1. Guía N°3<br />Series de potencias<br />n=1∞-1n+1 x2n-12n-1!<br />an=x2n-12n-1! <br />an+1=x2(n+1)-12(n+1)-1! <br />an+1=x2n+2-12n+2-1!= x2n+12n+1! <br />an+1=x2nx2n+1(2n-1)! <br />limn->∞x2nx2n+12n-1!x2nx-12n-1!= x2n+1x-1= x22n+1 <br />limn->∞x22n+1= x2limn->∞12n+1 <br />x2.0=0 Entonces CONVERGE<br />n=1∞x-3n4n.3n <br />an=x-3n4n.3n <br />an+1=x-3n+14n+1.3n+1=x-3n+14n+4.3n+1 <br />an+1=x-3n.x-34n+1.3n.3 <br />limn->∞x-3n.x-34n+1.3n.3x-3n4n.3n <br />limn->∞x-3n.x-3.4n.3n4n+1.3n.3.x-3n <br />limn->∞x-3.4n4n+1.3=x-3.n3n+1 <br />x-33limn->∞nn+1=x-33.0=0 <br />0<1 por lo tanto n=1∞x-3n4n.3n CONVERGE<br /> <br />n=1∞-1n+1xnn(lnn)2<br />an=xnn(lnn)2 <br />an+1=xn+1(n+1)(ln(n+1))2 <br />=limn->∞xn x(n+1)(ln(n+1))2 xnn(lnn)2 <br />= limn->∞xn(n+1)(ln(n+1))2xn x(n(lnn))2 <br />=-1<x<1 <br />=c=0 y r=1 <br />Si x= -1 entonces DIVERGE por el criterio de la integral<br />n=1∞4n+1x2nn+3<br />an=4n+1x2nn+3 <br />an+1=4n+2x2n+1n+4 <br />an+1=16*4nx2nx2n+4 <br />= limn->∞4n+1x2nn+316*4nx2nx2n+4 <br /> = limn->∞4x2 (n+3)n+4<br />4x2<1 <br />=-12<x<12 <br />=c=0 y r=12 <br />n=1∞(-1)n+1xn<br />an=xn <br />limn->∞xn = ∞ Como el imite tiende a infinito entonces n=1∞(-1)n+1xn<br /> DIVERGE<br />