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

theorem
  (for n holds s.n = 1/(n*(n+1)*(n+2)*(n+3))) implies for n holds
  Partial_Sums(s).n = 1/18-1/(3*(n+1)*(n+2)*(n+3))
proof
  defpred X[Nat] means Partial_Sums(s).$1 = 1/18-1/(3*($1+1)*($1+2)
  *($1+3));
  assume
A1: for n holds s.n = 1/(n*(n+1)*(n+2)*(n+3));
A2: for n st X[n] holds X[n+1]
  proof
    let n;
    assume Partial_Sums(s).n = 1/18-1/(3*(n+1)*(n+2)*(n+3));
    then Partial_Sums(s).(n+1) =1/18-1/(3*(n+1)*(n+2)*(n+3)) + s.(n+1) by
SERIES_1:def 1
      .=1/18-1/(3*(n+1)*(n+2)*(n+3)) +1/((n+1)*(n+1+1)*(n+1+2)*(n+1+3)) by A1
      .=1/18-(1/(3*(n+1)*(n+2)*(n+3))-1/((n+1)*(n+2)*(n+3)*(n+4)))
      .=1/18-(1/(3*(n+1)*(n+2)*(n+3))-(1*3)/((n+1)*(n+2)*(n+3)*(n+4)*3)) by
XCMPLX_1:91
      .=1/18-(1/((3*(n+2)*(n+3))*(n+1))-1*3/((3*(n+2)*(n+3))*((n+1)*(n+4))))
      .=1/18-(1/((3*(n+2)*(n+3))*(n+1))-1/(3*(n+2)*(n+3))*(3/((n+1)*(n+4))))
    by XCMPLX_1:76
      .=1/18-(1/(3*(n+2)*(n+3))*(1/(n+1))-1/(3*(n+2)*(n+3))*(3/((n+1)*(n+4))
    )) by XCMPLX_1:102
      .=1/18-1/(3*(n+2)*(n+3))*(1/(n+1)-3/((n+1)*(n+4)))
      .=1/18-1/(3*(n+2)*(n+3))*(1/(n+4)) by Lm9
      .=1/18-1/(3*(n+1+1)*(n+1+2)*(n+1+3)) by XCMPLX_1:102;
    hence thesis;
  end;
  Partial_Sums(s).0 = s.0 by SERIES_1:def 1
    .= 1/(0*(0+1)*(0+2)*(0+3)) by A1
    .=1/18-1/(3*(0+1)*(0+2)*(0+3)) by XCMPLX_1:49;
  then
A3: X[0];
  for n holds X[n] from NAT_1:sch 2(A3,A2);
  hence thesis;
end;
