reserve k, m, n, p, K, N for Nat;
reserve i for Integer;
reserve x, y, eps for Real;
reserve seq, seq1, seq2 for Real_Sequence;
reserve sq for FinSequence of REAL;

theorem Th14:
  n>0 implies 0<=bseq(k).n & bseq(k).n<=1/(k!) & bseq(k).n<=eseq.k
  & 0<=cseq(n).k & cseq(n).k<=1/(k!) & cseq(n).k<=eseq.k
proof
  defpred P[Nat] means bseq($1).n<=1/($1!);
  assume
A1: n>0;
  then
A2: n ^ (-k)>0 by POWER:34;
A3: now
    let k;
    assume
A4: P[k];
    thus P[k+1]
    proof
      per cases;
      suppose
A5:     k<n;
        (1/(k+1))*(bseq(k).n)<=(1/(k+1))*(1/(k!)) by A4,XREAL_1:64;
        then
A6:     (1/(k+1))*(bseq(k).n)<=1/((k+1)!) by Th11;
        aseq(k).n>=0 by A5,Th13;
        then
A7:     (1/(k+1))*(bseq(k).n)*(aseq(k).n)<=(1/((k+1)!))*(aseq(k).n ) by A6,
XREAL_1:64;
        aseq(k).n<=1 by A5,Th13;
        then
A8:     1/((k+1)!)*(aseq(k).n)<=1/((k+1)!) by XREAL_1:153;
        bseq(k+1).n=(1/(k+1))*(bseq(k).n)*(aseq(k).n) by A1,Th6;
        hence thesis by A7,A8,XXREAL_0:2;
      end;
      suppose
        k>=n;
        then
A9:     k+1>n by NAT_1:13;
        bseq(k+1).n = (n choose (k+1))*(n ^ (-(k+1))) by Def2
          .= 0*(n ^ (-(k+1))) by A9,NEWTON:def 3
          .= 0;
        hence thesis;
      end;
    end;
  end;
A10: bseq(k).n=(n choose k)*(n ^ (-k)) by Def2;
  hence 0<=bseq(k).n by A2;
A11: P[0] by Th10,NEWTON:12;
  for k holds P[k] from NAT_1:sch 2(A11,A3);
  hence
A12: bseq(k).n<=1/(k!);
  hence bseq(k).n<=eseq.k by Def5;
  hence thesis by A10,A2,A12,Th3;
end;
