無窮級數(shù)期末復習題高等數(shù)學下冊(上海電機學院)

1、第十一章無窮級數(shù)參考答案第十一章無窮級數(shù)一、選擇題(A)1.在下列級數(shù)當中，絕對收斂的級數(shù)是(D)(B)(C)2.n 0 n!2nX在-oo<x<+oo的和函數(shù)f X (A )(A)(B)(C)(D)3 .下列級數(shù)中收斂的是(B )(A)(B)(C)(D)lim Un 04 . nUn是級數(shù)n1收斂的(A)(C)充分條件充要條件(B)必要條件(D)無關(guān)條件2Unn1收斂的充分必要條件是limun0(A)nu(B)limnUn1(C)6.下列級數(shù)中，limsnn存在(Sn=U1+U2+Un)發(fā)散的級數(shù)是(D)(A)(C)(D)Un1cos-(B)第十一章無窮級數(shù)參考答案15 -n1x

2、1n5n的收斂區(qū)間是（B）(A)(0,2)(B)0,2(C)0,2(D)0,2nx的和函數(shù)是（8.n1n!x(A)ex(B)e1x（C）e9.下列級數(shù)中發(fā)散的是（(D)(A).一nsin2(B)n1(C)(D)n110.塞級數(shù)n1的收斂區(qū)間是(A)1,1(B)2,4(C)2,4(D)2,411.在下列級數(shù)中發(fā)散的是（D3,on(A)n12(B)n111.nn(C)n12n1(D)n13n(n1)_112.哥級數(shù)n02nn_2n1!x的和函數(shù)是（D(C)(B)cosxln1(D)sinx13.1x1n,n1的收斂區(qū)間是（5n14.15.(A)(C)在下列級數(shù)當中,(0,2)0,2絕對收斂的級數(shù)是

3、（1(A)n1J2n(B)(B)0,2(D)0,2（C）卜列級數(shù)中不收斂的是（(D)1)n1B.1)nC.n1)n-132nD.1)n16.在下列級數(shù)中發(fā)散的是（(A)(B)116132(C)0.0010.00130.001(D)17.哥級數(shù)ln(n1)Yn的收斂區(qū)間是（x(C)1,1(B)(-1,1)1,1(D)1,118.下列級數(shù)中條件收斂的是（B）1)n1(j)n3B.(1)n1、.nC.1)n1(D.(_1)n1nn112(x19.哥級數(shù)一n1a的收斂區(qū)間是(1nBB.D.C.20.在下列級數(shù)中,(3,1)222,2)條件收斂的是(A)(B)m1nTn(C)n12n1(D)in的和S=

4、(D3(B)(A)22(D)(C)522.設(shè)f(x)是周期為2的周期函數(shù)，他在)上的表達式為f(x)=x,若D2cf(x)的傅立葉級數(shù)展開式為-0-2(-ncosnxbnsinnx),貝U-nn1D. 02n12n_1A.-(1)B.-(1)C.-(1)nnn23.設(shè)f(x)是周期為2的周期函數(shù)，他在)上的表達式為f(x)=x2,若f(x)的傅立葉級數(shù)展開式為a02(-ncosnxbnsinnx),貝UbnAA.0B.4(n1)nC.;(1)n1D.4(1尸二、填空題n 2nx的和函數(shù)是0n!nn2x2.哥級數(shù)n0的收斂半徑為n13.若交錯級數(shù)n 11UnUn收斂絕對收斂，則正項級數(shù)n1n.n

5、4 .騫級數(shù)1a的和函數(shù)是n0n!15 .帚級數(shù).(3)nxn的收斂半徑是n12nnn1、6 .級數(shù)1二的收斂半徑R3。n13三、計算題nnX41 .求哥級數(shù)1的收斂區(qū)間。nn1ng3limn1n3ngn13n1limn3x43,1x7n當x1時，1nLL1,發(fā)散nn1ng3n1nnnn741當x7時，1-收斂n1ng3nn1n所以，級數(shù)的收斂區(qū)間為1,7ln(n1)n1一，sn1x2 .求帚級數(shù)n1n1的收斂域;limnlnn1n2gn1lnn2ln(n1)n1x當x1時，級數(shù)n1n1=1n(n1)1n1收斂n1n1ln(n1)n1x當x1時，級數(shù)n1n1=1n(n1)發(fā)散n1n1收斂域為1

6、,13 .求哥級數(shù)nln(n 1) n 1x1 n 1的收斂半徑.解：因為limnan 1an=limnln(n 2)n 2ln(n 1)=limnln(n2)ln(n 1)所以原哥級數(shù)的收斂半徑為：14 .將函數(shù)f(x)x展開為x的哥級數(shù).3 2x1一 x-3 11-2xx12x22x2x2nx 3n5 .判斷級數(shù)n1s)n的斂散性.由根值判別法lim n2n 1可知原級數(shù)收斂。6 .將函數(shù)f x1 L ,，一2展開成關(guān)于xx的哥級數(shù)。解:1)nxn逐項求導，得(1 x)2n n 1(1) nx1(1)n 1(n 1)xn ,0所以(1)n(n1)xn,x1,1.2f x7.將函數(shù)xX3展開

7、為x的哥級數(shù);2 2n上X1XXccfxx3,33 1x3n0338.將函數(shù)x2ln(12x)展開成x的哥級數(shù)，并討論收斂區(qū)間;n 1 n x1 n 1234xxx.QIn1xx一一L234x1,1x2ln 1 2x2x2x2x2x L34n 2x24xn 2n32x2x9 .將函數(shù)xe展開成x的哥級數(shù)，并討論收斂區(qū)間;、，234.n12xxxxn1xx111!2!34!n1!n1n110 .判斷級數(shù)1lnn-2的斂散性n1n1lnn212nlnn212nln1n1一 1 一，一，而 -2收斂，所以n 1 nn 1n 1 xn _ n 12nn! n 0 n!n1lnn-2-1絕對收斂.所以原級數(shù)收斂n11 .求級數(shù)nJx的和函數(shù);n02nn!n11Xn0n1!n!2nn1x1xn1n1!2n0n!2nnx1x1x2non!2non-!2xe2nx1x=_12non!212.判斷任意項級數(shù)33312 3442431 n34n的斂散性，并指出是否絕對收斂;將原級數(shù)各項加絕對值后的級數(shù)為33工3 L42433nn l利用比值判別法有4limn3 nn 14n 1