. Kho st tnh n iu ca dy

an =n!

(2n + 1)!!, n

1,

v xc nh gii hn ca n.

2.1.12. Hy xc nh tnh hi t hay phn k ca dy

an =(2n)!!

(2n + 1)!!, n 1.

2.1.13. Chng minh s hi t ca cc dy sau

an = 1 +1

22+

1

32+ ...

1

n2, n N.(a)

an = 1 + 122

+ 133

+ ... 1nn

, n N.(b)

2.1.14. Cho dy {an} c s hng tng qut

an =1

n(n + 1)+

1(n + 1)(n + 2)

+ ... +1

(2n 1)2n , n N.

Chng minh rng dy hi t.


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2.1. Dy n i u 25

2.1.15. Cho p

N, a > 0 v a1 > 0, nh ngha dy

{an

}bi

an+1 =1

p

(p 1)an + a

ap1n

, n N.

Tm limn

an.

2.1.16. Dy {an} c cho theo cng thc truy hi

a1 =

2, an+1 =

2 +

an vi n 1.

Chng minh dy {an} hi t v tm gii hn ca n.2.1.17. Dy {an} c xc nh theo cng thc truy hi

a1 = 1, an+1 =2(2an + 1)

an + 3vi n N.

Thit lp s hi t v tm gii hn ca dy {an}.2.1.18. Tm cc hng sc > 0 sao cho dy {an} c nh ngha bi cng thctruy hi

a1 = c2 , an+1 = 12 (c + a

2n) vi n Nl hi t. Trong trng hp hi t hy tm lim

nan.

2.1.19. Cho a > 0 c nh, xt dy {an} c xc nh nh sau

a1 > 0, an+1 = ana2n + 3a

3a2n + avi n N.

Tm tt c cc s a1 sao cho dy trn hi t v trong nhng trng hp hytm gii hn ca dy.

2.1.20. Cho dy {an} nh ngha truy hi bi

an+1 =1

4 3an vi n 1.

Tm cc gi tr ca a1 dy trn hi t v trong cc trng hp hy tmgii hn ca dy.


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26 Chng 2. Dy s thc

2.1.21. Cho a l mt s c nh bt k v ta nh ngha

{an

}nh sau:

a1 R v an+1 = a2n + (1 2a)an + a2 vi n N.Xc nh a1 sao cho dy trn hi t v trong trng hp nh th tm gii hnca n.

2.1.22. Cho c > 0 v b > a > 0, ta nh ngha dy {an} nh sau:

a1 = c, an+1 =a2n + ab

a + bvi n N.

Vi nhng gi tr ca a, b v c dy trn s hi t ? Trong cc trng hp hy

xc nh gii hn ca dy.

2.1.23. Chng minh rng dy {an} c nh ngha bi cng thc

a1 > 0, an+1 = 61 + an7 + an

, n N

hi t v tm gii hn ca n.

2.1.24. Cho c 0 xt {an} c cho hi cng thca1 = 0, an+1 =

c + an, n N.

Chng minh rng dy hi t v tm gii hn ca n.

2.1.25. Kho st s hi t ca dy c cho bi cng thc

a1 =

2, an+1 =

2an, n N.2.1.26. Cho k N, kho st s hi t ca dy {an} c cho bi cng thctruy hi sau

a1 =k

5, an+1 =k

5an, n N.2.1.27. Kho st s hi t ca dy {an} sau

1 a1 2, a2n+1 = 3an 2, n N.2.1.28. Vi c > 1, nh ngha dy {an} v {bn} nh sau:

a1 =

c(c 1), an+1 =

c(c 1) + an, n 1,(a)b1 =

c, bn+1 =

cbn, n 1.(b)

Chng minh rng c hai dy u c gii hn l c.


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2.1. Dy n i u 27

2.1.29. Cho a > 0 v b > 0, nh ngha dy

{an

}bi

0 < a1 < b, an+1 =

ab2 + a2n

a + 1vi n 1.

Tm limn

an.

2.1.30. Chng minh s hi t ca dy {an} c cho bi cng thc truy hi

a1 = 2, an+1 = 2 +1

3 + 1an

vi n 1

v tm gii hn ca n.

2.1.31. Dy {an} c cho bia1 = 1, a2 = 2, an+1 =

an1 +

an, vi n 2.

Chng minh dy trn b chn v tng ngt. Hy tm gii hn ca dy ny.

2.1.32. Dy {an} c xc nh theo cng thc truy hia1 = 9, a2 = 6, an+1 =

an1 +

an, vi n 2.

Chng minh rng dy trn b chn v gim ngt. Tm gii hn ca dy ny.

2.1.33. Dy {an} v {bn} c cho bi cng thc0 < b1 < a1, an+1 =

an + bn2

v bn+1 =

anbn vi n N.

Chng minh rng {an} v {bn} cng tin ti mt gii hn. (Gii hn ny cgi l trung bnh cng - nhn ca a1 v b1).

2.1.34. Chng minh rng c hai dy {an} v {bn} xc nh theo cng thc

0 < b1 < a1, an+1 =a2n + b

2n

an + bnv bn+1 =

an + bn2

vi n N

u n iu v c cng gii hn.

2.1.35. Hai dy truy hi {an} v {bn} c cho bi cng thc

0 < b1 < a1, an+1 =an + bn

2v bn+1 =

2anbnan + bn

vi n N.

Chng minh tnh n iu ca hai dy trn v ch ra rng c hai dy u tinti trung bnh cng - nhn ca a1 v b1. (Xem bi ton 2.1.33).


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28 Chng 2. Dy s thc

2.1.36. Chng minh s hi t v tm gii hn ca dy

{an

}vi

an =n + 1

2n+1

2

1+

22

2+ ... +

2n

n

vi n N.

2.1.37. Gi s c mt dy b chn {an} tho mn

an+2 1

3an+1 +

2

3an vi n 1.

Chng minh rng dy trn hi t.

2.1.38. Cho

{an

}v

{bn

}nh ngha bi:

an =

1 +

1

n

n, bn =

1 +

1

n

n+1vi n N.

S dng bt ng thc lin h gia trung bnh cng, nhn v iu ho chngminh rng

(a) an < bn vi n N.(b) dy {an} tng ngt,

(c) dy {bn} gim ngt,Chng minh rng {an} v {bn} c cng gii hn, c gi l se ca Euler.2.1.39. Cho

an =

1 +x

n

nvi n N.

(a) Chng t rng nu x > 0 th dy {an} b chn v tng ngt.(b) Gi sx l mt s thc tu . Chng minh rng dy {an} b chn v tng

ngt vi n >

x.

ex c nh ngha l gii hn ca dy ny.

2.1.40. Gi s c x > 0, l N v l > x. Chng minh rng dy {bn} vi

bn =

1 +x

n

l+nvi n N,

l dy gim ngt.


37/364

2.1. Dy n i u 29

2.1.41. Thit lp tnh n iu ca cc dy

{an

}v

{bn

}, vi

an = 1 +1

2+ ... +

1

n 1 ln n vi n N,

bn = 1 +1

2+ ... +

1

n 1 +1

n ln n vi n N.

Chng minh rng c hai dy trn cng tin n cng mt gii hn , gi lhng s Euler.Gi . S dng bt ng thc (1 + 1

n)n < e < (1 + 1

n)n+1, (suy ra t 2.1.38).

2.1.42. Cho x > 0 v t an = 2n

x, n

N. Chng t rng dy

{an

}b

chn. ng thi chng minh rng dy ny tng ngt nu x < 1 v gim ngtnu x > 1. Tnh lim

nan.

Hn na, t

cn = 2n(an 1) v dn = 2n

1 1

an

vi n N.

Chng minh rng {cn} l dy gim, cn {dn} l dy tng v c hai dy cngc chung gii hn.


38/364

30 Chng 2. Dy s thc

2.2 G ii hn. T nh cht ca dy hi t

2.2.1. Tnh:

limn

n

12 + 22 + ... + n2,(a)

limn

n + sin n2

n + cos n,(b)

limn

1 2 + 3 4 + ... + (2n)n2 + 1

,(c)

limn

(

2

3

2)(

2

5

2)...(

2

2n+1

2),(d)

limn

n

2n

,(e)

limn

n!

2n2,(f)

limn

1n

1

1 +

3+

13 +

5

+ ... +1

2n 1 + 2n + 1

,(g)

limn

1

n2 + 1+

2

n2 + 2+ ... +

n

n2 + n

,(h)

limn

n

n3 + 1

+2n

n3 + 2

+ ... +nn

n3 + n .(i)2.2.2. Cho s > 0 v p > 0. Chng minh rng

limn

ns

(1 +p)n= 0.

2.2.3. Cho (0, 1), tnh

limn

((n + 1) n).

2.2.4. Cho Q, hy tnhlimn

sin(n!).

2.2.5. Chng minh rng khng tn ti limn

sin n.

2.2.6. Chng minh rng vi mi s v t , limn

sin n khng tn ti.


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2.2. Gii hn. Tnh cht ca dy hi t 31

2.2.7. Vi

R, hy tnh

limn

1

n

a +

1

n

2+

a +

2

n

2+ ... +

a +

n 1n

2.

2.2.8. Gi s an = 1 vi mi n v limn

an = 1. Cho k nguyn dng, hy

tnh

limn

an + a2n + ... + a

kn k

an 1 .

2.2.9. Tnh

limn 11.2.3 + 12.3.4 + ... + 1n.(n + 1)(n + 2) .2.2.10. Tnh

limn

nk=2

k3 1k3 + 1

.

2.2.11. Tnh

limn

ni=1

ij=1

j

n3.

2.2.12. Tnh

limn

1 2

2.3

1 2

3.4

...

1 2

(n + 1).(n + 2)

.

2.2.13. Tnh

limn

nk=1

k3 + 6k2 + 11k + 5

(k + 3)!.

2.2.14. Cho x = 1 v x = 1, hy tnh

limn

n

k=1 x2k1

1 x2k.

2.2.15. Vi gi tr x R no th gii hn

limn

nk=0

(1 + x2k

).

tn ti v tm gi tr ca gii hn ny.


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32 Chng 2. Dy s thc

2.2.16. Tm tt c x

R sao cho gii hn

limn

nk=0

1 +

2

x2k + x2k

.

tn ti v tm gi tr ca gii hn ny.

2.2.17. Vi gi tr x R no thi gii hn

limn

nk=1

(1 + x3k

+ x2.3k

).

tn ti v tm gi tr ca gii hn ny.2.2.18. Tnh

limn

1.1! + 2.2! + ... + n.n!

(n + 1)!.

2.2.19. Vi x R no sao cho ng thc sau

limn

n1999

nx (n 1)x =1

2000

c thc hin

2.2.20. Cho a v b sao cho a b > 0, nh ngha dy {an} nh sau:a1 = a + b, an = a1 ab

an1, n 2.

Hy xc nh s hng thn ca dy v tnh limn

an.

2.2.21. nh ngha dy {an} bia1 = 0, a2 = 1 v an+1 2an + an1 = 2 vi n 2.

Hy xc nh s hng thn ca dy v tnh limn

an.

2.2.22. Cho a > 0, b > 0, xt dy {an} cho bi

a1 =ab

a2 + b2v

an =aan1

a2 + a2n1, n 2.

Tm s hng th n ca dy v tnh limn

an.


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2.2.23. Cho dy truy hi

{an

}nh ngha bi

a1 = 0, an =an1 + 3

4, n 2.

Tm s hng th n v gii hn ca dy.

2.2.24. Hy xt tnh hi t ca dy cho bi

a1 = a, an = 1 + ban1, n 2.2.2.25. Ta inh ngha dy Fibonacci {an} nh sau:

a1 = a2 = 1, an+2 = an + an+1, n 1.Chng minh rng

an =n n

,

trong v l nghim ca phng trnh x2 = x + 1. Tnh limn

n

an.

2.2.26. Cho hai dy {an} v {bn} theo cng thc sau:a1 = a, b1 = b,

an+1 =an + bn

2 , bn+1 =

an+1 + bn2 .

Chng minh rng limn

an = limn

bn.

2.2.27. Cho a {1, 2,..., 9}, hy tnh

limn

a + aa + ... +

n s hngaa...a

10n.

2.2.28. Tnh

limn nn 1n .2.2.29. Gi s rng dy {an} hi t ti 0. Hy tm lim

nann.

2.2.30. Cho p1, p2,...,pk v a1, a2,...,ak l cc s dng, tnh

limn

p1an+11 +p2a

n+12 + ... +pka

n+1k

p1an1 +p2a

n2 + ... +pka

nk

.


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34 Chng 2. Dy s thc

2.2.31. Gi s rng limn an+1an = q. Chng minh rng:

(a) Nu q < 1 th limn

an = 0,

(b) Nu q > 1 th limn

|an| = .

2.2.32. Gi s c limn

n|an| = q. Chng minh rng:

(a) Nu q < 1 th limn

an = 0,

(b) Nu q > 1 th limn

|an| = .

2.2.33. Cho l mt s thc v x (0, 1), hy tnh

limn

nxn.

2.2.34. Tnh

limn

m(m 1) ... (m n + 1)n!

xn, vi m N v |x| < 1.

2.2.35. Gi s limn

an = 0 v {bn} mt dy b chn. Chng minh rng

limn

anbn = 0.

2.2.36. Chng minh rng nu limn

an = a v limn

bn = b th

limn

max {an, bn} = max {a, b} .

2.2.37. Cho an

1 vi n

N v lim

nan = 0. Cho p

N, hy tm

limn

p

1 + an.

2.2.38. Gi s c dy dng {an} hi t ti 0. Cho s t nhin p 2, hy xcnh

limn

p

1 + an 1an

.


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2.2.39. Cho cc s dng a1, a2,...,ap, hy tnh

limn

p

(n + a1)(n + a2)...(n + ap) n

.

2.2.40. Tnh

limn

1

n2 + 1+

1n2 + 2

+ ... +1

n2 + n + 1

.

2.2.41. Cho a1, a2,...,ap l cc s dng, hy tm

limn

nan1 + an2 + ... + anpp

.

2.2.42. Tnh

limn

n

2sin2

n1999

n + 1+ cos2

n1999

n + 1.

2.2.43. Tnhlimn

(n + 1 + n cos n)1

2n+n sinn .

2.2.44. Tnh

limn

nk=1

1 +

kn2

1 .2.2.45. Hy xc nh

limn

nk=1

3

1 +

k2

n3 1

.

2.2.46. Cho cc s dng ak, k = 1, 2,...,p, hy tnh

limn

1p

pk=1

nakp .2.2.47. Cho (0, 1). Hy tnh

limn

n1k=0

+

1

n

k.


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36 Chng 2. Dy s thc

2.2.48. Cho s thc x

1, hy chng t rng

limn

(2 n

x 1)n = x2.


limn

(2 n

n 1)nn2

= 1.

2.2.50. Trong nhng dy di y, dy no l dy Cauchy ?

an =tan1

2

+tan2

22

+ ... +tan n

2n

,(a)

an = 1 +1

4+

22

42+ ... +

n2

4n,(b)

an = 1 +1

2+

1

3+ ... +

1

n,(c)

an =1

1.2 1

2.3+ ... + (1)n1 1

n(n + 1),(d)

an = 1q1 + 2q

2 + ... + nqn,(e)

vi |q| < 1, |k| M, k = 1, 2,...,an =

1

22+

2

32+ ... +

n

(n + 1)2.(f)

2.2.51. Cho dy {an} tho mn iu kin|an+1 an+2| < |an an+1|.

vi (0, 1). Chng minh rng {an} hi t .2.2.52. Cho dy {an} cc s nguyn dng, nh ngha

Sn =1

a1+

1

a2+ ... +

1

an

v

n =

1 +

1

a1

1 +

1

a2

...

1 +

1

an

.

Chng minh rng nu {Sn} hi t th{ln n} cng hi t.2.2.53. Chng minh rng nu dy {Rn} hi t n mt s v t x (nh nghatrong bi ton 1.1.20) th n l dy Cauchy.


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2.3. nh l Toe plitz, nh l Stolz 37

2.2.54. Cho mt dy cp s cng

{an

}vi cc s hng khc 0, hy tnh

limn

1

a1a2+

1

a2a3+ ... +

1

anan+1

.

2.2.55. Cho mt dy cp s cng {an} vi cc s hng dng, hy tnh

limn

1n

1

a1 +

a2+

1a2 +

a3

+ ... +1

an +

an+1

.

2.2.56. Tnh

(a) limn

n( n

e 1), (b) limn

e1n + e

2n + ... + e

nn

n.

2.2.57. Cho dy {an} nh ngha nh sau:

a1 = a, a2 = b, an+1 = pan1 + (1 p)an, n = 2, 3,...

Xc nh xem vi gi tr a, b v p no th dy trn hi t.

2.2.58. Cho

{an

}v

{bn

}nh ngha bi

a1 = 3, b1 = 2, an+1 = an + 2bn v bn+1 = an + bn.

Hn na cho

cn =anbn

, n N.

Chng t rng |cn+1

2| < 12|cn

2|, n N.(a)

Tnh limn

cn.(b)

2.3 nh l T oeplitz, nh l Stolz v ng dng

2.3.1. Chng minh nh l Toeplitz sau v php bin i chnh qui t dy sangdy.


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38 Chng 2. Dy s thc

Cho

{cn,k : 1

k

n, n

1

}l mt bng cc s thc tho mn:

cn,k n

0 vi mi k N,(i)n

k=1

cn,k n

1,(ii)

tn ti hng sC > 0 sao cho vi mi s nguyn dng n th(iii)n

k=1

|cn,k| C.

Khi vi mi dy hi t{

an}

th dy bin i{

bn}

c cho bi cng thc

bn =n

k=1

cn,kak, n 1, cng hi t v limn

bn = limn

an.


an = a th

limn

a1 + a2 + ... + ann

= a.

2.3.3.

(a) Chng minh rng gi thit (iii) trong nh l Toeplitz (bi ton 2.3.1) c

th b qua nu tt c cn,k l khng m.

(b) Cho {bn} l dy c nh ngha trong nh l Toeplitz (xem bi 2.3.1) vicn,k > 0, 1 k n, n 1. Chng minh rng nu lim

nan = + th

limn

bn = +.


an = + th

limn

a1 + a2 + ... + ann

= +.

2.3.5. Chng minh rng nu limn an = a th

limn

na1 + (n 1)a2 + ... + 1.ann2

=a

2.

2.3.6. Chng minh rng nu dy dng {an} hi t ti a thlimn

n

a1...an = a.


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2.3.7. Cho dy dng

{an

}, chng minh rng nu lim

n

an+1an

= a th

limn

n

an = a.

2.3.8. Cho limn

an = a v limn

bn = b. Chng minh rng

limn

a1bn + a2bn1 + ... + anb1n

= ab.

2.3.9. Cho {an} v {bn} l hai dy tho mn

bn > 0, n N, v limn (b1 + b2 + ... + bn) = +,(i)

limn

anbn

= g.(ii)

Chng minh rng

limn

a1 + a2 + ... + anb1 + b2 + ... + bn

= g.

2.3.10. Cho {an} v {bn} l hai dy tho mn

bn > 0, n N, v limn (b1 + b2 + ... + bn) = +,(i)limn

an = a.(ii)

Chng minh rng

limn

a1b1 + a2b2 + ... + anbnb1 + b2 + ... + bn

= a.

2.3.11. S dng cc kt qu ca bi trc, hy chng minh nh l Stolz.

Cho

{xn

},

{yn

}l hai dy tho mn:

{yn} tng thc s ti + ,(i)limn

xn xn1yn yn1 = g.(ii)

Khi

limn

xnyn

= g.


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40 Chng 2. Dy s thc

2.3.12. Tnh

limn

1n

1 +

12

+ ... +1n

,(a)

limn

n

an+1

a +

a2

2+ ... +

an

n

, a > 1,(b)

limn

1

nk+1

k! +

(k + 1)!

1!+ ... +

(k + n)!

n!

, k N,(c)

limn

1n

1n

+1

n + 1... +

12n

,(d)

limn1k + 2k + ... + nk

nk+1 , k N,(e)limn

1 + 1.a + 2.a2...nan

nan+1, a > 1,(f)

limn

1

nk(1k + 2k + ... + nk) n

k + 1

, k N.(g)

2.3.13. Gi s rng limn

an = a. Tm

limn

1n

a1 +a2

2+

a33

+ ... +an

n.

2.3.14. Chng minh rng nu dy{an} tho mnlimn

(an+1 an) = a,

thlimn

ann

= a.

2.3.15. Cho limn

an = a. Hy tnh

limnan1 + an12 + ... + a12n1 .2.3.16. Gi s rng lim

nan = a. Hy tnh

limn

an1.2

+an12.3

+ ... +a1

n.(n + 1)

,(a)

limn

an1

an121

+ ... + (1)n1 a12n1

.(b)


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2.3.17. Cho k l mt s t nhin c nh bt k ln hn 1. Hy tnh

limn

n

nk

n

.

2.3.18. Cho mt cp s cng dng {an}, tnh

limn

n(a1...an)1n

a1 + a2 + ... + an.

2.3.19. Cho dy

{an

}sao cho dy

{bn

}vi bn = 2an + an

1, n

2, hi t

ti b. Hy xt tnh hi t ca {an} .2.3.20. Cho dy {an} tho mn lim

nnxan = a vi s thc x no . Chng

minh rng

limn

nx(a1.a2...an)1n = aex.

2.3.21. Tnh

limn

1 + 12

+ ... + 1n1 +

1n

ln n,(a)

limn

1 + 13 + 15 + ... + 12n1ln n

.(b)

2.3.22. Gi s{an} tin ti a. Chng minh rng

limn

1

ln n

a11

+a22

+ ... +ann

= a.

2.3.23. Tnh

(a) limn

n!nnen

1n, (b) lim

n

(n!)3n3nen

1n,

(c) limn

(n!)2

n2n

1n

, (d) limn

n3n

(n!)3

1n

,

(e) limn

k

nn

n!, k N.


50/364

42 Chng 2. Dy s thc


an = a th

limn

1

ln n

nk=1

akk

= a.

2.3.25. Cho dy {an}, xt dy {An} cc trung bnh cng An = a1+a2+...+ann .Chng minh rng nu lim

nAn = A th

limn

1

ln n

nk=1

akk

= A.

2.3.26. Chng minh iu ngc li ca nh l Toeplitz trong 2.3.1.

Cho {cn,k : 1 k n, n 1} l mt bng s thc bt k. Nu vi midy {an} hi t bt k, dy bin i {bn} cho bi cng thc

bn =n

k=1

cn,kak, n 1

cng hi t n cng mt gii hn th

cn,k n

0 vi mi k N,(i)n

k=1

cn,k n

1,(ii)

tn ti hng sC > 0 sao cho vi mi s nguyn dng n, ta c(iii)n

k=1

|cn,k| C.

2.4 im gii hn. G ii hn trn v gii hn

d-i2.4.1. Cho {an} l dy tho mn {a2k} , {a2k+1} v {a3k} hi t.

(a) Chng minh rng dy {an} cng hi t.(b) Liu t s hi t ca hai trong ba dy con trn c suy ra c s hi t ca

{an}?


51/364

2.4. i m gi i hn. Gi i hn tr n v gii hn di 43

2.4.2. T s hi t ca tt c cc dy con ca dy

{an

}di dng

{as.n

}, s >

1, c suy ra c s hi t ca {an}?2.4.3. Cho {apn} , {aqn} , . . . , {asn} l cc dy con ca dy {an} sao cho{pn} , {qn} , . . . , {sn} ri nhau tng cp v hp thnh dy {n}. Chng minhrng nu S, Sp, Sq, . . . , S s tng ng l cc tp cc im gii hn (1) ca ccdy {an} , {apn} , {aqn} , . . . , {asn} th

S = Sp Sq ... Ss.Chng minh rng nu mi dy con {apn} , {aqn} ,..., {asn} hi t ti a th dy

{an

}cng hi t ti a.

2.4.4. nh l trn (bi ton 2.4.3) c ng trong trng hp s lng cc dycon l v hn khng ?

2.4.5. Chng minh rng, nu mi dy con {ank} ca dy {an} u cha mtdy con

ankl

hi t ti a th dy {an} cng hi t ti a.

2.4.6. Xc nh tp cc im gii hn ca dy {an}, vi

an =

4(1)n + 2,(a)

an =

1

2 n 2 3 n 13 n 3 3 n 13 ,(b)an =

(1 (1)n)2n + 12n + 3

,(c)

an =(1 + cos n) l n 3n + ln n

ln 2n,(d)

an =

cosn

3

n,(e)

an =2n2

7

2n2

7

.(f)

2.4.7. Tm tp hp cc im gii hn ca dy {an} cho bi cng thcan = n [n], Q,(a)an = n [n], Q,(b)an = sin n, Q,(c)an = sin n, Q.(d)

(1)Cn gi l cc gii hn ring hay cc im t ca dy.


52/364

44 Chng 2. Dy s thc

2.4.8. Cho

{ak

}l mt dy sinh ra t cch nh s mt-mt bt k cc phn

t ca ma trn { 3n 3m} , n, m N. Chng minh rng mi s thc ul im gii hn ca dy ny.

2.4.9. Gi s{an} l dy b chn. Chng minh rng tp cc im gii hnca n l ng v b chn.

2.4.10. Xc nh limn

an v limn

an vi:

an =2n2

7

2n2

7

,(a)

an = n 1n + 1

cos n3

,(b)

an = (1)nn,(c)an = n

(1)nn,(d)

an = 1 + n sinn

2,(e)

an =

1 +

1

n

n(1)n + sin n

4,(f)

an =n

1 + 2n(1)n ,(g)

an = 2cos 2n3

n ,(h)an =

ln n (1 + cos n)nln 2n

.(i)

2.4.11. Tm gii hn trn v gii hn di ca cc dy sau:

an = n [n], Q,(a)an = n [n], Q,(b)an = sin n, Q,(c)

an = sin n, Q.(d)2.4.12. Vi dy {an} bt k chng minh rng:

(a) nu tn ti k N sao cho vi mi n > k, bt ng thc an A lun ngth lim

nan A,

(b) nu vi mi k N tn ti nk > k ank A th limn

an A,


53/364


(c) nu tn ti k

N sao cho bt ng thc an

a ng vi mi n > k th

limn

an a,

(d) nu vi mi k N tn ti nk > k sao cho ank a th limn

an a.2.4.13. Gi s dy {an} tn ti gii hn trn v gii hn di hu hn. Chngminh rng

(a) L = limn

an khi v ch khi

(i) Vi mi > 0 tn ti k N sao cho an < L + nu n > k v

(ii) Vi mi > 0 v k N tn ti nk > k sao cho L < ank(b) l = lim

nan khi v ch khi

(i) Vi mi > 0 tn ti k N sao cho an > l nu n > k v(ii) Vi mi > 0 v k N tn ti nk > k sao cho ank < l +

Hy pht biu nhng khng ng tng ng cho gii hn trn v gii hn trongtrng hp v hn.

2.4.14. Gi s tn ti mt s nguyn n0 sao cho vi n n0, an bn. Chngminh rng

limn

an limn

bn,(a)

limn

an limn

bn.(b)

2.4.15. Chng minh cc bt ng thc sau (tr trng hp bt nh +v + ):

limn

an + limn

bn limn

(an + bn) limn

an + limn

bn

limn

(an + bn) limn

an + limn

bn.

Hy a ra mt s v d sao cho du trong cc bt ng thc trn cthay bng du < .

2.4.16. Cc bt ng thc sau

limn

an + limn

bn limn

(an + bn),

limn

(an + bn) limn

an + limn

bn.


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46 Chng 2. Dy s thc

c ng trong trng hp c v hn dy khng ?

2.4.17. Ly {an} v {bn} l c c dy s khng m. Chng minh rng (trtrng hp 0.(+) v (+).0) cc bt ng thc sau:

limn

an limn

bn limn

(an bn) limn

an limn

bn

limn

(an bn) limn

an limn

bn.

Hy a ra mt s v d sao cho du trong cc bt ng thc trn cthay bng du < .

2.4.18. Chng minh rng iu kin cn v mt dy {an} hi t l cgii hn trn v gii hn di hu hn vlimn

an = limn

an.

Chng minh rng bi ton vn ng cho trng hp cc dy phn k ti v +.2.4.19. Chng minh rng nu lim

nan = a, a R th

limn

(an + bn) = a + limn

bn,

limn

(an + bn) = a + limn

bn,


an = a, a R, a > 0, v tn ti mt snguyn dng n0 sao cho bn 0 vi n n0, khi

limn

(an.bn) = a. limn

bn,

limn

(an.bn) = a. limn

bn,


limn

(an) = limn

an, limn

(an) = limn

an.

2.4.22. Chng minh rng vi dy s dng {an} ta c

limn

1

an

=

1

limn

an,


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limn 1an = 1limn(a

n) .

( y 1+ = 0,

10+

= +.)2.4.23. Chng minh rng nu dy {an} l dy s dng tho mn

limn

(an) limn

1

an

= 1,

th dy {an} hi t.

2.4.24. Chng minh rng nu {an} l dy tho mn vi bt k dy {bn} ,limn

(an + bn) = limn

an + limn

bn,

vlimn

(an + bn) = limn

an + limn

bn.

th dy {an} hi t.2.4.25. Chng minh rng, nu {an} l mt dy dng tho mn vi bt k dydng {bn},

limn(an bn) = limnan limnbn.hoc

limn

(an.bn) = limn

an limn

bn,

v vy {an} hi t.2.4.26. Chng minh rng vi bt k dy dng {an},

limn

an+1an

limn

n

an limn

n

an limn

an+1an

.

2.4.27. Cho dy {an} , ly dy {bn} xc nh nh saubn =

1

n(a1 + a2 + ... + an), n N.

Chng minh rng

limn

an limn

bn limn

bn limn

an.


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48 Chng 2. Dy s thc


limn

(max {an, bn}) = max

limn

an, limn

bn

,(a)

limn

(min{an, bn}) = min

limn

an, limn

bn

,(b)

Kim tra cc bt ng thc sau:

limn

(min {an, bn}) = min

limn

an, limn

bn

,(a)

limn(max {an, bn}) = max limnan, limnbn(d)c ng khng?

2.4.29. Chng minh rng mi dy s thc u cha mt dy con n iu.

2.4.30. S dng kt qu bi trc chng minh nh l Bolzano-Weierstrass:

Mi dy s thc b chn u cha mt dy con hi t.

2.4.31. Chng minh rng vi mi dy s dng {an},

limn

a1 + a2 + ... + an + an+1an

4.

Chng minh rng 4 l nh gi tt nht.

2.5 Cc bi ton hn hp


an = + hay limn

an = th

limn

1 +

1

an

an= e.

2.5.2. Vi x R chng minh rng

limn

1 +

x

n

n= ex.


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2.5. Cc bi ton hn hp 49

2.5.3. Vi x > 0 hy kim chng bt ng thc

x

x + 2< ln(x + 1) < x.

(S dng o hm ) chng minh rng bt ng thc tri c th mnh hn nhsau

x

x + 2 0.


limn

n( n

a

1) = ln a, a > 0,(a)

limn

n( nn 1) = +.(b)

2.5.5. Ly {an} l dy s dng vi cc s hng khc 1, chng minh rng nulimn

an = 1 th

limn

ln anan 1 = 1.

2.5.6. Ly

an = 1 +1

1!+

1

2!+ ... +

1

n!, n N.

Chng minh rng

limn

an = e v 0 < e an < 1nn!

.


limn

1 +

x

1!+

x2

2!+ ... +

xn

n!

= ex.


limn

1

n+

1

n + 1+ ... +

1

2n

= ln 2,

(a)

limn

1

n(n + 1)+

1(n + 1)(n + 2)

+ ... +1

2n(2n + 1)

= ln2.

(b)


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50 Chng 2. Dy s thc

2.5.9. Tm gii hn ca dy

{an

},trong

an =

1 +

1

n2

1 +

2

n2

...

1 +

n

n2

, n N.

2.5.10. Ly {an} l dy c xc nh qui np nh sau

a1 = 1, an = n(an1 + 1) vi n = 2, 3,...

Tnh

limn

n

k=1 1 +1

ak .2.5.11. Chng minh rng lim

n(n!e [n!e]) = 0.

2.5.12. Cho cc s dng a v b, chng minh rng

limn

n

a + n

b

2

n=

ab.

2.5.13. Cho {an} v {bn} l cc dy dng tha mn

limn

ann = a, limn

bnn = b, trong a, b > 0,

v gi s cc s dng p, q tha mn p + q = 1. Chng minh rng

limn

(pan + qbn)n = apbq.

2.5.14. Cho hai s thc a v b, xc nh dy {an} nh sau

a1 = a, a2 = b, an+1 =n 1

nan +

1

nan1, n 2.

Tm limn

an.

2.5.15. Cho {an} l mt dy c xc nh nh sau

a1 = 1, a2 = 2, an+1 = n(an + an1), n 2.

Tm cng thc hin ca cc s hng tng qut ca dy.


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2.5.16. Cho a v b xc nh

{an

}nh sau

a1 = a, a2 = b, an+1 =1

2nan1 +

2n 12n

an, n 2.

Tm limn

an.

2.5.17. Cho

an = 3 n

k=1

1

k(k + 1)(k + 1)!, n N.

(a) Chng minh rng limn

an

= e.

(b) Chng minh rng 0 < an e < 1(n+1)(n+1)! .2.5.18. Tnh lim

nn sin(2n!e).

2.5.19. Gi s{an} l dy tho mn an < n, n = 1, 2,..., v limn

an = +.Hy xt tnh hi t ca dy

1 ann

n, n = 1, 2,....

2.5.20. Gi s dy {bn} dng hi t ti +. Xt tnh hi t ca dy1 +

bnn

n, n = 1, 2, ....

2.5.21. Cho dy truy hi {an} nh ngha nh sau0 < a1 < 1, an+1 = an(1 an), n 1,

chng minh rng

limn

nan = 1,(a)

limn

n(1 an)ln n

= 1,(b)

2.5.22. Xt dy truy hi {an} nh sau0 < a1 < , , an+1 = sin an, n 1.

Chng minh rng limn

nan =

3.


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52 Chng 2. Dy s thc

2.5.23. Cho

a1 = 1, an+1 = an + 1nk=1

ak

, n 1.

Chng minh rng

limn

an2 ln n

= 1.

2.5.24. Cho {an} nh saua1 > 0, an+1 = arctan an, n 1,

tnh limn

an.

2.5.25. Chng minh rng dy qui

0 < a1 < 1, an+1 = cos an, n 1,hi t ti nghim duy nht ca phng trnh x = cos x.

2.5.26. nh ngha dy {an} nh saua1 = 0, an+1 = 1 sin(an 1), n 1

Tnh

limn

1n

nk=1

ak.

2.5.27. Cho {an} l dy cc nghim lin tip ca phng trnh tan x =x, x > 0. Tm lim

n(an+1 an).

2.5.28. Cho |a| 2

v a1 R. Nghin cu tnh hi t ca dy {an} cho bicng thc sau:

an+1 = a sin an, n 1.2.5.29. Cho a1 > 0, xt dy

{an

}cho bi

an+1 = ln(1 + an), n 1.Chng minh rng

limn

nan = 2,(a)

limn

n(nan 2)ln n

=2

3.(b)


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2.5.30. Cho dy

{an

}nh sau

a1 = 0 v an+1 =

1

4

an, n 1.

Hy nghin cu tnh hi t ca dy.

2.5.31. Cho a1 > 0, nh ngha dy {an} nh sau:an+1 = 2

1an , n 1.Kho st tnh hi t ca dy.

2.5.32. Tm gii hn ca dy cho bi

a1 =

2, an+1 = 2an2 , n 1.


(an an2) = 0 th

limn

an an1n

= 0.

2.5.34. Chng minh rng nu vi dy dng {an} bt k tho mn

limn

n1 an+1an tn ti (hu hn hoc v hn) th

limn

ln 1an

ln n

cng tn ti v c hai gii hn bng nhau.

2.5.35. Cho a1, b1 (0, 1), Chng minh rng dy {an} v {bn} cho bi cngthc

an+1 = a1(1 an bn) + an, bn+1 = b1(1 an bn) + bn, n 1hi t v tm gii hn ca chng.

2.5.36. Cho a v a1 dng, xt dy {an} nh sauan+1 = an(2 nan), n = 1, 2,...

Kho st s hi t ca dy.


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54 Chng 2. Dy s thc

2.5.37. Chng minh rng nu a1 v a2 l hai s dng v

an+2 =an + an+1, n = 1, 2,...

th dy {an} hi t. Tm gii hn ca dy.2.5.38. Gi sf : Rk+ R l mt hm tng vi mi bin v tn ti a > 0sao cho

f(x,x,...,x) > x vi 0 < x < a,

f(x,x,...,x) < x vi x > a.

Cho cc s dng a1, a2, . . . , ak, nh ngha dy truy hi {an} nh sau:an = f(an1, an2,...,ank), vi n > k.

Chng minh rng limn

an = a.

2.5.39. Cho a1 v a2 l hai s dng. Xt tnh hi t ca dy {an} c nhngha truy hi nh sau

an+1 = aneanan1 vi n 1.

2.5.40. Cho a > 1 v x > 0 , nh ngha {an} bi a1 = ax, an+1 =aan , n N. Hy xt tnh hi t ca dy.2.5.41. Chng minh rng

2 +

2 + ... +

2

n - cn

= 2 cos

2n+1.

S dng kt qu trn tnh gii hn ca dy truy hi cho bi

a1 =

2, an+1 =

2 + an, n 1.2.5.42. Cho {n} l dy sao cho cc s hng ch nhn mt trong ba gi tr1, 0, 1. Thit lp cng thc

12 + 22 + + n2 = 2sin

4

nk=1

12...k2k1

, n N.v chng t rng dy

an = 1

2 + 2

2 + + n

2

hi t.


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2.5.43. Tnh

limn

arctan

1

2+ arctan

1

2.22+ ... + arctan

1

2n2

.

2.5.44. Tnh limn

sin(

n2 + n).

2.5.45. Xt tnh hi t ca dy truy hi di y

a1 =

2, a2 =

2 +

3, an+2 =

2 +

3 + an vi n 1.


limn

1 + 2

1 + 3

1 + ...

1 + (n 1)1 + n = 3.

2.5.47. Cho a > 0, cho dy {an} bi

a1 < 0, an+1 =a

an 1 vi n N.

Chng minh rng dy trn hi t ti nghim m ca phng trnh x2 + x = a.

2.5.48. Cho a > 0, xt dy {an} :

a1 > 0, an+1 =a

an + 1vi n N.

Chng minh rng dy hi t ti nghim dng ca phng trnh x2 + x = a.

2.5.49. Cho {an} l dy truy hi cho bi cng thc sau

a1 = 1, an+1 =2 + an1 + an

vi n N.

Chng minh rng {an} l dy Cauchy v tm gii hn ca n.2.5.50. Chng minh rng dy nh ngha bi

a1 > 0, an+1 = 2 +1

an, n N,

l dy Cauchy v tm gii hn ca dy.


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56 Chng 2. Dy s thc

2.5.51. Cho a > 0, nh ngha

{an

}nh sau:

a1 = 0 an+1 =a

2 + anvi n N.

Hy xt tnh hi t ca dy {an} .2.5.52. Gi s rng a1 R v an+1 = |an 21n| vi n N. Hy xt tnhhi t ca dy v trong trng hp hi t hy tm gii hn .


(a) Nu 0 < a < 1 th

limn

n1j=1

jaj

n j = 0,

(b) Nu 0 < a < 1 th

limn

nann

j=1

1

jaj=

1

1 a,

(c) Nu b > 1 th

limnn

bn

n

j=1

bj1

j =

1

b 1 .

2.5.54. Tnh

limn

sin

n + 1+ sin

n + 2+ ... + sin

2n

.

2.5.55. Tnh

limn

n

k=1 1 +k2

cn3 , vi c > 0,(a)limn

nk=1

1 k

2

cn3

, vi c > 1.(b)

2.5.56. Xc nh

limn

n3n

n!

nk=1

sink

n

n.


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2.5.57. Cho dy

{an

}nh ngha theo cng thc sau:

an =n

k=0

n

k

1, n 1,

Chng minh rng limn

an = 2.

2.5.58. Tm gi tr sao cho dy

an =

1

1

n

1

2

n

...

1

n 1

n

, n 2,

hi t.

2.5.59. Vi x R, nh ngha {x} = x [x]. Tnh limn

(2 +

3)n

.

2.5.60. Cho {an} l mt dy dng v t Sn = a1 + a2 + ... + an, n 1.Gi s ta c

an+1 1Sn+1

((Sn 1)an + an1), n > 1.

Hy tnh limn

an.

2.5.61. Cho {an} l dy dng tho mnlimn

ann

= 0, limn

a1 + a2 + ... + ann

< .

Tnh

limn

a21 + a22 + ... + a

2n

n2.

2.5.62. Xt hai dy dng {an} v {bn} tho mn

limn

an

a1 + a2 + ... + an

= 0 limn

bn

b1 + b2 + ... + bn

= 0.

nh ngha dy {cn} nh sau:

cn = a1bn + a2bn1 + ... + anb1, n N.

Chng minh rng

limn

cnc1 + c2 + ... + cn

= 0.


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58 Chng 2. Dy s thc

2.5.63. Tnh

limn

1 + 1nn2 en.

2.5.64. Gi s dy {an} b chn trn v tho mn iu kin

an+1 an > 1n2

, n N.

Hy thit lp s hi t ca dy {an} .2.5.65. Gi s dy {an} b chn tho mn iu kin

an+12n

2 an, n N.Hy thit lp s hi t ca dy {an} .2.5.66. K hiu l v L tng ng l gii hn di v gii hn trn ca dy{an} . Chng minh rng nu lim

n(an+1 an) = 0 th mi im trong khong

m(l, L) l im gii hn ca {an} .2.5.67. K hiu l v L tng ng l gii hn di v gii hn trn ca dy{an} . Gi s rng vi mi n, an+1 an > n, vi n > 0 v lim

nn = 0.

Chng minh rng mi im trong khong m(l, L) l im gii hn ca

{an

}.

2.5.68. Cho {an} l dy dng v n iu tng. Chng minh rng tp ccim gii hn ca dy

ann + an

, n N,l mt khong, khong ny suy bin thnh mt im trong trng hp hi t.

2.5.69. Cho a1 R, xt dy {an} nh sau:

an+1 =

an2

nu n chn,1+an

2nu n l.

Tm cc im gii hn ca dy trn.

2.5.70. Liu 0 c phi l mt im gii hn ca dy {n sin n} ?2.5.71. Chng minh rng vi dy dng {an} ta c

limn

a1 + an+1

an

n e.


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2.5.72. Chng minh kt qu tng qut ca bi ton trn: Cho s nguyn dng

p v dy dng {an}, Chng minh rng

limn

a1 + an+p

an

n ep.

2.5.73. Chng minh vi dy dng {an} ta c

limn

n

1 + an+1

an 1

1.

Chng minh 1 l hng s tt nht c th c ca bt ng thc trn.

2.5.74. Cho

an =

1 +

1 + ... +

1

n - cn

Tm limn

an.

2.5.75. Cho {an} l dy vi cc phn t ln hn 1 . Gi s ta c

limn

lnln ann

= ,

Xt dy {bn} nh sau:

bn =

a1 +

a2 + ... +

an, n N.

Chng minh rng nu < ln 2 th{bn} hi t, ngc li nu < ln 2 th dyphn k ti .2.5.76. Gi s cc s hng ca dy ca dy {an} tho mn iu kin

0

an+m

an + am vi n, mR.

Chng minh rng gii hn limn

ann

tn ti.

2.5.77. Gi s cc s hng ca dy ca dy {an} tho mn iu kin0 an+m an am vi n, m R.

Chng minh rng gii hn limn

n

an tn ti.


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60 Chng 2. Dy s thc

2.5.78. Gi s cc s hng ca dy ca dy

{an

}tho mn iu kin

|an| 1,an + am 1 an+m an + am + 1

vi n, m N.

(a) Chng minh rng gii hn limn

ann

tn ti.

(b) Chng minh rng nu gii hn limn

ann

= g th

ng

1

an

ng + 1 vi n

N.

2.5.79. Cho {an} l dy dng v n iu tng tho mn iu kinan.m nam vi n, m N.

Chng minh rng nu sup

ann

: n N < + th dy ann

hi t.

2.5.80. Cho hai s dng a1 v a2, chng minh dy truy hi {an} cho bi

an+2 =2

an+1 + anvi n N

hi t.

2.5.81. Cho b1 a1 > 0, xt hai dy {an} v {bn} cho bi cng thc truyhi:

an+1 =an + bn

2, bn+1 =

an+1bn vi n N.

Chng minh rng c hai dy u hi t ti cng mt gii hn.

2.5.82. Cho ak,n, bk,n, n N, k = 1, 2,...,n, l hai bng tam gic cc s thcvi bk,n = 0. Gi s rng ak,nbk,n n 1 u i vi k, c ngha l vi mi > 0,lun tn ti mt s dng n0 sao cho

ak,nbk,n 1 < vi mi n > n0 v k = 1, 2,...,n. Chng minh rng nu lim

n

nk=1

bk,n tn ti

th

limn

nk=1

ak,n = limn

nk=1

bk,n.


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2.5.83. Cho a

= 0, tm

limn

nk=1

sin(2k 1)a

n2.

2.5.84. Vi a > 0, tnh

limn

nk=1

a

k

n2 1

.

2.5.85. Tnh

limn

nk=1

1 + kn2 .2.5.86. Vi p = 0 v q > 0, hy tnh

limn

nk=1

1 +

kq1

nq

1p

1

.

2.5.87. Cho cc s dng a, b v d vi b > a, tnh

limna(a + d)...(a + nd)

b(b + d)...(b + nd) .


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71/364

Ch-ng 3

Chui s thc

Tm tt l thuyt

Cho chui hnh thc

n=1

an.(A )

an c gi l s hng thn hay s hng tng qut ca chui (A).Dy cc tng ring ca chui (A) c nh ngha l

sn =n

k=1

ak, n N.

sn c gi l tng ring th n ca chui (A).

Ni rng chui (A) hi t v c tng bng s, nu

limn

sn = s.

Trong trng hp ny, phn d ca chui (A) c nh ngha l

rn = s sn =

k=n+1

ak, n N.

Ni rng chui (A) phn k, nu gii hn ni trn khng tn ti

63


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64 Chng 3. Chui s thc

iu kin cn chui (A) hi t l

limn

an = 0.

iu kin cn v chui (A) hi t l: vi > cho trc, tnti n N sao cho

n+pk=n

ak

< , n > n, p N.

(A) c gi l chui dng nu an 0 vi mi n. Tiu chun so snh. Cho hai chui dng (A) v (B)

n=1

bn.(B)

Gi san bn n N.

Khi ,nu chui (B) hi t, th chui (A) cng hi t;

nu chui (A) phn k, th chui (B) cng phn k.

c bit, nu

limn

anbn

= k = 0,

th hai chui (A), (B) cng hi t hoc cng phn k.

Tiu chun t s (D'Alembert). Cho chui dng (A).Nu

limn

an+1an

< 1,

th chui (A) hi t.

Nulimn

an+1an

> 1,

th chui (A) phn k.


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Tm tt l thuy t 65

c bit, gi s tn ti gii hn

a = limn

an+1an

,

khi , nu a < 1 th chui (A) hi t; nu a > 1 th chui (A) phn k.

Tiu chun cn (Cachy). Cho chui dng (A). Gi s tn ti giihn

c = limn

n

an,

khi , nu c < 1 th chui (A) hi t; nu c > 1 th chui (A) phn k.

Tiu chun Raabe. Cho chui dng (A).Nulimn

n

an

an+1 1

> 1,

th chui (A) hi t.

Nu

limn

n

an

an+1 1

< 1,

th chui (A) phn k.

c bit, gi s tn ti gii hn

r = limn

n(an

an+1 1)

khi , nu r > 1 th chui (A) hi t; nu r < 1 th chui (A) phn k.

Ni rng chui (A) hi t tuyt i, nu chui (gm cc tr stuyt i)

n=1

|an|

hi tu.Chui hi t tuyt i th hi t. iu ngc li, ni chung, khng ng.

Ni rng chui (A) hi t c iu kin hay bn hi t, nuchui n hi t nhng khng hi t tuyt i.

Chui an du l chui c dng

b1 b2 + b3 + (1)n1 + , bn 0.


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Tiu chun Leibniz ni rng, nu dy s

{bn

}n iu gim v hi

t v0 th chui an du hi t. Php bin i Abel Cho hai chui bt k (A) v (B). t

An =n

k=1

ak, Bn =n

k=1

bk, Cn =n

k=1

akbk.

Khi ta c

Cn = anBn n1k=1

(ak+1 ak)Bk.

Tiu chun Abel. Cho hai chui bt k (A) v (B). X t chui (C)nh sau

n=1

anbn.(C)

Nu chui (B) hi t v dy {an} n iu v b chn th chui (C) hi t. Tiu chun Dirichlet. Nu dy {An} b chn, dy {bn} n iu

v c gii hn bng 0 th chui (C) hi t.


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3.1. Tng ca chui 67

3.1 T ng ca chui

3.1.1. Tm cc chui v tng ca chng nu dy {Sn} cc tng ring ca chngc cho nh sau:

(a) Sn =n + 1

n, n N, (b) Sn = 2

n 12n

, n N,

(c) Sn = arctan n, n N, (d) Sn = (1)n

n, n N,

3.1.2. Tm tng ca cc chui

(a)

n=1

2n + 1

n2(n + 1)2 , (b)

n=1

n

(2n 1)2(2n + 1)2 ,

(c)

n=1

n n2 1n(n + 1)

, (d)

n=1

1

4n2 1 ,

(e)

n=1

1

(

n +

n + 1)

n(n + 1).

3.1.3. Tnh cc tng sau

ln1

4+

n=1 ln (n + 1)(3n + 1)n(3n + 4) ,(a)n=1

ln(2n + 1)n

(n + 1)(2n 1) .(b)

3.1.4. Tm tng ca cc chui

n=1

1

n(n + 1) . . . (n + m), m N,(a)

n=11

n(n + m), m N,(b)

n=1

n2

(n + 1)(n + 2)(n + 3)(n + 4).(c)

3.1.5. Tnh

(a)

n=1

sinn!

720, (b)

n=1

1

n

ln n

n ln n

.


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3.1.6. Tnhn=1

sin 12n+1

cos 32n+1

.

3.1.7. Tm n=0

1

n!(n4 + n2 + 1).


n=1n

3

5

. . .

(2n + 1)

=1

2.

3.1.9. Gi s{an} l mt dy tho mnlimn

((a1 + 1)(a2 + 1) . . . (an + 1)) = g, 0 < g +.

Chng minh rng

n=1

an(a1 + 1)(a2 + 1) . . . (an + 1)

= 1 1g

.

( qui c: 1

= 0).

3.1.10. Dng kt qu trong bi ton trc, tm tng ca cc chui

n=1

n 1n!

,(a)

n=1

2n 12 4 6 . . . 2n ,(b)

n=2

1n2

1 1

22

1 1

32

...

1 1

n2

.(c)

3.1.11. Gi {an} l dy cho bia1 > 2, an+1 = a

2n 2 vi n N.

Chng minh rng

n=1

1

a1 a2 . . . an =a1

a12 4

2.


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3.1. Tng ca chui 69

3.1.12. Vi b > 2, kim tra rng

n=1

n!

b(b + 1) . . . (b + n 1) =1

b 2 .

3.1.13. Cho a > 0 v b > a + 1, chng minh ng thc

n=1

a(a + 1) . . . (a + n 1)b(b + 1) . . . (b + n 1) =

a

b a 1 .

3.1.14. Cho a > 0 v b > a + 2, kim tra ng thc sau

n=1

a(a + 1) . . . (a + n 1)b(b + 1) . . . (b + n 1) =

a(b 1)(b a 1)(b a 2) .

3.1.15. Cho

n=1

1an

l chui phn k vi cc s hng dng. Cho trc b > 0,

tm tng

n=1

a1 a2 . . . an(a2 + b)(a3 + b) . . . (an+1 + b)

.

3.1.16. Tnh n=0

(1)n cos3 3nx

3n.

3.1.17. Cho cc hng s khc khng a, b v c, gi s cc hm f v g tho mniu kin f(x) = af(bx) + cg(x).

(a) Chng minh rng nu limn

anf(bnx) = L(x) tn ti th

n=0

an

g(bn

x) =

f(x)

L(x)

c .

(b) Chng minh rng nu limn

anf(bnx) = M(x) tn ti th

n=0

ang(bnx) =M(x) af(bx)

c.


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3.1.18. Dng ng nht thc sin x = 3 sin x3

4sin3 x

3, chng minh rng

n=0

3n sin3x

3n+1=

x sin x4

,(a)

n=0

1

3nsin3

x

3n+1=

3

4sin

x

3.(b)

3.1.19. Dng ng nht thc cot x = 2 cot(2x) + tan x vi x = k 2

, k Z,chng minh rng

n=01

2n

tanx

2n

=1

x 2cot(2x).

3.1.20. Dng ng nht thc arctan x = arctan(bx) + arctan (1b)x1+bx2

, thitlp cc cng thc sau:

n=0

arctan(1 b)bnx

1 + b2n+1x2= arctan x vi 0 < b < 1,(a)

n=0

arctan(b 1)bnx

1 + b2n+1x2= arccot x vi x = 0 v b > 1.(b)

3.1.21. Cho {an} l dy Fibonacci c xc nh bia0 = a1 = 1, an+1 = an + an1, n 1

v t Sn =n

k=0

a2k . Tm

n=0

(1)nSn

.

3.1.22. Vi dy Fibonacci {an} trong bi trn, tnh

n=0

(1)nanan+2

.

3.1.23. Vi dy Fibonacci {an} trong bi trn, xc nh tng

n=1

arctan1

a2n.


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3.1. Tng ca chui 71

3.1.24. Tm tng

(a)

n=1

arctan2

n2, (b)

n=1

arctan1

n2 + n + 1,

(c)

n=1

arctan8n

n4 2n2 + 5 .

3.1.25. Cho {an} l dy dng phn k ti v cng. Chng minh rng

n=1arctan

an+1 an1 + anan+1

= arctan1

a1.

3.1.26. Chng minh rng vi bt k hon v no ca cc s hng ca chuidng, tng ca chui nhn c khng thay i.

3.1.27. Chng minh ng nht thc

n=1

1

(2n 1)2 =3

4

n=1

1

n2.


n=11

n2

=2

6

,(a)

n=1

1

n4=

4

90,(b)

n=0

(1)n 12n + 1

=

4(c)

3.1.29. Cho dy {an} c xc nh bia1 = 2, an+1 = a

2n an + 1 vi n 1.

Tm n=1

1an .

3.1.30. Cho dy {an} c xc nh nh sau

a1 > 0, an+1 = lnean 1

anvi n 1,

v t bn = a1 a2 . . . an. Tmn=1

bn.


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3.1.31. Cho dy

{an

}c xc nh bi

a1 = 1, an+1 =1

a1 + a2 + . . . + an

2 vi n 1.

Tm tng ca chui

n=1

an.

3.1.32. Tm tng ca cc chui sau

n=1(1)n1 1

n,(a)

n=1

(1)n1 2n + 1n(n + 1)

,(b)

n=1

1

x + 2n 1 +1

x + 2n 1

x + n

, x = 1, 2, . . . .(c)

3.1.33. Tnh n=1

(1)n1 ln

1 +1

n

.

3.1.34. Tnh n=1

(1)n1 ln

1 1(n + 1)2

.

3.1.35. Xc nh tng ca cc chui

n=1

1

n ln

1 +

1

n

.

3.1.36. Gi s hm f kh vi trn (0, +), sao cho o hm f ca n niu trn mt khong con (a, +), v limx f(x) = 0. Chng minh rng giihn

limn+

1

2f(1) + f(2) + f(3) + . . . + f(n 1) + 1

2f(n)

n1

f(x)dx

tn ti. Xt c c trng hp c bit ca khi hm f(x) c dng f(x) = 1

xv

f(x) = ln x.


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3.1. Tng ca chui 73

3.1.37. Xc nh tng ca chui

n=1

(1)n ln nn

.

3.1.38. Tm n=1

n ln

2n + 1

2n 1 1

.

3.1.39. Cho trc s nguyn k 2, chng minh rng chui

n=1 1(n 1)k + 1 + 1(n 1)k + 2 + . . . + 1nk 1 xnk

hi t i vi duy nht mt gi tr ca x. Tm gi tr ny v tng ca chui.

3.1.40. Cho dy {an} c xc nh bi

a0 = 2, an+1 = an +3 + (1)n

2,

tnh

n=0(1)[n+12 ]

1

a2n 1 .3.1.41. Chng minh rng tng ca cc chui

(a)

n=1

1

n!, (b)

n=1

1

(n!)2

l v t.

3.1.42. Cho {n} l dy vi n nhn hai gi tr 1 hoc 1. Chng minh rng

tng ca chui

n=1 nn! l s v t.3.1.43. Chng minh rng vi mi s nguyn dng k , tng ca chui

n=1

(1)n(n!)k

l v t.


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3.1.44. Gi s rng

{nk

}l dy n iu tng cc s nguyn dng sao cho

limk

nkn1n2 . . . nk1 = +.

Chng minh rngi=1

1ni

l v t.

3.1.45. Chng minh rng nu {nk} l dy cc s nguyn dng tho mn

limk

nkn1n2 . . . nk1 = + v limk

nknk1

> 1,

thi=1

1ni

l v t.

3.1.46. Gi s rng {nk} l dy n iu tng cc s nguyn dng sao cho

limk2k

nk = . Chng minh rngk=1 1nk l v t.

3.1.47. Gi s chui

n=1

pnqn

, pn, qn N l chui hi t v gi s

pnqn 1

pn+1qn+1 1

pnqn

.

K hiu A l tp tt c cc s n sao cho bt ng thc trn c du > . Chngminh rng

n=1

pnqn

v t khi v ch khi A l v hn.

3.1.48. Chng minh rng vi mi dy tng ngt cc s nguyn dng {nk},tng ca chui

k=1

2nknk!

l v t.


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3.2. Chui dng 75

3.2 Chui d-ng

3.2.1. Cc chui sau hi t hay phn k

(a)

n=1

(

n2 + 1 3

n3 + 1), (b)

n=1

(n2 + 1

n2 + n + 1)n

2

,

(c)

n=2

(2n 3)!!(2n 2)!! , (d)

n=1

(n

n + 1)n(n+1),

(e)

n=1 1 cos1

n , (f)

n=1(n

n

1)n,

(g)n=1

( n

a 1), a > 1.

3.2.2. Kim tra s hi t ca cc chui sau y

(a)

n=1

1

nln

1 +

1

n

, (b)

n=2

1n

lnn + 1

n 1 ,

(c)

n=1 1n2 ln n, (d)

n=2 1(ln n)ln n ,(e)

n=2

1

(ln n)lnlnn.

3.2.3. Chon=1

an,n=1

bn l cc chui dng tho mn

an+1an

bn+1bn

vi n n0.

Chng minh rng nu

n=1

bn hi t, th

n=1

an cng hi t.


(a)

n=1

nn2

enn!, (b)

n=1

nn

enn!.


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3.2.5. Tm gi tr ca cc chui sau hi t

(a)

n=1

n

a 1 , a > 1, (b) n=1

n

n 1 ,(c)

n=1

1 +

1

n

n+1 e

, (d)

n=1

1 n sin 1

n

.

3.2.6. Chng minh rng nu chui dng

n=1

an hi t th

n=1

(aan 1) , a > 1

cng hi t.

3.2.7. Kho st s hi t ca cc chui sau

(a)n=1

ln

cos1

n

, (b)

n=1

ea lnn+bc lnn+d ,a,b,c,d R

(c) n=1

n2n

(n + a)(n+b)(n + b)(n+a), a, b > 0.

3.2.8. Gi s chui

n=1

an vi cc s hng khng m hi t. Chng minh rng

n=1

anan+1 cng hi t. Chng minh rng iu ngc li l khng ng, tuy

nhin nu dy {an} n iu gim th iu ngc li ng.

3.2.9. Gi s rng chui dng

n=1 an phn k. Nghin cu s hi t cc chuisau y(a)

n=1

an1 + an

, (b)

n=1

an1 + nan

,

(c)

n=1

an1 + n2an

, (d)n=1

an1 + a2n

.


85/364

3.2. Chui dng 77

3.2.10. Gi s chui dng

n=1 an phn k, k hiu dy cc tng ring ca nl {Sn} . Chng minh rng

n=1

anSn

phn k,

v n=1

anS2n

hi t.

3.2.11. Chng minh rng vi cc gi thit nh ca bi trc, chui

n=2

an

SnSn1

hi t vi mi > 0.

3.2.12. Chng minh rng cc gi thit cho bi tp 3.2.10 , chui

n=2anSn

hi t nu > 1 v phn k nu 1.

3.2.13. Cho chuin=1

an hi t, k hiu rn =

k=n+1

ak, n N l dy ccphn d ca n. Chng minh rng

n=2

anrn1

phn k,(a)

n=2an

rn1hi t.(b)

3.2.14. Chng minh rng vi cc gi thit c cho bi trc , chui

n=2

anrn1

hi t nu < 1 v phn k nu 1.


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3.2.15. Chng minh rng vi gi thit nh bi 3.2.13, chui

n=1 an+1 ln2 rnhi t.

3.2.16. Cho chui dng

n=1

an. Gi s rng

limn

n lnan

an+1= g.

Chng minh rng

n=1

an hi t nu g > 1 v phn k nu g < 1 (k c trng

hp g = + v g = ). Hy a v d chng t rng khi g = 1 th takhng th a ra kt lun c.

3.2.17. Nghin cu s hi t ca cc chui sau y

(a)

n=1

1

2n

, (b)

n=1

1

2lnn,

(c)

n=1

1

3ln n, (d)

n=1

1

alnn, a > 0,

(e)

n=2

1

alnlnn, a > 0.

3.2.18. Kho st s hi t ca chui

n=1

a1+12+...+

1n , a > 0.

3.2.19. Dng kt qu ca bi ton 3.2.16, chng minh dng gii hn ca Tiuchun Raabe.Cho an > 0, n N, t

limn

n

an

an+1 1

= r.

Chng minh rng n=1

an hi t nu r > 1 v phn k nu r < 1.

3.2.20. Cho dy {an} c xc nh bi

a1 = a2 = 1, an+1 = an +1

n2an1 vi n 2.

Nghin cu s hi t ca chui

n=1

1an

.


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3.2. Chui dng 79

3.2.21. Cho a1 v l cc s dng. Dy

{an

}c xc nh nh sau

an+1 = anean , vi n = 1, 2, . . . .

Hy xc nh v chuin=1

an hi t.

3.2.22. Xc nh a chui

n=1

n!

(a + 1)(a + 2) . . . (a + n)

hi t.3.2.23. Cho a l s dng tu v {bn} l dy s dng hi t ti b. Nghincu s hi t ca chui

n=1

n!an

(a + b1)(2a + b2) . . . (na + bn) .

3.2.24. Chng minh rng nu dy cc s dng {an} tho mnan+1

an= 1

1

n n

n ln n

,

trong n > 1, thn=1

an hi t. Mt khc, nu

an+1an

= 1 1n

nn ln n

,

trong n < 1, thn=1

an phn k. (Tiu chun Bertrand.)

3.2.25. Dng tiu chun Bertrand v Raabe chng minh tiu chun Gauss.

Nu dy cc s dng {an} tho mnan+1

an= 1

n n

n,

trong > 1, v {n} l dy b chn, thn=1

an hi t khi > 1 v phn

k nu 1.


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3.2.26. Kho st s hi t ca chui

n=1

( + 1) . . . ( + n 1)n!

(+ 1) . . . (+ n 1)(+ 1) . . . (+ n 1)

y , v l cc hng s dng.

3.2.27. Tm gi tr ca p chui

n=1

(2n 1)!!

(2n)!!

phi t.

3.2.28. Chng minh tiu chun c c ca Cauchy.

Cho {an} l dy n iu gim cc s khng m. Chng minh rng chui

n=1

an hi t khi v ch khi chui

n=1

2na2n hi t.


(a)

n=21

n(ln n), (b)

n=31

n

ln n

lnln n.

3.2.30. Chng minh nh l Schlomilch (suy rng ca nh l Cauchy, xembi tp 3.2.28).

Nu {gk} l dy tng ngt cc s nguyn dng sao cho vi c > 0 no vvi mi k N, gk+1 gk c(gk gk1) v vi dy dng {an} gim ngt,ta c

n=1

an < khi v ch khi

n=1

(gk+1 gk)agk < .

3.2.31. Cho {an} l dy n iu gim cc s dng. Chng minh chuin=1 an

hi t khi v ch khi cc chui sau hi t

(a)

n=1

3na3n, (b)

n=1

nan2, (c)

n=1

n2an3,

(d) S dng tiu chun trn hy nghin cu s hi t ca cc chui trong bitp 3.2.17.


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3.2. Chui dng 81

3.2.32. Gi s{

an}

l dy dng. Chng minh rng chui

n=1 an hi t nulimn

(an)1

lnn 1 th chui hi t, cn nu g < 1 th chui phn k( y g c th bng ).

Cho v d chng t rng trong trng hp g = 1 th cha th c kt lun g.

3.2.58. Chng t rng tiu chun Raabe (xem 3.2.19) v tiu chun cho trongbi tp 3.2.16 l tng ng. Hn na, chng t rng khng nh trong bitp trn l mnh hn cc tiu chun .

3.2.59. Nghin cu s hi t ca chui

n=1

an vi cc s hng c cho bi :

a1 =

2, an =

2

2 +

2 + . . . +

2

(n 1) - cn, n 2.

3.2.60. Cho {an} l mt dy n iu gim ti 0. Chng t rng nu dy sc s hng tng qut l

(a1 an) + (a2 an) + . . . + (an1 an)

b chn th chui

n=1

an phi hi t.


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3.2.

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