reserve i,n,m for Nat,
        r,s for Real,
        A for non empty closed_interval Subset of REAL;

theorem Th16:
  Partial_Sums(Leibniz_Series).1 = 2/3 &
    (n is odd implies
      Partial_Sums(Leibniz_Series).(n+2) =
      Partial_Sums(Leibniz_Series).n + 2/(4*n^2 + 16*n +15))
proof
  set L=Leibniz_Series,P=Partial_Sums L;
  set n1=n+1,n2=n+2;
A1: P.0 = L.0 by SERIES_1:def 1
       .= (-1) |^0 * (1|^(2*0+1))/ (2*0+1) by Def2
       .= 1;
  L.1 = (-1) |^1 * (1|^(2*1+1))/ (2*1+1) by Def2
       .= -1/3;
  then P.(0+1) = -1/3+ 1 by A1,SERIES_1:def 1;
  hence P.1 = 2/3;
  assume
A2:  n is odd;
A3:P.(n1+1) = P.n1 + L.n2 by SERIES_1:def 1
           .= P.n + L.n1 + L.n2 by SERIES_1:def 1;
A4:L.n1 = (-1) |^n1 * (1|^(2*n1+1))/ (2*n1+1) by Def2
       .= 1*(1|^(2*n1+1))/ (2*n1+1) by A2,POLYFORM:8
       .=(1*(2*n2+1))/((2*n1+1)*(2*n2+1)) by XCMPLX_1:91;
  L.n2 = (-1) |^n2 * (1|^(2*n2+1))/ (2*n2+1) by Def2
      .= (-1)*(1|^(2*n2+1))/ (2*n2+1) by A2,POLYFORM:9
      .=-1/(2*n2+1)
      .=-(1*(2*n1+1))/((2*n2+1)*(2*n1+1)) by XCMPLX_1:91;
  hence thesis by A3,A4;
end;
