高等数学无穷级数上课习题与答案第一次作业??写出级数√??????????2+24+√2468+?的一般项。.2246+??????1??(??2)解:一般项为u??=(2??)!!∞2nn!2nn!。2.已知级数∑收敛,试求极限limnnn→∞nnn=1解:由级数收敛必要条件可知2nn!limnn=0...
1时,级数∑(n+1)p收敛,知所求级数绝对收敛。n=1∞()????23.鉴别级数∑-1??的敛散性。2??2-??=1(-1)nn2≠0,由级数收敛的必要条件知级数发散。解:因为lim-nn→∞2n2∞14.鉴别级数∑??????(????+??????)的敛散性。??=1∞1∞1为交错级数,解:级数∑sin(nπ+()nsinlnn)=∑lnnn=1-1n=1111因为limsinlnn=limsinlnn?lnn=∞,n→∞1n→∞11nlnn?n∞1发散,知所求级数不绝对收敛。由∑nn=1又因为limsin11lnn单一递减,n→∞=0,且sinlnn由莱布尼茨定理知所求级数条件收敛。??(??+1)2??n!鉴别级数∑(-1)2????的敛散性。??=1解:因为limun+12n=2||=lim1<1n→∞unn→∞e(1+n)n(n+1)2nn!所以级数∑(-1)2n绝对收敛。n=1n∞(-1)??-1??????6.鉴别级数∑??的敛散性。??=1lnn1∞n-1lnn()解:由un=>(n≥3)知∑-1非绝对收敛。nnn=1n设f(x)=lnx′1-lnxx,因为f()=()xx2<0x>e,当x>e时,f(x)单一减少,所以un>un+1又因为limf(x)=limlnx1=0,x=limx→∞x→∞x→∞x所以limun=0,由莱布尼兹定理知所求级数条件收敛。n→∞第四次作业1.判断级数1111?3+3?5+5?7+?的敛散性,若收敛求其和。解:因为lim1[(1-11-11-11????=lim)+(3)+?+(2??-12??)]=2??→∞??→∞235+11所以级数收敛,且S=2∞∞若级数∑un收敛于S,求级数∑(un+un+1)的和。n=1n=1∞∞‘解:设Sn为∑un前n项和,Sn为∑(un+un+1)前n项和n=1n=1S‘n=(u1+u)+(u2+u)+?+(un+un+1)23=2(u1+?+un)-u1+un+1=2Sn-u1+un+1∞因为∑un收敛,所以limSn=S,且limun+1=0n→∞n→∞n=1故limS‘n=2S-u1n→∞∞3.当k>0时,鉴别级数∑(-1)??k+n的敛散性。n2n=1k+n,因为limun=limk+n?2=limk+n解:设un=n2n=1n→∞1?nn→∞1?nn→∞n∞1由∑发散,知所求级数不绝对收敛。n=1n设f(x)=k+x′()=-2kx()所以un单一减少x2,因为fxx4<0x>0,又由于limk+nn2=0,故所求级数条件收敛。n→∞∞sinna鉴别级数∑2的敛散性。nn=1sinna1∞1∞sinna解:因为|2|≤2,由级数∑2收敛,知级数∑|2|收敛nnnn=1nn=1∞sinna绝对收敛。所以级数∑n2n=14??5.鉴别级数∑5??-3??的敛散性。n=1????+5??-3??1-(3??41=4lim5)??=<1解:因为lim+??+1=4lim3??→∞??????→∞5??→∞5??1-335-(5)4??所以级数∑??-??收敛。53n=1∞()??6.鉴别级数∑(-1)??+1n+1的敛散性。??+1n=12n解:设u??=(n+1)??u??1(1+1??=????+1,因为lim?=lim)22n??→∞??→∞2??1??∞∞1而∑发散,故∑un发散,即原级数不绝对收敛。nn=1n=1u??+1????+1??+2??+1??2+2????+1(=(<1,同时,=(??+)??+))u??11??2+2??+1即u且limu11??>u??+1lim[(1+)]=0n??→∞??=2????→∞??∞(n+1)n故级数∑(-)n+1条件收敛。12nn+1n=1第五次作业∞????(??+1)1.求幂级数∑????+1的收敛域。??+1??=1??1??????+2??+11解:??=lim|??+()=1,进而??==1????|=lim??+2?????(??+1)????→∞??→∞∞()lnn+1发散当x=1时,级数∑n+1n=1∞当x=-1时,级数∑(-1)n+1ln(n+1)收敛n+1n=1故收敛区间为[-1,1)∞??2.求幂级数∑(-1)??-1(2??-3)的收敛域。??=1解:令t=2x-3,∞????,??=lim??+lim2??-1=1原级数为∑??=1(-1)??-12??-1|??1|=2??+1→∞??→??????∞故R=1∞1当t=1时,∑(-1)n-11收敛,2n-n=1∞(-)n∞1当时,∑()n-1t=-11n=1-12n-1=∑(-2n-1)发散n=1故-1