09_ISMCHAP9 - INSTRUCTOR'S SOLUTIONS MANUAL SECTION...

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INSTRUCTOR’S SOLUTIONS MANUALSECTION 9.1(PAGE 478)CHAPTER 9.SEQUENCES, SERIES,AND POWER SERIESSection 9.1Sequences and Convergence(page 478)1.2n2n2+1=22n2+1=1,85,95, . . .is bounded,positive, increasing, and converges to 2.2.2nn2+1=1,45,35,817, . . .is bounded, positive,decreasing, and converges to 0.3.4(1)nn=5,72,133, . . .is bounded, positive, andconverges to 4.4.sin1n=sin 1,sin12,sin13, . . .is bounded,positive, decreasing, and converges to 0.5.n21n=n1n=0,32,83,154, . . .is boundedbelow, positive, increasing, and diverges to infinity.6.enπn=eπ,eπ2,eπ3, . . .is bounded, positive,decreasing, and converges to 0, sincee< π.7.enπn/2=eπn. Sincee/π >1, the sequenceis bounded below, positive, increasing, and diverges toinfinity.8.(1)nnen=1e,2e2,3e3, . . .is bounded, alternat-ing, and converges to 0.9.{2n/nn}is bounded, positive, decreasing, and convergesto 0.10.(n!)2(2n)!=1n+12n+23n+3· · ·n2n12n.Also,an+1an=(n+1)2(2n+2)(2n+1)<12. Thus the sequence(n!)2(2n)!is positive, decreasing, bounded, and convergentto 0.11.{ncos(nπ/2)} = {0,2,0,4,0,6, . . .}is divergent.12.sinnn=sin 1,sin 22,sin 33, . . .is bounded and con-verges to 0.13.{1,1,2,3,3,4,5,5,6, . . .}is divergent.14.lim52n3n7=lim5n237n= −23.15.limn24n+5=limn4n1+5n= ∞.16.limn2n3+1=lim1n1+1n3=0.17.lim(1)nnn3+1=0.18.limn22n+11n3n2=lim12nn+1n21n21n3= −13.19.limenenen+en=lim1e2n1+e2n=1.20.limnsin1n=limx0+sinxx=limx0+cosx1=1.21.limn3nn=lim1+3nn=e3by l’Hˆopital’sRule.22.limnln(n+1)=limx→∞xln(x+1)=limx→∞11x+1=limx→∞x+1= ∞.23.lim(n+1n)=limn+1nn+1+n=0.24.limnn24n=limn2(n24n)n+n24n=lim4nn+n24n=lim41+14n=2.25.lim(n2+nn21)=limn2+n(n21)n2+n+n21=limn+1n1+1n+11n2=lim1+1n1+1n+11n2=12.347
SECTION 9.1(PAGE 478)R. A. ADAMS: CALCULUS26.Ifan=n1n+1n, thenliman=limn1nnnn+1n=lim11nnlim1+1nn=e1e=e2(by Theorem 6 of Section 3.4).27.an=(n!)2(2n)!=(1·2·3· · ·n)(1·2·3· · ·n)1·2·3· · ·n·(n+1)·(n+2)· · ·2n=1n+1·2n+2·3n+3· · ·nn+n12n.Thus liman=0.28.We have limn22n=0 since 2ngrows much faster thann2and lim4nn!=0 by Theorem 3(b). Hence,limn22nn!=limn22n·22nn!=limn22nlim4nn!=0.29.an=πn1+22n0<an< (π/4)n. Sinceπ/4<1,therefore(π/4)n0 asn→ ∞. Thus liman=0.30.Leta1=1 andan+1=1+2anforn=1,2,3, . . ..Then we havea2=3>a1. Ifak+1>akfor somek,thenak+2=1+2ak+1>1+2ak=ak+1.

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