數(shù)學(xué)分析簡(jiǎn)明教程答案10數(shù)項(xiàng)級(jí)數(shù)

yaaxax2axn 則必有ai0,i1,2,證明：若yaaxax2axn微分方程的一個(gè)解 y'a2ax3ax2na y''2a6axn(n1)a

xy''2ax6ax2n(n1)a xyaxax2ax3a xy''y'xya(4aa)x(9aa)x2(n2a )xn1axn

n2a n

試確定系數(shù)aa,a,axn (1x2)y''2xy'l(l1)y

nanxn1 n(n1)anxn2

1)1)axn2n(n axn2naxnl(ln 2a1a ,n2,3,a (n y(x)a1l(l1)x2l(l2)(l1)(l3)x40 ax(l1)(l2)x3(l1)(l3)(l2)(l4)x51

1

(5n4)(5n1)

5n45nn1 1 1 1 5n 6 11 16 5n11lim 5n 5n1 1 2n 2n n1 1 1 1

2n 3 5 7 2n 2n1 1lim 2n 2 2n1 n1

11 2

3n

2n;

2n

2n 2n 設(shè)S ,則S ,于是S S

n

1 2于是S

rnsinnx,

n解：記Sn kn

rk12cosxsinnknrk1sin(k1)xsin(kk

Sn

1r22rcos

rnsinnxlimn

rsinx 1r22rcosxrncosnx,

n解：記Sn rkcoskx.kn2rcosxSnrk12cosxcosknrk1cos(k1)xcos(kk

Sn

1r22rcos rncosnxlimSn .

1r2rcos 10, n2n 11 由于級(jí)數(shù)1

1 3n

n1 cos limcos . nn(n(nn證明定理10.2若級(jí)數(shù)unvn收斂,則級(jí)數(shù)unvn也收斂,且unvnun

可設(shè)limUnulimVn 收斂,

limUnVnlimUnlimVnu即

unvnun

,n0,1,

k

k

k其中k00,k0k1k2knkn1Un1Un

uk1uk2

1 2

那么對(duì)于級(jí)數(shù)un有0,NkN10,當(dāng)nN時(shí),對(duì)于p0un1unpUN1UNp 由于

發(fā)散

n2

n2

n1

由 1

nn

(2n

由 收斂那么可以知道原級(jí)數(shù)收斂 n1;n ;

由于limn 1,那么可以知道原級(jí)數(shù)發(fā)散 n2n sin 由于sin,而級(jí)數(shù)收斂

n1

,

由于 1,而級(jí)數(shù)

n11 1 n1

1 由于

,

n1nn nnn n1

由于 2n ,而級(jí)數(shù)收斂 2n

n1

nlnnnlnn 1n

nlnn2 2

n,而級(jí)數(shù)n收斂 n n32sinn3

由于2sinnn,而級(jí)數(shù)n收斂 n 3(1)nsin 由于3(1)sin n,而級(jí)數(shù)收斂,故原級(jí)數(shù)收斂 n1

1sin2n

由于limun1lim(n1)n!

1 n1

由于sin2n1而級(jí)數(shù)1收斂 n12n1cos1 由于lim2

1cos11,而級(jí)數(shù)

n

n n1 由于lim cos

1 n3 由于 n3

1

1,而級(jí)數(shù)

nn1 n n1n

由于limn2ln(1n)01,

1收斂,

3n1 sin1arcsin1 由于limnsin1arcsin1

t t n1 由于narctann,而級(jí)數(shù)

n

n111n

1

111nn2n 4n11nn2n 4n2n1 1 11n21n2n4 n1 n1 ee

n limun1lim(n1)n!n

(n1)nlim11n

n

n n

11,

nnu

n limn1 n1

n n(n n

(n nn limn1lim lim n1 n n(n nn

n

n1n limn n1limn n11limn n1

n n(n n n (n 1n1

p§ 11xxe

21(17 n 1nn

由于

limen nelim 21, n n 22n12 由于limn2n1

232n232n2n12

1313

由于

n1

n13n2n (1x)(1x2)(1x3)(1xn),(x解：當(dāng)x0時(shí),

00,收斂

(1x)(1x2)(1x3)(1xn

(1(1x)(1x2)(1x3)(1xn

33535735793579111 4 7 710

nn

31 解：由于N9,當(dāng)nN11,而且

nln n1 n1(lnn)ln 解：由于Nee21,當(dāng)n 1

e2ln

1n而2n 1

解:由于1

1,

n12ln eln n111

解:由于1

1 ,1

n13ln n1 n1n1;n;n13n113n解：由于limlnn0,故N0,當(dāng)nNn

nlnn,即 1

3 3 3

解：由于

0,而級(jí)數(shù)n2收斂,因此級(jí)數(shù)3n

lnnnp解：當(dāng)p1時(shí),有l(wèi)nn1,而調(diào)和級(jí)數(shù)發(fā)散,因此次級(jí)數(shù)發(fā)散 npln

0;由于級(jí)數(shù)

n

n1

發(fā)散p綜上可知級(jí)數(shù)

p n2nplnn

解：當(dāng)p1時(shí),級(jí)數(shù)nplnnnlnn;函數(shù)f(xxlnx在區(qū)間[2,) 然有f(n) ;極限lim dxlimt1dlnxlnlnx,那么由柯西積分判

t2xln t2ln 由于當(dāng)p1時(shí), npln

nplnn

p,而級(jí)數(shù)nnnn

綜上可知級(jí)數(shù) 1ne1nn1

1 1 1 1 1 1 lnn

on2n2

n2

1

e en

e 1

n 1 12non

1

p 1e e n

e1

e po p n

n1

n

n

n1

n

lnp1 n n 1lnp1sin2 n 1sin2osin22p n1sin2osin22p n1

o2p n2 因此它與n2

同階無(wú)窮小量,于是當(dāng)p1時(shí),級(jí)數(shù)收斂;當(dāng)p1

nplnnnnn

n

pln

p

1nn

nln n1n1 nn

n n1 n n因此于是級(jí)數(shù)的通項(xiàng)是與1p同階的無(wú)窮小量,于是當(dāng)p0時(shí),級(jí)數(shù)收斂;

nn4n2nbn

nna n2nnna n2nna4n2nna2n2n na4n2nbnan2nb(2a1)nba2 nana4n2nna n2n因此當(dāng)a11是同階無(wú)窮小量,故此時(shí)收斂;1 解：對(duì)于函數(shù)f(x) x(lnx)

,我們知道它在區(qū)間[2,)上連續(xù)且單調(diào)遞減,顯然有f(nun 極限 f(x)dx t

t

2x(lnx) dxlim dxlnlnx,那么由柯西積分判別法可tt2x(lnx)

t2xln 1 n(lnn)

dx t

lnx1plimlnx1 ln21

t2x(lnx) 1

2

1 1

p.p 解：函數(shù)f(x) 在區(qū)間[2,)上連續(xù)且單調(diào)下降,顯然有f(n)u;我們以xlnxlnln

limtf(x)dxlim dxlnlnlnxt t2xlnxlnln ,1

解：當(dāng)0時(shí),由于

n2nln n2nlnnplnln 收斂, q nlnnlnln nln n2nln若q1時(shí),由于函數(shù)f(x

lim

dx

1lnlnlnx1q 1lnlnln21qt2xlnxlnln 1

2q2 nnlnnplnln

.(2n1)!!(2n)!!p是實(shí)數(shù)n1 (2n1)!!解：此級(jí)數(shù)的通項(xiàng)為un(2n)!! (2n1)!!p(2n2)!! 2n2n

(2n)!! (2n1)!! 2n1 2n2 limn n1limn n

2n1 2nlimn 2nn 1 limn1 o 1n 2n n

(1)(n1)1 (1)(n1) (n n1n1n

(1)(n1)(n)(n nn n1n1 limn n1limn n nn limn

nnn 1x1lim1

11x1x1x11因此當(dāng)時(shí)級(jí)數(shù)收斂,時(shí)級(jí)數(shù)發(fā)散。當(dāng)

unn

n

n2k1 nn2k a2a nn

x xN1

n nM

n n 2 2設(shè)

n

0,因此0,N20,當(dāng)nN2時(shí) 綜上0,Nmax(N1N20,當(dāng)nNaa12a2nana12a2N11aN1N12aN2N13aN3a1a12a2N1nN1

nMa

N1

aN

1于是可知lima12a2nan1 2a1,nk2,k1,2設(shè) ,k1, k

收斂

n1

n1

kk

k1kn級(jí)數(shù)2nnk

k

k klimk2 0;因此由海涅原理可知極限limna不存在,于是可知limnak k2 n設(shè)a0,且liman1l.求證 l,n n

nn

n a3n證明lim a1 ;由于an0可知 因此可知

nn

a2

a3a2n

a1,nk2,k1,反之不成立

,

1,而liman1nn 1,k1, nnn k

n2nn2證明：取un 2那么1

2

1n

n

nn1 n

nn收斂那么可知必有

22n1 n22n證明：取un an!,那limun1lim2n n

n1 n12設(shè)a0,且數(shù)列na有界證明：級(jí)數(shù)a2 證明：由于數(shù)列na有界,因此我們可設(shè)M0,使得 M,n.那么可知aM, M ann2,我們知道級(jí)數(shù)n2收斂,于是根據(jù)比較判別法可知級(jí)數(shù)an

an

n1an

14設(shè)limal

,因此根據(jù)比較判別法可知級(jí)數(shù)22

n1l1,

nanl l 時(shí)

nan發(fā)散 時(shí)會(huì)有什么結(jié)論(1)若l1,那么0,使得l12;由于limal根據(jù)極限的保號(hào)性可知N,當(dāng)n時(shí),an1

n1N11 1n n n n1n n n n1 nN nN nN

1收斂,因此可知級(jí)數(shù)a n n(2若l1,那么0,使得l12;由于limal根據(jù)極限的保號(hào)性可知N,當(dāng)n時(shí),an1

n1N11 1n1 n1 nN2 n1 nN2

1發(fā)散,因此可知級(jí)數(shù)

1nN2

n1

nan可能收斂也可能發(fā)散。我 數(shù)列an ,顯然有限limalim1plnlnnn n n 但是對(duì)于級(jí)數(shù) ,可知 n(lnn)p兩邊取對(duì)數(shù)易知等式成立;n1n1 n1 p

pn 時(shí),級(jí) a收斂, n

nnn

,n

n1

lnnsinn是一個(gè)交錯(cuò)級(jí)數(shù),且lnnsin lnnsin 而級(jí) 必是收斂的那么這兩個(gè)級(jí)數(shù)的和也是收斂的即原級(jí)數(shù)收斂k 111 n

n1n1n1n1n1nnn n n n111 1111 n n1 n

111 n ,

n

nnnn nn1 nn n(1)n(1)n1(1)n11nnn

o nn nn

o nn nn

nn

(1)n 1nn2n

o1nn 1 n我們知道級(jí)數(shù) ,on都收斂,而級(jí)數(shù)發(fā)散,因此原級(jí)數(shù)發(fā)散nsin

n21;

n2 n2n21(1)nsinn2

n2n(1)n

n2n2

n可以證得從n2開(kāi)始n2n2

單調(diào)下降趨于零

n22 2

,而級(jí)數(shù)

必是收斂的;因此我們知道原級(jí)數(shù)絕對(duì)收斂,

(1)n,p1當(dāng)p011

1sinnn1

sin

,而級(jí)數(shù)

11 n1(1)ncos

n ncos解：我 級(jí)數(shù)n

n ncos S S 2 2knk

k

k k k 2sin1cos2k2sin1cos2ksin(2k1)sin(2k1)sin1sin(2n1

k k

nnk,

kcos2ksin1sin(2n1) 2 2

cosk

ncos4n同理可以知道級(jí)數(shù)

2n

sin2n n

nsin2n

(1)n1cos2n ncos我們知道級(jí)數(shù)

與級(jí)數(shù)

(1)nsin,x 可得：

1 (1)nsinx

o (1)no1 n

(1)nx,(1)n

1

n,

n

n n(1)n解：由

1 1 1 233nn2 233nn2 1 2nnnn對(duì)于N取n3N,p6N,

nn1npn1np1n11np

2n2n9N

2np2np23N29N3

n

1an

n

1a

1,由比較判別法可知原級(jí)數(shù)絕對(duì)收斂1 n隨n的變化單調(diào)有界,1

nsinn1 n sinn1 sinncos1cosnsin cosnsin n n n n cos1sin

當(dāng)n2時(shí) n, n單調(diào)趨于零,而sinn,cosn易證得有界,那么由狄理克雷判別法 sinncos cosnsin可知級(jí)數(shù) n, n都收斂,于是原級(jí)數(shù)收斂 nn解：取Nx1,則當(dāng)nN時(shí),(1)n

,n

sin(2n

n!,而級(jí)數(shù) n1

sinnx,0xn 解：級(jí)數(shù)sinnx的部分和為Snsinkx k 2sinxSn2sinxsinkxcos(k1)xcos(k1)x1cos(nk

ksinS1cos(n1)x sin 2sinn而數(shù)列1單調(diào)下降趨于零,nnsin2 n cos 容易證明級(jí)數(shù)

cosnx,0x

n1

n

,而級(jí)數(shù)nn

當(dāng)p0時(shí),由于limcosnx0,可知級(jí)數(shù)發(fā)散那么根據(jù)比較判別法可知當(dāng)p0

p綜上級(jí)數(shù)

p

;n解：可以證明N,當(dāng)nN

nn nn

4 455

sinn

,我們知道級(jí)數(shù) 收斂,因此由比較判別法可以知n1

nsinnsinnn n 1 n 1n nsinn

nno

n(1)n

no

n nn

n 1由萊布尼茲判別法可知級(jí)數(shù)

no n n(1)n 1

1 解：可以知道1n

n 1

1,pn1np(1)n (1)n 1 p 1 np1n1pon1p np 1n1

p,p (1)n 1 p p (1)n nn n . .

p知

np同階的無(wú)窮小量,因此級(jí)數(shù) p

1n1npp p 收斂,而數(shù)列1單調(diào)有界,那么由阿

p p

1 p 當(dāng)p0易知 p

n2nsin2n;n

n2 n2 n那么當(dāng)xkk時(shí)總有102sin2x,于是有n 4 當(dāng)xkk3時(shí),由于lim1n2sin2xn0, 4

n

2nsin2n nn4當(dāng)xk n4

nn

nx n1an

,limaan

x x解：當(dāng)xa時(shí),級(jí)數(shù)發(fā)散。此時(shí)lim nan

lim na

xxa時(shí),級(jí)數(shù)絕對(duì)收斂。此時(shí)有極限的保號(hào)性可知N,當(dāng)nN時(shí),0,1xx Nx x Nx n1an n1an nN1an n1an nN

1 當(dāng)xa時(shí),不能判斷級(jí)數(shù)的斂散性。例如an ,由于lim n11

e1

n 級(jí)數(shù)發(fā)散;但是對(duì)于a

1有l(wèi)ima1;x

1

pn

a

n

n1n

n1n

n,r

n0,故此時(shí)級(jí)數(shù)發(fā)散。當(dāng)r0,1

nnnnnn

xe

(n1)n故

nnnn

xex當(dāng)xe時(shí),由于lim n nln ,pn 2k1 np 2n2p3n3p4n4p 2k12k1

2kn2kpnn1

nn (1)n 1nn p p p o nnn (1)n1 n (1)n1 n nn n sinnn nn1np 4sinn sinn sinn sinn sinn sin2n; o1 o1;npsinn

np

n2 np

4 aaqaq2aqn,q1,aA,n0,1, 解：對(duì)于0,取N

1q

1,則對(duì)于p0及nN 1 1111111

1

1對(duì)于Nn3Np3N16an1an2an3an

an1an2an3

16N16N3N1313N

3N 3N 3N 3N 3N 6N 6N 16N N1116N3N 6N

6N 求證：若級(jí)數(shù)aa0n12,)收斂,則級(jí)數(shù)a2收斂但是反之不成立 證明：若級(jí)數(shù)a收斂,則必有l(wèi)ima0,那么N,當(dāng)nN時(shí)有 1,即此時(shí)有a2a成立n

n nn

a2a2a2a2 nN

nNnn 反之不成立,例如級(jí) n2收斂,但是級(jí)

n1 n若級(jí)數(shù)a收斂,且limn1,問(wèn)是否能斷定b也收斂？研究例子a ,ba nnn

答：不能,如題中所給例子：顯然有l(wèi)im

lim

n1lim(1)n ba

n1

ancnbn,n1,,證明：由題意可知對(duì)于0,N0,當(dāng)nN時(shí),對(duì)于pan1an2anbn1bn2bn由于ancnbnn1,2,,

an1an2anpcn1cn2cnpbn1bn2bnpcn1cn2cncn1cn2cn

an1an2anbn1bn2bn

1,

1,

n,

nx收斂,則當(dāng)xx0n

nx也收斂;n

nx0發(fā)散,則當(dāng)xx0n

證明：i.由于xx0可設(shè)xx0其中0;an

an1

n11

n1nx0 n1nx0

由于級(jí)數(shù)nx0收斂,而數(shù)列n單調(diào)有界,則由阿貝爾判別法可知級(jí)數(shù)nx

,數(shù)

nx0收 n1

其中Bn bnn,那

k k k k kanbnna limna

可設(shè)nanM成立。由于anan1是絕對(duì)收斂,因此0N0,當(dāng)nN,pn有k

akak1.同樣由于bn收斂0,N0,當(dāng)nN,p0

nk

M nn n

k

n akbkk

akak1BkanpBnpknp1a M

k

akak1biMMk

k

1M 求證：若級(jí)數(shù)a2和b2都收斂 ab,ab2,

n

n1

aaab nn

a2a2 a2

n n 于是可知級(jí)數(shù)ab收斂,那么可知級(jí)數(shù)ab2a2n n

x取ann,可知級(jí)數(shù)ann2,收斂那么級(jí)數(shù)anbnnx

x

n

設(shè)正項(xiàng)數(shù)列n單調(diào)上升且有界求證： n1 n1

1xn

xn1xn nn0nnn1 n1 nn0nn取取

1,則blimxx,由正項(xiàng)數(shù)列x單調(diào)上升且有界可知x 1 單

xnx

xn1那么由阿貝爾判別法可知級(jí)數(shù)anbn 1 n1 n1nk

Ankan1an2ankSnkSn,k0,1,2,,p,An

an2M

2M成立; nn n

k

n akbkbkbk1AnkAnpbnp2M

bkbk1bnpk

k k 由于級(jí)數(shù)bk收斂那么由柯西收斂原理可知0,N10,當(dāng)nN1p0np1 n k

bkk

bkbk14M又由于limb0,可知0,N0,當(dāng)nNbn 綜上可知0,NmaxN1N20,當(dāng)nNn n akbk2Mbkbk1bnp2M4M4Mk k

akbkSkbkSnbnSkbkk

k

kanbnSnbnlimSnbn 又因?yàn)镾n有界且limbn0可知limSnbn0,于是有anbn 設(shè)a收斂,且limna0,求證na 收斂,且有na a

k k k 由于limnan0,可知nanan1anbn 若an0,則a1a1a2a2a3a3收斂若an0,則a1a1a2a2a3a3收斂 若a收斂,則(1na收斂 答：錯(cuò);例如

(1)n,則級(jí)數(shù)a收斂;

(1)

n

n n1 答：正確;由a2收斂可知lima20,于是可知N,當(dāng)nN時(shí) 1;與此同此0,N,

nn n

nN時(shí),p0有

成立。那么0,NmaxNN,當(dāng)nNp0 k

np

n n nk

k

a2 k

kkk

nn答：錯(cuò)誤;調(diào)和級(jí)數(shù)

1發(fā)散但是顯然有l(wèi)im1n1 n 答：錯(cuò)誤;例如取an

n(1)nnnn,bn 1n(1)nnnn且有l(wèi)im

nnnnn

n 1 nnanbn n

n nn

n1

n原理可知對(duì)于0,N0,當(dāng)nN時(shí),p0有k

Mn n k

Mk

MM

(1)n 答：錯(cuò)誤;如級(jí)數(shù) 發(fā)散

nn1 n1n若an收斂an0,則limnan

p p

1

解：由于級(jí)數(shù)

在p 0,N n

于是可知有l(wèi)im 0成立n(n1) (n2) (2n)p lim1 1;n p2n 解：由于級(jí)數(shù)n在p1時(shí)收斂那么由柯西收斂原理可知對(duì)于0,N0,當(dāng)nN

于是可知有l(wèi)im1 10成立n p2n

證明：由級(jí)數(shù)an收斂根據(jù)柯西收斂原理可知0,N10,對(duì)于nN1pnk

k

那么對(duì)于0,K2N1,當(dāng)kK

,于是有l(wèi)imkak0,即limnan§5

k anananan證明：取a

,a

那么0aa,0aa;

a anananan

為S*那么顯然有0,

0,當(dāng)nNSS

; S*SS*SSSS*

S

S*

S*SSS0

S*SS*SSS

2k 2k 由an收斂可知必有l(wèi)iman0,那么0,N20,當(dāng)nN2時(shí)有an

S*S

2k 2k 2k 2k n

n

312 111進(jìn)行加括號(hào)得312

1 11 35 2 35

14n14n14n11314n1314n14n1取

,,1212

14n14n11112 14n14n