reserve n for Nat,
  a,b for Real,
  s for Real_Sequence;

theorem
  (for n holds s.n = (2*n-1)/(n*(n+1)*(n+2))) implies for n st n>=1
  holds Partial_Sums(s).n = 3/4-2/(n+2)+1/(2*(n+1)*(n+2))
proof
  defpred X[Nat] means Partial_Sums(s).$1= 3/4-2/($1+2)+1/(2*($1+1)*($1+2));
  assume
A1: for n holds s.n = (2*n-1)/(n*(n+1)*(n+2));
  then
A2: s.0 = (2*0-1)/(0*(0+1)*(0+2)) .= 0 by XCMPLX_1:49;
A3: for n be Nat st n>=1 & X[n] holds X[n+1]
  proof
    let n be Nat;
    assume that
    n>=1 and
A4: Partial_Sums(s).n = 3/4-2/(n+2)+1/(2*(n+1)*(n+2));
    n+1>=1+0 by NAT_1:11;
    then
A5: n+1>0 by NAT_1:13;
    n+2>=2 by NAT_1:11;
    then n+2>0 by XXREAL_0:2;
    then
A6: (n+1)*(n+2)>0 by A5,XREAL_1:129;
    n+3>=3 by NAT_1:11;
    then
A7: n+3>0 by XXREAL_0:2;
    Partial_Sums(s).(n+1) = 3/4-2/(n+2)+1/(2*(n+1)*(n+2))+ s.(n+1) by A4,
SERIES_1:def 1
      .= 3/4-2/(n+2)+1/(2*(n+1)*(n+2)) + (2*(n+1)-1)/((n+1)*((n+1)+1)*((n+1)
    +2)) by A1
      .= 3/4-(2/(n+2)-1/(2*((n+1)*(n+2))))+(2*n+1)/(((n+1)*(n+2))*(n+3))
      .= 3/4-((2*(n+1))/((n+2)*(n+1))-1/(2*((n+1)*(n+2)))) +(2*n+1)/(((n+1)*
    (n+2))*(n+3)) by A5,XCMPLX_1:91
      .= 3/4-((2*(n+1)*2)/(2*(n+2)*(n+1))-1/(2*((n+1)*(n+2)))) +(2*n+1)/(((n
    +1)*(n+2))*(n+3)) by XCMPLX_1:91
      .= 3/4-((4*(n+1))-1)/(2*((n+2)*(n+1))) +(2*n+1)/(((n+1)*(n+2))*(n+3))
    by XCMPLX_1:120
      .= 3/4-((4*n+3)*(n+3))/(2*((n+2)*(n+1))*(n+3)) +(2*n+1)/(((n+1)*(n+2))
    *(n+3)) by A7,XCMPLX_1:91
      .= 3/4-((4*n+3)*(n+3))/(2*((n+2)*(n+1))*(n+3)) +((2*n+1)*2)/(((n+1)*(n
    +2))*(n+3)*2) by XCMPLX_1:91
      .= 3/4-(((4*n+3)*(n+3))/(2*((n+2)*(n+1))*(n+3)) -((2*n+1)*2)/(((n+1)*(
    n+2))*(n+3)*2))
      .= 3/4-((4*n+3)*(n+3)-((2*n+1)*2))/(2*((n+2)*(n+1))*(n+3))by XCMPLX_1:120
      .= 3/4-(4*(n+1)*(n+2)-(n+1))/(2*((n+2)*(n+1))*(n+3))
      .= 3/4-((4*((n+1)*(n+2)))/(2*(n+3)*((n+2)*(n+1))) -(n+1)/(2*((n+2)*(n+
    1))*(n+3))) by XCMPLX_1:120
      .= 3/4-(4/(2*(n+3))-1*(n+1)/(2*(n+2)*(n+3)*(n+1))) by A6,XCMPLX_1:91
      .= 3/4-((2*2)/((n+3)*2)-1/(2*(n+2)*(n+3))) by A5,XCMPLX_1:91
      .= 3/4-(2/((n+1+2))-1/(2*(n+1+1)*(n+1+2))) by XCMPLX_1:91;
    hence thesis;
  end;
  Partial_Sums(s).(1+0)=Partial_Sums(s).0+s.(1+0) by SERIES_1:def 1
    .= s.0 + s.1 by SERIES_1:def 1
    .=(2*1-1)/(1*(1+1)*(1+2)) by A1,A2
    .= 3/4-2/(1+2)+1/(2*(1+1)*(1+2));
  then
A8: X[1];
  for n be Nat st n>=1 holds X[n] from NAT_1:sch 8(A8,A3);
  hence thesis;
end;
