2019高等数学辅导讲义练习题详解7第七章解答

【解】应选如果(|an||bn|)|an||an||bn |bn||an||bn| an和bn都收敛 anbncn知,0bnancnan又an与cn(bnan 收敛,又bnbnanan而anbnn nn1n

n(n|an

f

|f(c M Mn(n

n(n

n3/则anln(enp)

~1

lnen(1

nln(1

ln(enp)

发散.级数ln(enp是一个交错级数，且unln(enp 0,则级数ln(enp【解】应选（D）.limanbn1liman0和limbn0 则级数an和bn n【解】应选（C）.如an n )(aa)(aa)L(11)(11)

1111

(n

(n

)a，lim 0,

则级数(ntann)a2n与a2n同敛散，而由于正项级数an收敛，则级数a2n

n

敛，故级数(1)n

a

n n 级数un和级数un1逐项相加所得级数(unun1 a 得级 n1 则级数

anan2

n

1,a2b2M2

，则级数a2b2n

n

n 显然级数

a1a2a3a4La)就是由收敛级数(1)n1a加括号得来的，则级数

2n

(a2n1a2n D

nsinnasin1~1

nsin

n

1同敛散，则11

n1 3 .由级数 2

021,即12,故2n

即lim 存在,而级数np收敛，则级数an收敛n

间(3,1)内绝对收敛，从而幂级数a(x1)nx间(3,1)内绝对收敛，从而幂级数a(x1)nx0对收敛，即级数nn敛

【解】应选（A）显然

1.nnnn(x

na3a1,a3与n

n2xa)nn2x1)2的收敛半径为1，1ln10，则n2xa)nxln k【解】应选（C）.由数列{an}单调减少，liman0,Snk

正项级数a发散而交错级数(1)na收敛，即幂级数a(x1)nx n

x0处收敛，则幂级数an(x1)n的收敛域为np2.由limnp(e11)a1limnp1a1,即limann n级数an与p1同敛散，若anp11,pn 【解】应填(2,4由于幂级数na(x1)n1可看做幂级数axn n

【解】应填(1,5].若幂级数a(x2)nx0处收敛，则幂级数a(x3)nx n

处收敛，若幂级数a(x2)nx4发散，则幂级数a(x3)nx1 n

定理知，幂级数an(x3)n的收敛域为【解】应填(2,0).由limalnnlim 1知a

0,又正项数列{an

1ln

n0

(1)n

故幂级数 n(x1)的收敛半径也为1，则其收敛区间为 【解】应填(2,0由级数(1)na条件收敛可知，幂级数(1)na(x1)nx n

x0为幂级数(1)nan(x1)n(1)n n1.而幂级数 n(x1)n与幂级数(1)nn

(x1n(1)n

n

n(x1)的收敛区间为 【解】应填[ 2].由幂级数anxn在x2时条件收敛可知，该幂级数收敛半 2,且a2n收敛.则当x2时幂级数axn收敛，从而可知幂级数ax2n n

22x22时收敛，即当 x 时幂级数anx2n收敛，由an2n收敛可知，幂22 2数anx2n在x 处收敛，则幂级数anx2n的收敛域为2

1

[(3)n(2)n 3 1

313 (n 则

x3(3)n

2n

(3)n

2n

x3 (3)n2n1a(nn1a(nn

3n(2)n

3n(2)n【解】(1）由于limn

na1时,原级数为,则当1时原级数发散,当1nnn41nn41x40

2,n2n2(3）由于0

0n3[n3[2(1)n

2

n3[2n

敛 (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),故级数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,收敛域为(,).设y(x)(4n)!，则 (x)y(x),y(0)1,y(0)y(0y(00y(4)(xy(xr41 i.则该方程的通解为yCexCexCcosxCsinx,利 y(0)1y(0)y(0)y(0)0y1(exex2cos4

x2n3n(2n

x2n2

n(n1)(2n

x1即1x1x1时，级数为n(2n1)显然收敛，故原幂级数的收敛域为(1)n1x2n

(1)n1

(1)n1x2n

f(x),x因

n(2n1)

n(2n1)，设

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 从而s(x)2x2arctanxxln(1x2 x 2n【解】令S(x)(2n1)(2n ，S(x)

(1)n1 x 2n2n

(1)n1x2n1xarctan2n 上式两端从0到1积分得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（万元1 S1.05 A20094203980（万元3980（万元 【解】（1）yaxnynaxn1,yn(n1)axn n

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

n

(n1)(n2)an22(n2)an0,从而an2n1an,n

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

n 从而y n

x2n1

x2n1x

(x2

xex2

an

2n

3

【解】(Ⅰ)由题设得a2n(2n)!a2n1(2n1)!所以幂级数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的特征根为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记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) 故幂级数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 【证】由limfx)2f(0)0,f(0)2 f( f(内f(x)0，f(x)单调增 1单调减.由f(0)0f( f( f(

由lim 2知limf()0,则交错级数(1)f()收敛 1

()f而|1)nf()f

f(n

2故(1)nf

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

nn0f(t)dtn

nf2 n2 n2(01 1f2n

令

f(x)dxA，则n ，

n(0【证】由an1bn1可知，bnbn1，则b1Lbn1bn， b

a1a由a)anbn可知，若bn收敛,则an

a由 )anbn可知，若an发散,则bn发散a 则 22 L2n 1u2n12u2n322u2n5L2n1u31 32n1u

n1故级数

5

n1【解】应填a1.a

x2cos2xdx【解】应选（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