第 9 章 無窮級數 (Infinite Series)

1 9 (Infinite series ) .. Taylor Taylor .. (Sequences) (sequence) N f, f(n) an fa1, a2, a3, ..g fang fang1n=1 a1 (first term),an n [ ] (1)an=pn,fang=f1,p2,p3, .. ,pn, ..g (2)bn= ( 1)n+11n,fbng=f1, 12,13, 14, ..g (3)cn=n 1n,fcng=f0,12,23,34, .. ,n 1n, ..g 89 9 (4)dn= ( 1)n+1,fdng=f1, 1,1, 1, .. ,( 1)n+1, ..g , , , (recursion formula) (1)a1= 1,an=an 1+ 1 (2)a1= 1,an=nan 1 (3) :x0= 1,xn+1=xn (sinxn x2ncosxn 2xn) sinx x2= 0 (4)Fibonacci :a1= 1,a2= 1,an+1=an+an 1 (1) fang 8 >0,9N n > N jan Lj< , fang (limit) L limn!1an=L n!1,an!L (2) , (converge), (diverge) (3) fang M, N, 8n > N)an> M, fang (diverges to infinity) limn!1an=1, an!

2 1 (4) m, N, 8n > N)an< m, fang (divergesto negative infinite) limn!1an= 1, an! 1 (1)limn!1k=k (2)limn!11n= 0 (3) r >0, limn!11nr= 0 (4)limn!1pn=1 frng f1, 1,1, 1, ..( 1)n+1, ..g [ ] , f1, 2,3, 4,5, 6, ..g f1,0,2,0,3,0,4, ..g fang fbng ,c (1)limn!1(an+bn) = limn!1an+ limn!1bn (2)limn!1(an bn) = limn!1an limn!1bn (3)limn!1c an=c limn!1an (4)limn!1anbn= limn!1an limn!1bn , 90 9 (5) limn!1bn6= 0, limn!1anbn=limn!1anlimn!1bn (6) p2R an>0, limn!1apn= limn!1an p ( p <0, limn!1an6= 0 ) :(1)limn!1( 5n2) (2)limn!1(n 1n) (3)limn!14 7n6n6+3 (4)limn!1(n pn+ 1pn+ 3) (1) fang,fbng N n,an bn , , limn!1an limn!1bn.(2)( , , Sandwich Theorem, Squeeze Theorem) fang,fbng,fcng N n,an bn cn limn!

3 1an= limn!1cn=L, limn!1bn=L (1) limn!1janj= 0, limn!1an= 0 (2) jbnj cn, cn!0, bn!0 (3) limn!1an= 0, fbng , limn!1anbn= 0 [ ] limn!1janj6= 0, (1) :(1)limn!1cosnn(2)limn!1( 1)n1n ( , The continuous function theorem for sequences). fang , an!L f(x) , an , L , f(an)!f(L) [ ] : f(x) =bxc,an=n 1n, limn!1an= 1, limn!1f(an) = 06=f(1). :(1)limn!1qn+1n(2)limn!121n(3)limn!1sin n , 91 9 f(x) [n0,1) , fang an=f(n),8n n0, limx!1f(x) =L)limn!1an=L.[ ] limn!1sinn = 0, limx!1sinx :(1)limn!1lnnn (2)limn!1lnn25n2 (3)limn!12n5n (4)limn!1(1 +xn)n (5)limn!1(n+1n 1)n (6)limn!1x1n, (x >0) (7)limn!1xn, (jxj<1) (8)limn!1npn (9)limn!1npn2 (10)limn!1np3n (11)limn!1xnn! (12)limn!1n!nn :(1)limn!1tan 1nn (2)limn!]

4 1nsin1n (3)limn!1lnnln 2n (4)limn!1(ln(n+ 1) lnn) (1) an< an+18n 1, fang (2) an an+1,8n, fang (nondecreasing sequence) (3) an> an+18n 1, fang , 92 9 (4) an an+1,8n, fang (nondecreasing sequence) (5)fang , (monotonic) (6) N, an< an+1,8n > N, fang (ultimately increasing) (1) M, an M,8n, fang (bounded above), M (upper bound) (2) N, an N,8n, fang (bounded below), M (lowerbound) (3)fang , (bounded sequence) :(1)f1,2,3, .. , n, ..g(2)f12,23,34, .. ,nn+1, ..g,(3)f3,3,3, ..g (1) , , (2) 3n+5 nn2+1 ( monotonic sequence theorem) [ ] :(1) , f( 1)ng (2) , fng a1= 1,an+1= 3 1an, limn!1an fang , (a)a1= 2,an+1=12(an+ 6) (b)a1= 10,an+1=12(an+ 6) (c)a1= 2,an+1= 2 (an+ 6) [ ] whether the sequence converges or diverges.

5 If it converges, findthe limit.(a)an=( 1)n 1nn2+1(b)an=( 1)nn3n3+2n2+1(c)an=(2n 1)!(2n+1)!, , 93 9 (d)an=np21+3n,(e)an= 2 ncosn ,(f)an=sin 2n1+pn,(g)an=n!2n.[ ] (a)Determine whether the sequence defined as follows is convergent ordivergent:a1= 1, an+1= 4 anforn 1.(b)What happens if the first term isa1= 2?[ ] whether the sequence is increasing, decreasing, or not the sequence bounded?(a)an=2n 33n+4,(b)an=ne n,(c)an=n+ (Infinite series ) 1 1 + 1 1 + 1 1 + ?[ ] Guido Ubaldus , something has been created out of nothing (1) fang, a1+a2+a3+ +an+ (infiniteseries), an n (2) sn=nPk=1ak, fsng (sequence of partial sums), sn n (3) fsng , limn!1sn=s, 1Pn=1an (converges), s , 1Pn=1an=s fsng , (diverges) (4) 1Pn=1an , Rn=s sn n (remainder) (1) , , (2) , 1Pn=1an sn= 3 n2 n, an 1Pn=1an , 94 9 (geometric Series) 1Pn=1arn 1, r ,a6= 0 jrj<1, a1 r; jrj 1, (1)1Pn=119 13 n 1,(2)1Pn=1( 1)n5n4n,(3)5 103+209 4027+ ,(4)1Pn=122n31 n,(5) a , r (0< r <1) (6) c , 1Pn=2(1 +c) n= 2 ( , telescoping) fang, 1Pn=1(an an+1) limn!

6 1an , a1 limn!1an (1)1Pn=11n(n+1),(2)1Pn=1lnn+1n,(3)1Pn=13 n2+3n+1(n2+n)3 (1) 1Pn=1an , limn!1an= 0 (2)( ) limn!1an 0, 1Pn=1an [ ] (1) , 1Pn=11n (2) , (harmonic Series) 1Pn=11n , 95 9 (1)1Pn=1n2,(2)1Pn=1n+1n,(3)1Pn=1( 1)n+1,(4)1Pn=1n25n2+4 1Pn=1an(an6= 0) , 1Pn=11an ? 1Pn=1an,1Pn=1bn , (1)1Pn=1can=c1Pn=1an (2)1Pn=1(an+bn) =1Pn=1an+1Pn=1bn (3)1Pn=1(an bn) =1Pn=1an 1Pn=1bn (1)1Pn=1(anbn) =1Pn=1an1Pn=1bn :an=bn= 12 n (2) 1Pn=1an ,1Pn=1bn , 1Pn=1(an bn) (3) 1Pn=1an,1Pn=1bn ,1Pn=1(an bn) an= 1, bn= 1,8n (1)1Pn=13n 1 16n 1,(2)1Pn=142n,(3)1Pn=1h3n(n+1)+12ni , ( series with positive terms) 1Pn=1an , 96 9 [ ] whether the series is convergent or divergent.

7 If it is convergent,find its sum.(a)1Pn=1en3n 1,(b)1Pn=1n(n+2)(n+3)2,(c)1Pn=11+2n3n,(d )1Pn=1np2,(e)1Pn=1(cos 1)n,(f)1Pn=1(1en+1n(n+1)),(g)1Pn=1lnnn+1 ,(h)1Pn=13n(n+3),(i)1Pn=1(cos1n2 cos1(n+1)2).[ ] the values ofxfor which the series converges. Find the sum of theseries for those values ofx.(a)1Pn=0(x 2)n3n,(b)1Pn=0sinnx3n,(c)1Pn= (Integral Test) ( , integral test) fang f(x) [N,1) , f(n) =an,8n N, 1Pn=1an R1Nf(x)dx ( ) f(x) x 1 , an=f(n),1Pn=1an sn ,s , Rn=s sn, Z1n+1f(x)dx Rn Z1nf(x)dx, , 97 9 sn+Z1n+1f(x)dx s sn+Z1nf(x)dx. (1) 1Pn=11n2+1 (2) 1Pn=1lnnn (p- )1Pn=11np p >1 p , :(1)1Pn=31n(lnn)p (2)1Pn=1 pn 1n+1 (a) 10 1Pn=11n3, (b) , ?[ ] whether the series is convergent or divergent.]

8 (a)1Pn=1nn2+1,(b)1Pn=1n2e n3,(c)1Pn=1pn+4n2,(d)1Pn=11n(lnn)2,(e)1P n=1e1nn2.[ ] the values ofpfor which the series is convergent.(a)1Pn=31nln[ln(lnn)]p,(b)1Pn =1n(1 +n2)p.[ ] to three decimal places.[ ] many terms of the series1Pn=21n(lnn)2would you need to add to find itssum to within , 98 9 (Comparison Test) ( ) 1Pn=1an , :(1) 1Pn=1cn an cn,8n N, 1Pn=1an (2) 1Pn=1dn an dn,8n N, 1Pn=1an :(1)1Pn=155n 1 (2)1Pn=1lnnn (3)1Pn=112n2+4n+3 (4)1Pn=11n! (5)5 +23+17+ 1 +12+p1+14+p2+ +12n+pn+ 100 1Pn=11n3+1, ( , Limit Comparison Test) an, bn>0,8n > N (1) limn!1anbn=c >0, 1Pn=1an 1Pn=1bn (2) limn!1anbn= 0, 1Pn=1bn , 1Pn=1an (3) limn!1anbn=1, 1Pn=1bn , 1Pn=1an :(1)1Pn=12n+1n2+2n+1 (2)1Pn=12n2+3np5+n5 (3)1Pn=1lnnn32 , 99 9 (4)1Pn=11+nlnnn2+5 (5)1Pn=112n 1 (1) 1Pn=1an , 1Pn=1sin(an) ?

9 (2) 1Pn=1an 1Pn=1bn , 1Pn=1anbn ?[ ] whether the series is convergent or divergent.(a)1Pn=1n 1n2pn,(b)1Pn=13pnpn3+4n+3,(c)1Pn=1tan ,(d)1Pn=14n+13n 2,(e)1Pn=1n+4nn+6n,(f)1Pn=1(1 +1n)2e n,(g)1Pn=1e1nn,(h)1Pn=1n!nn,(i)1Pn=1sin( 1n),(j)1Pn=11n1+1n,(k)1Pn=11lnn,(l)1Pn=1 lnnenpn.[ ] the sum of first 10 terms to approximate the sum of the seriesPn!11pn4+ the error. , 100 9 (Ratio Test) ( , d Almbert) 1Pn=1an , limn!1an+1an= , :(1) <1, 1Pn=1an (2) >1 , 1Pn=1an (3) = 1, :(1)1Pn=1n4+3n3+2n2+4n+53n (2)1Pn=1(n52 2007n24+100n 90) lnnn32 (3)1Pn=1pn3+3n 1n2+5 (4)1Pn=1 n+5en (5)1Pn=1(2n)!n!n! (6)1Pn=1nnn! (7)1Pn=14nn!n!(2n)! (8)an= n2nn 12nn ` k , 1Pn=1(n!)`(kn)! (1) 1Pn=11n2n, s5 , (2) n , sn [ ] whether the series is convergence, or divergence.

10 (a)1Pn=1(2n)!(n!)2,(b)1Pn=12 4 6 (2n)n!. , 101 9 (Root Test) ( ) 1Pn=1an , limn!1npan= , :(a) <1, 1Pn=1an (b) >1 , 1Pn=1an (c) = 1, :(1)1Pn=1 11+n n (2)1Pn=1 2n+33n+2 n (3)1Pn=1n22n (4)1Pn=12nn2 (5)an= n2nn 12nn [ ] whether the series is convergence, or divergence.(a)1Pn=1(n2+12n2+1)n,(b)1Pn=1 (1 +1n)n2, (Alternating series ) bn 0,8n, 1Pn=1( 1)n+1bn 1Pn=1( 1)nbn (alternating series ) ( , Leibniz ) 1Pn=1( 1)n+1bn, bn 0 , (a)fbng (b)limn!1bn= 0 , 102 9 ( ) 1Pn=1( 1)n+1bn , L, sn Rn jRnj=jsn Lj bn+1 :(1)1Pn=1( 1)n+1n (2)1Pn=1( 1)n3n4n 1 (3)1Pn=1( 1)n+1n2n3+1 (4)1Pn=1ncosn 2n n ,bn=1n; n ,bn=1n2 1Pn=1( 1)n 1bn s8 1Pn=0( 1)n12n , 1Pn=1( 1)nn!