第十二章無窮級(jí)數(shù)習(xí)題解答

111

； 1 解因?yàn)榇思?jí)數(shù)為 13131 13

1

;1n 1n33n解unn

3n10(n由級(jí)數(shù)收斂的必要條件可知此級(jí)數(shù)發(fā)散習(xí)題12１.用比較審斂法或極限審斂法判別下列級(jí)數(shù)的斂散性

111

111

11

；解因?yàn)?/p>

1

1而級(jí)數(shù)

發(fā)散故所給級(jí)數(shù)發(fā)散 n

n1

sinsinsinsin

； sin 解

2n而級(jí) 1 n

1n1

(a0)當(dāng)0a1時(shí)u

lim

0 1

n斂的必要條件這時(shí)原級(jí)數(shù)發(fā)散11 aa1時(shí)a

n1

n1而級(jí)數(shù)n

n收斂級(jí)數(shù)收斂 所以原級(jí)數(shù)當(dāng)a1時(shí)收斂當(dāng)0a1時(shí)發(fā)散２.用比值審斂法判別下列級(jí)數(shù)的斂散性n3(2)n3u

(n1)2

1n 解limn1

1n

n3n n

2n；u

2n1(n

n limn1

n

(n

2n

nn1 2ntann12n

(n1)tan 解

un1

n1

1n

ntan

n

*3. 2n1n1

11所以級(jí)數(shù)收斂

nn2nn４. 3 3

3

2

n 4u

4

433(n1)4

n1

limn1lim

1n

3n4

n (4)n

nsin2n1sin

2n1解

3n1

3n12232n232nn

2nsin

2n

；解因?yàn)榧?jí)數(shù)的一般項(xiàng)un

10(n)nnnn

a

2a1

na

(a0,b0)解limanb1而級(jí)數(shù)

n1５.判斷下列級(jí)數(shù)是否收斂？如果是收斂的是絕對(duì)收斂還是條n

n

n3nn3nnn3n

n 解n

n

,3n3n

3 3所以級(jí)數(shù)n1是收斂的,從而原級(jí)數(shù)收斂并且絕對(duì)收斂3n(3)1

1

1

1313n

31

3解n

n n 2因?yàn)榧?jí)數(shù)n2

1

；111nnn解由于n

ln(n

級(jí)數(shù)斂

un

，且un1

，lim

(1)n1 . 2解級(jí)數(shù)的一般項(xiàng)為un(1)n1 ，因?yàn)?/p>

|u|nn

2nn

2nn

(2n)nn

，3，習(xí)題121.求下列冪級(jí)數(shù)的收斂

xx2 2

24

2解

24

n24R所以收斂區(qū)域?yàn)?,) 3 n解

n

n1

R3

n(n1) nx3時(shí)，級(jí)數(shù)nx3時(shí),級(jí)數(shù)

收斂.n以收斂區(qū)域?yàn)閇3,3

n

2x22 2

3 n2 3解

2

n2

n(n1) R1收斂區(qū)間為112 2 當(dāng)x2

,級(jí)數(shù)nn

收斂;

x2

,級(jí)數(shù)

11n

n21收斂；所以收斂區(qū)域?yàn)?/p>

n

x2n；2n1

nx2nx2nn2x2nn2n

2nx2n2nx2n2unun1

x由比值審斂法

1

x1時(shí)冪級(jí)數(shù)絕對(duì)收斂x21即|x1時(shí)冪級(jí)數(shù)發(fā)散R1x1時(shí)n n

n 級(jí)數(shù)n

2n1收斂x1時(shí)級(jí)數(shù)

收斂.2n以收斂域?yàn)閇1

n

2n12n

2n2這里級(jí)數(shù)的一般項(xiàng)為

2n1x2n2un1un12n2n(2n

22由比值審斂法2

1x

時(shí)冪級(jí)數(shù)絕對(duì)收斂1x222

2時(shí)冪級(jí)數(shù)發(fā)散故收斂半徑為R x

2時(shí)級(jí)數(shù)為2n

2n2

發(fā)散x

時(shí)冪級(jí)數(shù)成為n

n12n2

也是發(fā)散的所以收斂域?yàn)?

2)22

n

xn.n

1故收斂半徑為R1即當(dāng)nn1x51時(shí)級(jí)數(shù)收斂x51nnx5

nx4時(shí)冪級(jí)數(shù)為n

收斂x51x6時(shí)冪級(jí)數(shù)為

1發(fā)散所以收斂域?yàn)閚２.nnxn1n解S(x

S(x)nxn1nS(x)

x

n

n

0

d

0

x

n n 1 xn

(1

(1x1)n x4nn4n1n解S(x

x4nnS(x) 4n1n x

x4n1

x S(x)S(0)0S(x)dx04n1dx0 dn

nx 1dxx111 11 dx 01

0

1

21x21ln1x1arctanxx(1x1) 1 x提示x

0S(x)dxS(x)S(0)

xx35 35

2n

解S(x

S(x)2n1x

2nS(x)S(0)

x

x2n1x0S(x)dx

02n x x2n2dx

x1dx1ln1x(1x1)

01 1習(xí)題12-

exeshx 2

(a0)

ax

sin2x

xn 解(1)因?yàn)閑 ,(x)所以ex(1)n .n0 n exe

1

nxn故shx2

2n!

x2n

,(xn1(2n因?yàn)閘naxlna1x(lnaln1xln(1x)

aaaanxn1n

1x所

n1alna

(axa)n0(n1)an 由

x,ax

xln

(xlna)n

xn

x(,)因?yàn)閟in2x1(1cos2x11cos2x cosx

,x,,所n

1 n22n

n22n1

x 2

2n2

n

，xR因?yàn)閘n(1

x)n

n1n

,1

x1,所(1x)ln(1x)(1

nxn1n

nnxnnxn nn

n

n

nx

nnxnnxn nn

n

n

(1)n1

n1

nnxn

nn1 n

xn

n(n

(1x1) 解因?yàn)?1x)m1mxm(m1)x2m(m1)(mn1)xn

,(1x1)33 所 [1(x1)]21

(x1)2 (x 3313n 2 (x1)n

,(1x11) 13(x1)2

322

(x1)231(1)(3)(52n)(x1)n2n

x1n2

(x1)

(n!)2(n1)(n2)2n2

(0x2)

lgx

n1xn 由ln(1

(n

n

lgx

1ln10n

n1(xn

x2.４.f(xcosx展開成x解cosx

3 3

cosx33cosx3cos3sinx3sin 3 3cosx 3

sinx 31(1)n

3 (1)n

2(2n)!x3 2(2n1)!x3

1

3 2n13 (1)n

x

x

,(x)n

3

(2n

3 ５.f(x)1展開成(x3x

1 x3解f(x)

(1)n 31x3

3 1

x(x (x3n

3n 由1x310x63６.f(x)

x23x

展開成(x4解f(x

x23x

x1

x x

3(x

31xxx331x43

(x 2n0 32

n

3n

1

1x4 x 2(x 21

x4

n02

(x4)n

,x412 (x (xnf(x)x23x2n

n

n11 n02n0

3n1(x 由1x41及1x41知6x2 習(xí)題12(2)e（誤差不超過0.001解ex1x1x21xnx e111111e由 r

2! 1

n!

(n

(n

n

12

(n2)(n1)

14

3

3

n4

111

1

1

1.648e 2!e

3!

4!２.利用被積函數(shù)的冪級(jí)數(shù)展開式求下列定積分的近似值05 dx（誤差不超過0.000101解因?yàn)?

(1)nx4n,x11x x

01x4

0n5x55

9

，故0 d01

0.5

5

9

0.513

0 01

dx12

5

9

0.4940(1)yxyx 解ya0anxnynanxn1

n n n

nan xa0anx

n

n .所以

1，

1a0，

1，

1a0，a1

2

3a1 24

，

35

，，

135(2n 1

，所以 ya0x

x3

x2n (2n1)!!

22

24(2n)!!a0x

x3

x2n (2n1)!! x2 1x2

1xn (1a0)112 2!2

n!2 xa0(1a0)(1a0)e

x

x3

x2n (2n1)!!x(1a0)e

1x

x3x

nn

x2n (2n1)!!x2nyCe

(1)(2n1)!!４.yy2x3，

x

12解ya0ann

，由初始條件 1，x 1y2

nanxn1a1

nanxn1

a1nanxn1x3 anxnn

a1nann

x4n

an

2x211a1，2aa，

aa2，

a2a

1，

1所 a1，a1，

1，a9，

y11x1x21x3 習(xí)題12１.f(x2f(x展開成級(jí)數(shù)，如果f(x)在,上的表達(dá)式為：f(x)3x21（x解a (3x21)d a (3x21)cos cos23x2cosnxdx2cos 61x2d(sinnx)21n 6x2sinnx

xd(cosn

02xsinnxd

n2n2n2

00cosnxd00cosnxd

(1)n12,(n1,2,)b (3x21)sinnxdx0(n1,2,) 所以fx

1

1n

cosnx,(,)

f(x)axax，0x

x(a，b為常數(shù)，且ab解因?yàn)?/p>

axd

ab) a

nxx]ba[1(1)n]

b 0bxsinnxdx 所以f(x) f(x)(ab)[1(1)n](b (1)n1(a n1

cosnx

sinnx x nn 2４.設(shè)f(x)是以2為周期的函數(shù),它在,上的表達(dá)式為，－x－2 2 f(x)

x 將f(x)展開成級(jí)數(shù)

x2f(x為奇函數(shù)故an0(n0,1,2, 2

bnf(x)sinnxdx2xsinnxdx

sinnxdx

2 n2

2f(xx2n1)n0,1,2, nf(x) n1

n2

sinnx22５.f(x)

2

(x(2n1),n0,1,2,)（0x）解作奇延拓得F(x)

f0

0xx

.f

x到(,x0,F(x)f(xF(0)02

f(0)因?yàn)閍n

b xsinnxdx

nf(x)sinnx（0x）x0處收斂于0nn６.f(x2x2（0x）f(x作奇延拓展開成正弦級(jí)數(shù)，這時(shí)an0(n0,1,2,2

4x2

2xsinnxd0

dcos04 2

cos

xcosnxdx n (1)n14

8 xsin

1cosnx n2

0(1)n14

f(x奇延拓后在[0,x

n1

8

2x

n

n

f(x作偶延拓展開成余弦級(jí)數(shù)，這時(shí)bn0(n1,2,a 2x2dx42 a 2x2cosnxdx

n4x2sinnx

xd(cosn

02xsin

n28 (xcos sinnx(1)n ，(n0,nN).n2

0 f(x作偶延拓后在[0,2x2

22

cosnx，x[0,] n1習(xí)題12：f(x)1x21x122 22 解f(x)1x2為偶函數(shù)，所以bn0(n a22(1x2)dx42(1x2)dx4x1x3211 12

2an22

2(1x2)0

12

dx42(1x2)cos2nxd011

4n2

sin(2nx)4n22cos(2nx)8n33sin(由于f(x)在(,)f(x)111

x(,) 2

f(x)1

0x121x2 解a0f(xdx

xdx2dx1dx a1(1x2)cosn

xcosnxdx2cosnxdx1cos020 xsinnx 1

sinnx

n22

211(1)n

sin

n2

b1(1x2)sinn xsinnxdx2sinnxdx1sin

cosn

21(n1,2,) f(x在(,x2k,2k1k0,1,2,2f(x)4 2sinn 12cos 1(1)

2sinnxn1

n2

(3)f(x)2x1，3x 0x解a13f(x)dx1

(2x1)dx

3dx1 3

3

13f(x)cosnx 3 10(2x1)cosnxdx1

3

3 1

1 n(2x1)

3

3dxn

n213f(x)sinnx 3 10(2x1)sinnxdx1

3

3 6(1)n1(n1,2,)f(x在(,)x3(2k1),k

n1

2f(x) 2n1

n2

3x3(2k1),kf(x)

0x2ll

x2解f(x2

bn

2x dxl(lx)

d

l2l nx l

nx2 x

2 ln

l

n

l02

(lx)

nx l

lnxl

l24lsinnn2

n2

n為奇數(shù)當(dāng)n為偶又f(x)作延拓后在0l上連續(xù)，4l(1)k (2knf(x)2(2k1)2 n

,x[0,l]再將f(x)bn0(n1,2,)2 a0

2xdxl(lx)dx2

2

l

an

2x dxl(lx)

2 l l 2 l l nln22n22cosnln22n22(1)

n2

n為奇數(shù)當(dāng)n為偶 4l(1)k (2kn f(x)42(2k1)2 n

,x[0,l]1.填(1)對(duì)級(jí)數(shù)unlimun0

部分和數(shù)列sn有界是正項(xiàng)級(jí)數(shù)un收斂的充要條件 若級(jí)數(shù)unun

條件收斂，則級(jí)數(shù)

２nnnnnn分析注意到nn1(nnn nn 解因?yàn)? nn

發(fā)散，故

發(fā)散

n1

n解

(n2(n2

1,所以級(jí)數(shù)發(fā)散n

ncos22（3） n32n解

ncos2

n，而對(duì)于nn n limun1

11n

ncos2nn由比值判別法知n收斂，故 nn

收斂n1 n ln（4） lnn解因?yàn)?/p>

1ln10

而級(jí)數(shù)

發(fā)散故所給級(jí)數(shù)發(fā)散nn

n

n1

sn1

(a0,s0)nn解unn

ana1(1時(shí)，原級(jí)數(shù)收斂；當(dāng)a1(1nna1(1時(shí)，原級(jí)數(shù)為ss1nns1時(shí)，原級(jí)數(shù)收斂 設(shè)正項(xiàng)級(jí)數(shù)un和vn都收斂，證明(unvn)2也收斂n

n

由unlimun0lim

limun0n

n

比較判別法的極限形式知un2收斂.同理vn2也收斂 所以 2(unvn

n

(unvn)2u2v22uv2(u2v2) n 所以由比較判別法知(unvn)2收斂 ４.設(shè)級(jí)數(shù)un收斂，且limn1.問級(jí)數(shù)vn試說明理由

n

n

n分析對(duì)于此類問題如果結(jié)論成立就要給出證明，如果結(jié)論不解不一定 n設(shè)unn2vnn2n顯然滿足題設(shè)條件,并且vn收斂n設(shè)nn

n，n

(1)n

1n但是vn發(fā)散n５.討論下列級(jí)數(shù)的絕對(duì)收斂性與條件收斂性(1)n1n n分析pp取值進(jìn)行討論解當(dāng)p0時(shí)，limun0級(jí)數(shù)發(fā)散;當(dāng)0p1時(shí)，滿錯(cuò)unnp1un收斂，故級(jí)數(shù)絕對(duì)收斂n sin(1)n1 n1

sin n1

,

n

n

n1n1（3）(1)nn

n；n解 lnn2ln11ln11ulimun0

nnn

n1

ln(11uln11，

n1n n

n以u(píng)n發(fā)散；所以原級(jí)數(shù)條件收斂n（4）n

n(n1)!nn解un1

(n2)!

n2

n (n

(n

n

n

1

６.求下列極1n1

1k

1

1（1） 1

（2）lim2349827(2n)3nnnk13k k

分析21 1解 級(jí)數(shù)3k1k

k

k1 k1 123k1 k

11

1

ek

k3 k 21 1所以3k1k

收斂k

n1

1k設(shè)snk1

limsnsk13 k1n1

1k 1 lim nnk13k k

n lim2349827(2n s12n ，故s1s2s11n 3 3 n1s3

lim3

113n1223 n1223

n

n2 1于 lim2349827(2n)3n2

n

3nn

xn

3n1n

5n5R5n1 3n5nn15因?yàn)楫?dāng)x 時(shí)冪級(jí)數(shù)為5n

5nx1 n3n5n

11級(jí)數(shù)為n

5n收斂；所以收斂區(qū)間為

,55 1

1n1

xnnn

1分析an1n

, 收斂求解解

nn

1nn

xex用根值審斂法，當(dāng)ex1x1e當(dāng)e

1

1時(shí)，原級(jí)數(shù)發(fā)散；當(dāng)ex1e n2

0

n e

n ex1時(shí)發(fā)散.故冪級(jí)數(shù)的收斂區(qū)間為11 een(x1)nn解令tx 冪級(jí)數(shù)ntn的收斂區(qū)間nlim|an1|

1R1 nnt1nt1時(shí)，冪級(jí)數(shù)為n (1)nn,發(fā)散；故ntn的收斂區(qū)間為(1,1)，從而原冪級(jí)數(shù)斂區(qū)間為

n2

n 2n2n12nn解這里級(jí)數(shù)的一般項(xiàng)為n12nn

x2nunun1

x2n2

x2由比值審斂法2

1即|x

時(shí)冪級(jí)數(shù)絕對(duì)收斂2

12即x 2時(shí)冪級(jí)數(shù)發(fā)散故收斂半徑為R 222x22

時(shí)冪級(jí)數(shù)為n是發(fā)散的xn

時(shí)為n也是發(fā)散的所以收斂區(qū)間為(n

2)

n

2n12n

2(n1)

2n

解S(x)n

2n un12(un12(n1)2n(2n

12R2

x

2時(shí)，冪級(jí)數(shù)為(n2發(fā)散；級(jí)數(shù)

2n1

的收斂區(qū)間為(

2)

n1 由冪級(jí)數(shù)的逐項(xiàng)積分性知,x(2,

2n20S(x)dxn 2n

x2 所

S(x)dx

x2n1

2

2

n1 2 從 S(x)dx ，Sx

2. 2

2x2

(2x2

x2n1

2n1S(xn

(1)n12n 2n n2(nn2(n1)2n1(1)n1x2n1unun1

xR1x1時(shí),級(jí)數(shù)收斂.于是冪級(jí)數(shù)的收斂區(qū)間為[1,1由冪級(jí)數(shù)的逐項(xiàng)可導(dǎo)性知,x1,1)S(x)n

(1)n1x2(n1)n

(x2)n1 11x1所 S(x)

dxarctanx，x[1,1]01 n(x1)nnS(xlimn11x11n0x2時(shí)級(jí)數(shù)收斂.x0為(0,2

x2時(shí)級(jí)數(shù)是發(fā)散的 S(x)n(x1)n(x1)n(xn

n x1

x(x1)(x

(x

，x n

1x1

(2nn(n1)n解S(x

n(n

1n(n1)(nR1x1n設(shè)S(x) ，x[1,1]，由于S(0)0nn1n(n

xS(x)

xn，n1n(n

nn1n

，

xn1n

1

，x所 xS(x)x1dxln(1x)010x0

S(x)1(1x)ln(1x)xx1x1S(x)

1xln(1x),

xx[1,0)xn 解n2解

n1n1

n1(n

n1(n

1

1n2(n

n1(n

n0

n0

n0由于en!x1enn故所求數(shù)項(xiàng)級(jí)數(shù)的和為

n2e

n1(1)nn1(2n n n2n1n解n

(2n1)!

2(2n1 n 22

(2n)!

(2n因 sinx

x2n，(2n

cosxn

所 n

(2n

cos1n

n 故所求數(shù)項(xiàng)級(jí)數(shù)的和為n

(2n 10.x x21)解ln(x x21) ，ln(x x2

0x2 x