theorem
  seq is absolutely_summable implies a * seq is absolutely_summable
proof
A1: for n being Nat
  holds ||.a * seq.||.n >= 0 & ||.a * seq.||.n <= (|.a.| (#) ||.seq
  .||).n
  proof
    let n be Nat;
    ||.a * seq.||.n = ||.(a * seq).n.|| by BHSP_2:def 3;
    hence ||.a * seq.||.n >= 0 by BHSP_1:28;
    ||.a * seq.||.n = ||.(a * seq).n.|| by BHSP_2:def 3
      .= ||.a * seq.n.|| by NORMSP_1:def 5
      .= |.a.| * ||.seq.n.|| by BHSP_1:27
      .= |.a.| * ||.seq.||.n by BHSP_2:def 3
      .= (|.a.| (#) ||.seq.||).n by SEQ_1:9;
    hence ||.a * seq.||.n <= (|.a.| (#) ||.seq.||).n;
  end;
  assume seq is absolutely_summable;
  then ||.seq.|| is summable;
  then |.a.| (#) ||.seq.|| is summable by SERIES_1:10;
  then ||.a * seq.|| is summable by A1,SERIES_1:20;
  hence thesis;
end;
