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Guia1 lu

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Guia1 lu

  1. 1. Guía N°1<br />Luisa Fernanda Sánchez Gómez<br />CRITERIO DE SUMAS PARCIALES<br />Ejemplo para desarrollar en clase:<br />Determinar la convergencia de la serie: <br />a. <br />s1=12<br />s2=12+14=34<br />s3=12+14+18=78<br />s4=12+14+18+116=1516<br />s5=12+14+18+116+132=3132<br />sn=2n-12n<br />limn->∞2n-12n limn->∞1-12n 1-limn->∞12n <br />S=1<br />Por lo tanto CONVERGE y converge a 1<br />b. <br />s1=2<br />s2=2+32=54<br />s3=2+32+54=104<br />s4=2+32+54+98=298<br />s5=2+32+54+98+1716=7516<br />No se aproxima a un valor determinado, por lo tanto DIVERGE <br />Ejercicios para desarrollar fuera de la clase:<br />Determine por sumas parciales si las series convergen o divergen:<br />a. <br />s1=12=0,5<br />s2=12+13=56=0,83<br />s3=12+13+14=1312=1,08<br />s4=12+13+14+15=7760=1,28<br />s5=12+13+14+15+16=8760=1,45<br />No se aproxima a un valor determinado, por lo tanto DIVERGE <br />b. <br />s1=1<br />s2=1+14=54=1,25<br />s3=1+14+19=4936=1,36<br />s4=1+14+19+116=205144=1,4<br />s5=1+14+19+116+125=45193600=1,46<br />Se aproxima a 2, de manera que CONVERGE <br />c. <br />s1=110<br />s2=110+1100=11100=0,11<br />s3=110+1100+11000=1111000=0,11<br />s4=110+1100+11000+110000=111110000=0,11<br />s5=110+1100+11000+110000+1100000=11111100000=0,11<br />sn=10n<br />La suma parcial se aproxima a 0,12 entonces CONVERGE <br />SERIE GEOMETRICA<br />Ejercicios para resolver en clase:<br />a. <br />s1=1<br />s2=34<br />s3=916<br />s4=2764<br />s5=81256<br />r=34=0,75<1 Converge, entonces:<br />s=31-34=12<br />s=12<br />b. <br />s1=12<br />s2=14<br />s3=18<br />s4=116<br />s5=132<br />r=12=0,5<1 Converge, entonces:<br />s=11-12=2<br />s=2<br />Ejercicios para resolver fuera de la clase<br />a. <br />s1=1π<br />s2=eπ2<br />s3=eπ3<br />r=eπ= 0,86 < 1 CONVERGE, entonces:<br />s=1π1-eπ = 1ππ-eπ = 1π-e<br />s=1π-e=2,3 <br />b. <br /> n=0∞2n5n + n=0∞3n5n n=0∞(25)n + n=0∞(35)n<br />r=25= 0,4< 1 CONVERGE, entonces:r=35=0,6 < 1 CONVERGE, entonces:<br />s=11-25= 53 s=11-35= 52<br />SERIE TELESCOPICA<br />Ejercicios para desarrollar en la clase:<br />a. <br />1n+1(n+2)=An+1+Bn+2<br />1n+1(n+2)=An+2+B(n+1)n+1(n+2)<br />1=An+2A+Bn+B<br />1=n(A+B)+2A+B<br /> A+B=0 <br />2A+B=1 (-1)<br /> A+B=0 <br />-2A-B=-1 <br /> -A =-1<br />Entonces A=1 y B=1<br />n=1∞(1n+1-1(n+2))<br />limn->∞12-1n+2<br />S=12<br />La serie CONVERGE:<br />S1=(12-13)<br />S2=(12-13+13-14)=( 12-14)<br />S2=(12-13+13-14+14-15)=( 12-15)<br />b. <br />1nn+1(n+2)=An+B(n+1)+C(n+2)<br />1nn+1(n+2)=An+1+Bnn(n+1)+C(n+2)<br />1nn+1(n+2)=An+1+Bnn+2+C(n(n+1))n(n+1)(n+2)<br />1=(An+A+Bn)(n+2)+Cn2+Cn<br />1=An2+2An+An+2A+Bn2+2Bn+Cn2+Cn<br />1=n2(A+B+C)+n(3A+2B+C)+2A<br />A+B+C=0-2B – 2C = 1 <br />3A+2B+C=02B + C = 32<br /> -C = -12<br />2A=1<br /> A = 12C = 12<br />n=1∞12n+1(n+1)+12(n+2) <br />n=1∞12n+1(n+1)+12(n+2) <br />= 2n+2- 2n4n(n+2)<br />= 4n+44n2+ 8n<br />S1=12 - 12 + 16 = 16 <br />S2=12 - 12 + 16 + 14 -13 +18 = 524 <br />S3=12 - 12 + 16 + 14 -13 +18 + 16 - 14 + 110 = 9040 <br />CRITERIOS DE COMPARACION<br />Ejercicio para resolver en la clase:<br />a. DIVERGE<br />b. DIVERGE<br />

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