微积分全英微积分全英 (77).pdf

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1、Exercises Lesson（IV）for Infinite SeriesExample 1Show that if is a positive integer,then 1lim0.pnn=0Find the convergence region and sum function of power series(1)(3)=+nnnnx0Coefficient of power series(1)(3)=+：nnann1|(2)(4)limlim1|(1)(3)+=+nnnnannannRadius of convergence:1=R00when 1,series(1)(3)and(1

2、)(3)(1)are divergence.=+nnnxnnnnConvergence region is(1,1).Example 1Show that if is a positive integer,then 1lim0.pnn=0Find the convergence region and sum function of power series(1)(3)=+nnnnx0let()(1)(3),(1,1)=+nnS xnnxx1110000()(3)(2)+=+=+xnnnnnnS t dtnxnxx10,1+=nnxxxExample 1Show that if is a pos

3、itive integer,then 1lim0.pnn=0Find the convergence region and sum function of power series(1)(3)=+nnnnx221120002(2)(2)()1(1)+=+=+=xnnnnxxxnxntdtxx1100223()(2)23 (),(1,1)(1)1(1)+=+=+=nnnnS xnxxxxxxxxxxExample 2Show that if is a positive integer,then 1lim0.pnn=220Find the convergence region and sum fu

4、nction of power series.(1)(21)+=+nnxnn24222|(2)(23)lim|(1)(21)+=+nnnxnnxxnnwhen|1,absolute convergence;when|1,divergencexx01when 1,series is convergence.(1)(21)=+nxnnConvergence region is 1,1.Example 2Show that if is a positive integer,then 1lim0.pnn=220Find the convergence region and sum function o

5、f power series.(1)(21)+=+nnxnn220let (),1,1(1)(21)+=+nnxf xxnn2122002()2,()2,(1,1),211+=+nnnnxfxfxxxnx(0)0,(0)0=ff2002()()ln(1)ln(1),1=+xxfxft dtdttttExample 2Show that if is a positive integer,then 1lim0.pnn=220Find the convergence region and sum function of power series.(1)(21)+=+nnxnn00()()ln(1)l

6、n(1)=+xxf xf t dttt dt (1)ln(1)(1)ln(1),(1,1)=+xxxx x11(1)lim()2ln2,(1)lim()2ln2+=xxff xff x11(1)=nnnxExample 3Show that if is a positive integer,then 1lim0.pnn=111Find the sum function()of series(1)in interval(1,1)=nnnS xnx111()(1)=nnnS xnx21()1(1)=+xxxExample 4Show that if is a positive integer,th

7、en 1lim0.pnn=023Find(1)(21)!=+nnnn210sin(1)(|)(21)+=+！nnnxxxn20cos(1)(|)(2)=+！nnnxxxn0023(21)2(1)(1)(21)!(21)!=+=+nnnnnnnn0011(1)2(1)cos12sin1(2)!(21)!=+=+nnnnnnExample 5Show that if is a positive integer,then 1lim0.pnn=0Find the sum of the first n terms of sequence(1)in(0,).(2)!=+nnnxn20cos(1)(|)(2

8、)!=+nnnxxxn200()(1)(1)cos(2)!(2)!=nnnnnnxxxnnExample 6Show that if is a positive integer,then 1lim0.pnn=111Series satisfies 1,(1)(),proof:when|12+=+=+,nnnaananax1 is convergence,and find it sum=.nnna xFrom condition,0,na11|2limlim1,|1+=+nnnnnaan1The convergence radius of is 1=nnna x1when|1,is conver

9、gence.=nnnxa xExample 6Show that if is a positive integer,then 1lim0.pnn=111Series satisfies 1,(1)(),proof:when|12+=+=+,nnnaananax1 is convergence,and find it sum=.nnna x1when|1()=,nnnxS xa x1111()1(1)+=+nnnnnnS xna xnax111111()().22=+=+nnnnnnna xa xxS xS xExample 6Show that if is a positive integer

10、,then 1lim0.pnn=111Series satisfies 1,(1)(),proof:when|12+=+=+,nnnaananax1 is convergence,and find it sum=.nnna x11()()()22(1)11=CS xS xS xxxx1(0)02()2(1)1=SCS xxExample 7Show that if is a positive integer,then 1lim0.pnn=11Find the convergence interval of,and discuss the 3(2)convergence of the end p

11、oint of the interval.=+nnnnxn11121()3(2)13limlim23(2)(1)331()(1)3+=+nnnnnnnnnnnnits convergence radius is 3,and convergence interval is(3,3).13111when 3,is divergence,so it is divergence at 3;3(2)2=+nnnnxxnnnExample 7Show that if is a positive integer,then 1lim0.pnn=11Find the convergence interval o

12、f,and discuss the 3(2)convergence of the end point of the interval.=+nnnnxn(3)1121when 3,(1),3(2)3(2)=+nnnnnnnxnnn11121(1)and are convergence3(2)=+nnnnnnnnit is convergence at 3=.xExample 8Show that if is a positive integer,then 1lim0.pnn=211What is the convergence radius of?(3)2=+nnnnnx111(1)2(3)li

13、m|lim|2(3)+=+nnnnnnnnanan21()113lim|2332()3+=+nnnnn3=RExample 9Show that if is a positive integer,then 1lim0.pnn=11(2)is convergence at 0,and divergence at 4.Find the convergence domain of(3).=+=nnnnnnaxxxax00Series(2)whose convergence center is 2,=+=nnnaxxand it is convergence at 0 and divergence a

14、t 4.=xxits radius is 2,and convergence domain:(-4,0Example 9Show that if is a positive integer,then 1lim0.pnn=11(2)is convergence at 0,and divergence at 4.Find the convergence domain of(3).=+=nnnnnnaxxxax0(2)is convergence at 222=+nnnaxx0(3)is convergence at 232,=nnnaxx0so the convergence domain of(3)is(1,5=nnnaxExample 10Show that if is a positive integer,then 1lim0.pnn=121111(1)2,5,find=nnnnnnnaaa12121211115,(1)=2=nnnnnnnnnaaaa213=nna1538=+=nnaExercises IV