十章無窮級數(shù)自測試題參高數(shù)章

A級自測1．常數(shù)項(xiàng)級數(shù) 1ln1 （A． n A．條件收B．絕對收 C．發(fā)D1解：由萊布尼茨定理知級數(shù)ln1 n2．正項(xiàng)級數(shù)A級自測1．常數(shù)項(xiàng)級數(shù) 1ln1 （A． n A．條件收B．絕對收 C．發(fā)D1解：由萊布尼茨定理知級數(shù)ln1 n2．正項(xiàng)級數(shù)A．必要條2收斂是級 a收斂的（B．n2Da a收斂，則 lima0，由比較審斂法知級 收斂2nnn反之，若級 收斂，則正項(xiàng)級 a不一定收斂。如a1。故應(yīng)選（B2nnnn3．若常數(shù)項(xiàng)級數(shù)(u1u2(u3u4L(u2n1u2nL收斂，則級數(shù)u1u2u3u4Lu2n1u2nL級數(shù)u1u2u3u4Lu2n1u2nL級數(shù)u1u2u3u4Lu2n1u2nL不一定收級數(shù)u1u2u3u4Lu2n1u2nL必定發(fā)散A1n，故當(dāng)p1p1時，級數(shù)解：unp，原級數(shù)絕對收斂；當(dāng)0p1，即1p0時，級數(shù)級 u收斂，即原級數(shù)條件收斂；當(dāng)p0，即p0時，limu0，原級數(shù)發(fā)散nn（D1 n（a0）當(dāng)a 時收斂n12121 aa2 解： an2aanna2．已知冪級數(shù)anxna12a2x3a3x24a4x3的收斂半徑 ，和函數(shù) 解：顯然級數(shù)a2ax3ax24ax3L naxn1=(axn)=s(x)，21 aa2 解： an2aanna2．已知冪級數(shù)anxna12a2x3a3x24a4x3的收斂半徑 ，和函數(shù) 解：顯然級數(shù)a2ax3ax24ax3L naxn1=(axn)=s(x)，故冪級數(shù) 34nn3．冪級數(shù)的收斂域?yàn)閇2,2)n1nn1 ，limlim R2nnn n1nx2n所求冪級數(shù)的收斂域?yàn)閇2,2)x0x4．f(x，則其以2x1x2收斂 f(xxxf()f()121222 的收斂性，若收斂，求其和.n1(3n1)(3n11 1)n(3n1)(3n 33n 3ns1[(11)(11)(11)L11n 1(1111)，故lims ，原級數(shù)收斂且其和 n3 (p0)是否收斂111111p1發(fā)1p1pn211，而1發(fā)散，故1發(fā)散1散；當(dāng)0p11 11p收斂；當(dāng)0p111發(fā)散23．判別級數(shù) 是否收斂，其中l(wèi)imuna，un及a都為正數(shù)n1un22 a2時級數(shù)收斂；當(dāng)0a2a解：由題設(shè)知limnuna11，而1發(fā)散，故1發(fā)散1散；當(dāng)0p11 11p收斂；當(dāng)0p111發(fā)散23．判別級數(shù) 是否收斂，其中l(wèi)imuna，un及a都為正數(shù)n1un22 a2時級數(shù)收斂；當(dāng)0a2a解：由題設(shè)知limnuna2時，級數(shù)可能收斂也可能發(fā)散 4．判別級數(shù)nπn1sinn1sin1sinπ，此為正項(xiàng)級數(shù)， sinπ ，所11nnnπnππ 收斂，從而 絕對收斂nn2nnn1(x1．求冪級數(shù)nnx11，即2x0nnx2x0(2,0]1 11 2n1(1xs(x)dx1ln1x，(1x1)xs(x)s(0) s(x)dx1x01 10s(x1ln1x(1x1) 111．f(x展開成（x1）x2x1111f(x) )x2x 3x x3(1)n(x2x2 1n2(x1)n)x2x x12 1x3(x1)nx11，即0x2x x 1(x (1)n1111 f(x)x2x2)=(x1)（0x2(1(1)n(x2x2 1n2(x1)n)x2x x12 1x3(x1)nx11，即0x2x x 1(x (1)n1111 f(x)x2x2)=(x1)（0x2(1n x x n0dex1nx的冪級數(shù)，并求級數(shù)2．f(x)展開n1(nxe（x，e1（xxxexn!(n（xx0xf(x)d(ex1)d)nxn1（x，x0 (n n1(nxdex1nxexexf(1)1f(xn1(n x23f(x)2π為周期的周期函數(shù)，它在[ππ)上的表達(dá)式為：f(x3x21（πxπ）,f(xf(x的傅里葉f(x)dx (3x21)dx2(222a002f(x)cosnxdx (3x21)cosnx2an0 [n2xcosnx]12cosn12（n1,2,3,Lbn0(n1,2,f(x) ancosnxcosnx（x22 分4證明：因?yàn)閍nbn都收斂，故(bnananunbn0unanbnan，由比較審斂法可知正項(xiàng)級數(shù)(unan而unanunan，故級數(shù)unB級自測（04（B．證明：因?yàn)閍nbn都收斂，故(bnananunbn0unanbnan，由比較審斂法可知正項(xiàng)級數(shù)(unan而unanunan，故級數(shù)unB級自測（04（B．A．若limna0，則級 a收斂nB．若存在非零常數(shù)，使得limna，則級 a發(fā)散nnC．若級數(shù)a收斂，則limna02nnD若級 a發(fā)散，則存在非零常數(shù)，使得limnan11limnan0，但an解舉反例排除。取ann1nln1A，則級 a收斂，但limnan，排除（C2nn本題對（B）也可采用比較判別法的極限形式limna 0，而級注nn1/1n發(fā)散，因此級 a也發(fā)散n2．已知級數(shù)(1)n1an2,a2n15，則級數(shù)an等于（C．)... aa)a解 aaaa... 2nn1234 ana2n12a2n1(1)n1an2528.選（C11 （C．un1n11A．發(fā)D5111... 1 1 ，由limunn u2 u3 u4 un1 0，所以limSlim11 nx xun14．若級數(shù)an(an0)收斂，則有（B．a(chǎn)aaC nA(an)2發(fā)散Bn111... 1 1 ，由limunn u2 u3 u4 un1 0，所以limSlim11 nx xun14．若級數(shù)an(an0)收斂，則有（B．a(chǎn)aaC nA(an)2發(fā)散Bnnnn1111aannn解：由正項(xiàng)級數(shù)的收斂性知(a)2n2n2n知與aaan1收斂，而由x1n11nn1．若級數(shù)收斂，則a應(yīng)滿 a1a1時有l(wèi)imn10n2n1與2a222n2．設(shè)冪級數(shù)an(x1)n在x3處條件收斂，則該冪級數(shù)的收斂半徑為R 解：令tx，則原冪級數(shù)變 at，由題設(shè)它在t4處條件收斂，由Abel定nn收斂，與題設(shè)“在t4處條件收斂”矛盾。故只能有R4。3．函數(shù)f(x)x22x1展開成（x1）的冪級數(shù) f(14f(1)4f(1)2f(n)(10n3f(x)f(1)f(1)(x1)f(1)(x1)244(x1)(x1)2(x4．f(xπxx2(πxπa(acosnxbsinnx)0nn2則其中系數(shù)b3的值 1xsin3xdx31f(x)sin3xdxπ(πxx2)sin3xdx解3611nln1 nn11111n 11 11) o()]2解：u o(n n1n 1n1111 n o(n2，即lim334n n1n11)且 1，問級數(shù)(（02研設(shè)un0n1n un1若收斂，是絕對收斂還是條件收斂？（8分 limn11111nln1 nn11111n 11 11) o()]2解：u o(n n1n 1n1111 n o(n2，即lim334n n1n11)且 1，問級數(shù)(（02研設(shè)un0n1n un1若收斂，是絕對收斂還是條件收斂？（8分 limn111(1)(1)n11(1)n11(1).因?yàn)?1)n1n nn11111條件收斂 ()絕對收斂，所 條件收斂，從而nuu n1n1x4x29x316x4L(1)n1n2xnLxs(x)xs(t)dt 4x39x416x5Ln112 501x2(11)x3(21)x4(31)x5L(1)n1[(n1)1]xn1n2345x[x1x21x31x41x5L(1)n1 x3[12x3x2L(1)n(n1)xn21xn1nxln(1x)x3[(1)n1xn]xln(1x)x3( 1xln(1x)(11x（ x2(3故s(xxln(1x]1x（(11 (1 n2(n2111S(xS(xn n1.(n2n271 (x,nnn 11n xnnxxxdxxn1dxln(1而01n0 0x1從而S(x) [ln(1x)] [ln(1x)x 22x12(|x|1,x41 (n253S2 x84f1 (x,nnn 11n xnnxxxdxxn1dxln(1而01n0 0x1從而S(x) [ln(1x)] [ln(1x)x 22x12(|x|1,x41 (n253S2 x84fxx（06研）將函2xAB解：f(x(2x)(1 2 1A3A(1x)B(2x)令x3AB令xf(x)33B311131(2 (1 (1x[1211()n (1)nx13(1)n1xnxn32 2解a0 (x1)dx0 a (x1) (x1) xcos24n22n222n n2k4n21)1]n,(k1,812n8故展開式為 (2n222f(x)x1(0x2)在(2,0)上作偶延拓，得到[2,2]2n81x,x[0,f(2)f(2f(x(2n2228（04n并證明當(dāng)1時，級 nfn(xxnnx1.x0fxnxn1n0f(x在[0nn 0,fn nx 0存唯一正xn111故當(dāng)10xx（04n并證明當(dāng)1時，級 nfn(xxnnx1.x0fxnxn1n0f(x在[0nn 0,fn nx 0存唯一正xn111故當(dāng)10xxnnx10x0，知0 n. n nn 1而正項(xiàng)級數(shù) 收斂，所以當(dāng)1時，級 x收斂nnnnf10，證明級數(shù)f2．f(xx0的某鄰域內(nèi)具有二階連續(xù)導(dǎo)數(shù)且x nf(x)0f(0)f'(0)1：由xf(