Aloha :)
sin(2n+1nπ)=sin(n1⋅(2n+1)n1⋅nπ)=sin(2+n1π)→sin(2+0π)=1
n2+n+1−n2+1=(n2+n+1+n2+1)(n2+n+1a−n2+1b)(n2+n+1a+n2+1b)=n1+n1+n21+n1+n21(n2+n+1)a2−(n2+1)b2=n1+n1+n21+n1+n21n=1+n1+n21+1+n211→1+0+0+1+01=21
log(n+1)−2log(n)=log(n+1)−log((n)2)=log(n+1)−log(n)=log(nn+1)=log(1+n1)→log(1+0)=0
Die Folge (xn) konvergiert also gegen (1∣∣∣21∣∣∣0).