 reserve n,i,k,m for Nat;
 reserve p for Prime;

theorem Tele1:
  for n being Nat holds
    Reci-seq1.n = 1 / (n - 1 / 2) - 1 / (n + 1 / 2)
  proof
    let n be Nat;
    set a = 1 / 2;
    not 1 / 2 is Nat
    proof
      assume
A4:   1 / 2 is Nat;
      0 <= 1 / 2 & 1 / 2 <= 0 + 1;
      hence thesis by A4,NAT_1:9;
    end; then
A2: n - a <> 0;
    1 / (n - a) - 1 / (n + a)
      = (1 * (n + a)) / ((n - a) * (n + a)) - 1 / (n + a) by XCMPLX_1:91
     .= (1 * (n + a)) / ((n - a) * (n + a)) -
         (1 * (n - a)) / ((n + a) * (n - a)) by A2,XCMPLX_1:91
     .= ((n + a) - (n - a)) / ((n + a) * (n - a)) by XCMPLX_1:120
     .= 1 / (n ^2 - 1 / 4);
    hence thesis by MyDef;
  end;
