2019高等數(shù)學(xué)輔導(dǎo)講義練習(xí)題詳解7第七章解答

【解】應(yīng)選如果(|an||bn|)|an||an||bn |bn||an||bn| an和bn都收斂 anbncn知,0bnancnan又an與cn(bnan 收斂,又bnbnanan而anbnn nn1n

n(n|an

f

|f(c M Mn(n

n(n

n3/則anln(enp)

~1

lnen(1

nln(1

ln(enp)

發(fā)散.級(jí)數(shù)ln(enp是一個(gè)交錯(cuò)級(jí)數(shù)，且unln(enp 0,則級(jí)數(shù)ln(enp【解】應(yīng)選（D）.limanbn1liman0和limbn0 則級(jí)數(shù)an和bn n【解】應(yīng)選（C）.如an n )(aa)(aa)L(11)(11)

1111

(n

(n

)a，lim 0,

則級(jí)數(shù)(ntann)a2n與a2n同斂散，而由于正項(xiàng)級(jí)數(shù)an收斂，則級(jí)數(shù)a2n

n

斂，故級(jí)數(shù)(1)n

a

n n 級(jí)數(shù)un和級(jí)數(shù)un1逐項(xiàng)相加所得級(jí)數(shù)(unun1 a 得級(jí) n1 則級(jí)數(shù)

anan2

n

1,a2b2M2

，則級(jí)數(shù)a2b2n

n

n 顯然級(jí)數(shù)

a1a2a3a4La)就是由收斂級(jí)數(shù)(1)n1a加括號(hào)得來的，則級(jí)數(shù)

2n

(a2n1a2n D

nsinnasin1~1

nsin

n

1同斂散，則11

n1 3 .由級(jí)數(shù) 2

021,即12,故2n

即lim 存在,而級(jí)數(shù)np收斂，則級(jí)數(shù)an收斂n

間(3,1)內(nèi)絕對(duì)收斂，從而冪級(jí)數(shù)a(x1)nx間(3,1)內(nèi)絕對(duì)收斂，從而冪級(jí)數(shù)a(x1)nx0對(duì)收斂，即級(jí)數(shù)nn斂

【解】應(yīng)選（A）顯然

1.nnnn(x

na3a1,a3與n

n2xa)nn2x1)2的收斂半徑為1，1ln10，則n2xa)nxln k【解】應(yīng)選（C）.由數(shù)列{an}單調(diào)減少，liman0,Snk

正項(xiàng)級(jí)數(shù)a發(fā)散而交錯(cuò)級(jí)數(shù)(1)na收斂，即冪級(jí)數(shù)a(x1)nx n

x0處收斂，則冪級(jí)數(shù)an(x1)n的收斂域?yàn)閚p2.由limnp(e11)a1limnp1a1,即limann n級(jí)數(shù)an與p1同斂散，若anp11,pn 【解】應(yīng)填(2,4由于冪級(jí)數(shù)na(x1)n1可看做冪級(jí)數(shù)axn n

【解】應(yīng)填(1,5].若冪級(jí)數(shù)a(x2)nx0處收斂，則冪級(jí)數(shù)a(x3)nx n

處收斂，若冪級(jí)數(shù)a(x2)nx4發(fā)散，則冪級(jí)數(shù)a(x3)nx1 n

定理知，冪級(jí)數(shù)an(x3)n的收斂域?yàn)椤窘狻繎?yīng)填(2,0).由limalnnlim 1知a

0,又正項(xiàng)數(shù)列{an

1ln

n0

(1)n

故冪級(jí)數(shù) n(x1)的收斂半徑也為1，則其收斂區(qū)間為 【解】應(yīng)填(2,0由級(jí)數(shù)(1)na條件收斂可知，冪級(jí)數(shù)(1)na(x1)nx n

x0為冪級(jí)數(shù)(1)nan(x1)n(1)n n1.而冪級(jí)數(shù) n(x1)n與冪級(jí)數(shù)(1)nn

(x1n(1)n

n

n(x1)的收斂區(qū)間為 【解】應(yīng)填[ 2].由冪級(jí)數(shù)anxn在x2時(shí)條件收斂可知，該冪級(jí)數(shù)收斂半 2,且a2n收斂.則當(dāng)x2時(shí)冪級(jí)數(shù)axn收斂，從而可知冪級(jí)數(shù)ax2n n

22x22時(shí)收斂，即當(dāng) x 時(shí)冪級(jí)數(shù)anx2n收斂，由an2n收斂可知，冪22 2數(shù)anx2n在x 處收斂，則冪級(jí)數(shù)anx2n的收斂域?yàn)?

1

[(3)n(2)n 3 1

313 (n 則

x3(3)n

2n

(3)n

2n

x3 (3)n2n1a(nn1a(nn

3n(2)n

3n(2)n【解】(1）由于limn

na1時(shí),原級(jí)數(shù)為,則當(dāng)1時(shí)原級(jí)數(shù)發(fā)散,當(dāng)1nnn41nn41x40

2,n2n2(3）由于0

0n3[n3[2(1)n

2

n3[2n

斂 (4n (n1)由于1)nnn (n1)由于

【解】由yxy,y(0)1可知，y(0) y(0)2.由可1 1 1 1yn1nn2on2yn1nn2on2),故級(jí)數(shù)yn1n

n1 【解】fx)

( 5x

x (1x251x

251x5(1)n(31)(x4) (1x 【解】

1

ln[1(x1)2

22xn

(2x

x

(nx【解】

(x)16

21

x2

2

1)(2

(x

(nx4n1)1)(

(xf(x)4

xx4n1(0( x4nf(x)424n2(2

2n[1,1].Sxx24n21

(1)n1x2nS(x)1

2n

12xarctan2nS(x) 0

【解】易求得R,收斂域?yàn)?,).設(shè)y(x)(4n)!，則 (x)y(x),y(0)1,y(0)y(0y(00y(4)(xy(xr41 i.則該方程的通解為yCexCexCcosxCsinx,利 y(0)1y(0)y(0)y(0)0y1(exex2cos4

x2n3n(2n

x2n2

n(n1)(2n

x1即1x1x1時(shí)，級(jí)數(shù)為n(2n1)顯然收斂，故原冪級(jí)數(shù)的收斂域?yàn)?1)n1x2n

(1)n1

(1)n1x2n

f(x),x因

n(2n1)

n(2n1)，設(shè)

n(2n

(1,1f(x)

(1)n12nx2n1

(1)n1x2n1

n(2n

2nf(x)

(1)n1x2n2 1f(00,f(0)0f(x) xf(t)dtf(0)0

x01t

dt2arctanx f(x) xf(t)dtf(0)2xarctan 2tarctantx dt0 02xarctanxln(1x2 從而s(x)2x2arctanxxln(1x2 x 2n【解】令S(x)(2n1)(2n ，S(x)

(1)n1 x 2n2n

(1)n1x2n1xarctan2n 上式兩端從0到1積分得S(1)0xarctanxdx (2n1)(2n1) n1【解】 2(1

arctanx (arctanx)dx

2n

,x0

n02n (1)n1 于是f(x)12n1 2n1 12n1 (1)n1 x2n,xn11 因此14n22f(1)1]42A

A

10

200n

S(xnxnx1,1). n x S(x)xxx1xn

(1

11420（萬元1 S1.05 A20094203980（萬元3980（萬元 【解】（1）yaxnynaxn1,yn(n1)axn n

y2xy4y0并整理，得(n1)(n2)axn2naxn4axn 2a24a0于是

n

(n1)(n2)an22(n2)an0,從而an2n1an,n

n（2）y(0a00,y(0)a11a2n0,n a2n12na2n1L2n(2n2)L42a1

n 從而y n

x2n1

x2n1x

(x2

xex2

an

2n

3

【解】(Ⅰ)由題設(shè)得a2n(2n)!a2n1(2n1)!所以冪級(jí)數(shù)anx S(xanxn,所以S(x)nan S(xan(n1)xn2a

n(n

0n2 n S(x)axn2axnS n

即S(xS(x)(II)S(xS(x0的特征根為1和1,S(x)CexC S(0)

3,S(0)

11知，C11,C22.所以S(x) 2e(1【解】由于

n

1x1記S(x) anxn，S(x)

naxn1a

naxn11[1(n

n

(n

2n

11

axna] nn 2

na 1S(x)2 S

,S(0)1得S(x) 故冪級(jí)數(shù)anxn

1S(x

an

S(x

nan

(nS(x)(n

S(x) (1即S(x)S(x) 解方程S(x)Cex 1

S(0)a02，由S(0)2得C1,S(x)ex 11f

2 n1n(1)f n 【證】由limfx)2f(0)0,f(0)2 f( f(內(nèi)f(x)0，f(x)單調(diào)增 1單調(diào)減.由f(0)0f( f( f(

由lim 2知limf()0,則交錯(cuò)級(jí)數(shù)(1)f()收斂 1

()f而|1)nf()f

f(n

2故(1)nf

1n1【證】an0f(nx)dx,令nxt,則ana21(nf(t)dt)2

nn0f(t)dtn

nf2 n2 n2(01 1f2n

令

f(x)dxA，則n ，

n(0【證】由an1bn1可知，bnbn1，則b1Lbn1bn， b

a1a由a)anbn可知，若bn收斂,則an

a由 )anbn可知，若an發(fā)散,則bn發(fā)散a 則 22 L2n 1u2n12u2n322u2n5L2n1u31 32n1u

n1故級(jí)數(shù)

5

n1【解】應(yīng)填a1.a

x2cos2xdx【解】應(yīng)選（C）.f(x以周期2S(9)S(1)S(1)1 2a00(1x)dx2 n2a x)cosnxdx 2 0

n2k

f(x)8

(2k

0x2n1(2k 2

【解】a00(1x)dx22 an0(1x)cosnxdx

nf(x)13

0x2f(0)13

nn

f(x)1