theorem Th21:
  seq^\k is convergent implies seq is convergent
proof
  assume seq^\k is convergent;
  then consider g1 such that
A1: for p st 0<p ex n st for m st n<=m holds |.(seq^\k).m-g1.|<p;
  take g1;
  let p;
  assume 0<p;
  then consider n1 such that
A2: for m st n1<=m holds |.(seq^\k).m-g1.|<p by A1;
  take n=n1+k;
  let m;
  assume
A3: n<=m;
  then consider l be Nat such that
A4: 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 A4;
  then consider m1 be Element of NAT such that
A5: m1=m-k;
  now
    assume not n1<=m1;
    then m1+k<n1+k by XREAL_1:6;
    hence contradiction by A3,A5;
  end;
  then
A6: |.(seq^\k).m1-g1.|<p by A2;
  m1+k=m by A5;
  hence thesis by A6,NAT_1:def 3;
end;
