theorem
  seq^\k is convergent implies lim seq = lim (seq^\k)
proof
  assume
A1: seq^\k is convergent;
A2: now
    let p;
    assume 0<p;
    then consider n1 such that
A3: for m st n1<=m holds |.(seq^\k).m-(lim (seq^\k)).|<p by A1,SEQ_2:def 7;
    take n=n1+k;
    let m;
    assume
A4: n<=m;
    then consider l be Nat such that
A5: m=n1+k+l by NAT_1:10;
    reconsider l as Element of NAT by ORDINAL1:def 12;
    m-k=n1+l+k+-k by A5;
    then consider m1 such that
A6: m1=m-k;
    now
      assume not n1<=m1;
      then m1+k<n1+k by XREAL_1:6;
      hence contradiction by A4,A6;
    end;
    then
A7: |.(seq^\k).m1-(lim(seq^\k)).|<p by A3;
    m1+k=m by A6;
    hence |.seq.m-(lim(seq^\k)).|<p by A7,NAT_1:def 3;
  end;
  seq is convergent by A1,Th21;
  hence thesis by A2,SEQ_2:def 7;
end;
