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

theorem
  (for n holds s.n = 1/(n*(n+1)*(n+2))) implies for n holds Partial_Sums
  (s).n = 1/4-1/(2*(n+1)*(n+2))
proof
  defpred X[Nat] means Partial_Sums(s).$1 =1/4-1/(2*($1+1)*($1+2));
  assume
A1: for n holds s.n = 1/(n*(n+1)*(n+2));
A2: for n st X[n] holds X[n+1]
  proof
    let n;
    assume Partial_Sums(s).n =1/4-1/(2*(n+1)*(n+2));
    then Partial_Sums(s).(n+1) =1/4-1/(2*(n+1)*(n+2))+ s.(n+1) by
SERIES_1:def 1
      .=1/4-1/(2*(n+1)*(n+2))+1/((n+1)*(n+1+1)*(n+1+2)) by A1
      .=1/4-(1/((2*(n+1))*(n+2))-1/((n+1)*(n+2)*(n+3)))
      .=1/4-(1/((2*(n+2))*(n+1))-(1*2)/((n+2)*(n+1)*(n+3)*2)) by XCMPLX_1:91
      .=1/4-(1/((2*(n+2))*(n+1))-(1*2)/((n+2)*2*((n+1)*(n+3))))
      .=1/4-(1/((2*(n+2))*(n+1))-1/((n+2)*2)*(2/((n+1)*(n+3)))) by XCMPLX_1:76
      .=1/4-(1/(2*(n+2))*(1/(n+1))-1/(2*(n+2))*(2/((n+1)*(n+3)))) by
XCMPLX_1:102
      .=1/4-1/(2*(n+2))*(1/(n+1)-2/((n+1)*(n+3)))
      .=1/4-1/(2*(n+2))*(1/(n+3)) by Lm8
      .=1/4-1/(2*(n+1+1)*(n+1+2)) by XCMPLX_1:102;
    hence thesis;
  end;
  Partial_Sums(s).0 = s.0 by SERIES_1:def 1
    .=1/(0*(0+1)*(0+2)) by A1
    .=1/4-1/(2*(0+1)*(0+2)) by XCMPLX_1:49;
  then
A3: X[0];
  for n holds X[n] from NAT_1:sch 2(A3,A2);
  hence thesis;
end;
