 reserve n,s for Nat;

theorem Th84:
  Partial_Sums ReciTriangRS.n = 2 - 2 / (n + 1)
  proof
    defpred P[Nat] means
      Partial_Sums ReciTriangRS.$1 = 2 - 2/($1+1);
A1: P[0]
    proof
      ReciTriangRS.0 = 0;
      hence thesis by SERIES_1:def 1;
    end;
A2: for k being Nat st P[k] holds P[k+1]
    proof
      let k be Nat such that
A3:   P[k];
A4:  ReciTriangRS.(k+1) = 1 / Triangle (k+1) by Def13
          .= 2 / ((k + 1) * (k + 1 + 1)) by Th83
          .= 2 / (k * k + 3 * k + 2);
A5:  (k + 2) / (k + 2) = 1 by XCMPLX_1:60;
A6:  (k + 1) / (k + 1) = 1 by XCMPLX_1:60;
      Partial_Sums ReciTriangRS.(k+1) =
              2 * ((k + 2) / (k + 2)) - (2 / (k + 1)) * 1 +
             2 / ((k + 1) * (k + 2)) by A5,A3,A4,SERIES_1:def 1
           .= (2 * (k + 2)) / (k + 2) - (2 / (k + 1)) * ((k + 2) / (k + 2)) +
             2 / ((k + 1) * (k + 2)) by XCMPLX_1:74,A5
           .= ((2 * (k + 2)) / (k + 2)) * ((k + 1) / (k + 1)) -
              (2 * (k + 2)) / ((k + 1) * (k + 2)) + 2 / ((k + 1) * (k + 2))
                by A6,XCMPLX_1:76
           .= (2 * (k + 2) * (k + 1)) / ((k + 2) * (k + 1)) - (2 * (k + 2)) /
               ((k + 1) * (k + 2)) + 2 / ((k + 1) * (k + 2)) by XCMPLX_1:76
           .= (2 * k * k + 6 * k + 4 - (2 * k + 4)) / ((k + 2) * (k + 1))
                + 2 / ((k + 1) * (k + 2)) by XCMPLX_1:120
           .= (2 * k * k + 6 * k + 4 - (2 * k + 4) + 2) / ((k + 2) * (k + 1))
                by XCMPLX_1:62
           .= (2 * (k + 1)) * (k + 1) / ((k + 2) * (k + 1))
           .= (2 * (k + 2) - 2) / (k + 2) by XCMPLX_1:91
           .= 2 * (k + 2) / (k + 2) - 2 / (k + 2) by XCMPLX_1:120
           .= 2 - 2 / (k + 2) by XCMPLX_1:89;
      hence thesis;
    end;
    for k being Nat holds P[k] from NAT_1:sch 2(A1,A2);
    hence thesis;
  end;
