Homework Solutions 1.

  1. 1.

    For any n1, let q2n-1=n and q2n=n-1n.

  2. 2.

    For any n1, let qn=n2,

  3. 3.

    For any n1, let q2n-1=n, q2n=1n.

  4. 4.

    Fix ε>0. Then we have N1 so that if n,m>N, then |qn-qm|<ε/2 and |rn-rm|<ε/2. Then |(qn+rn)-(qm+rm)|<ε showing that {qn+rn}n=1 is a Cauchy-sequence.