let z be Nat; :: thesis: ( 2 <= z implies for k being Nat ex N being Nat st
for x being Nat st N <= x holds
1 / (z to_power x) < 1 / (x to_power k) )

assume P0: 2 <= z ; :: thesis: for k being Nat ex N being Nat st
for x being Nat st N <= x holds
1 / (z to_power x) < 1 / (x to_power k)

let k be Nat; :: thesis: ex N being Nat st
for x being Nat st N <= x holds
1 / (z to_power x) < 1 / (x to_power k)

consider N being Nat such that
P1: for x being Nat st N <= x holds
1 / (2 to_power x) < 1 / (x to_power k) by L2;
take N ; :: thesis: for x being Nat st N <= x holds
1 / (z to_power x) < 1 / (x to_power k)

now :: thesis: for x being Nat st N <= x holds
1 / (z to_power x) < 1 / (x to_power k)
let x be Nat; :: thesis: ( N <= x implies 1 / (z to_power x) < 1 / (x to_power k) )
assume N <= x ; :: thesis: 1 / (z to_power x) < 1 / (x to_power k)
then E1: 1 / (2 to_power x) < 1 / (x to_power k) by P1;
P3: 0 < 2 to_power x by POWER:34;
2 to_power x <= z to_power x
proof
now :: thesis: ( ( 2 = z & 2 to_power x <= z to_power x ) or ( 2 < z & 2 to_power x <= z to_power x ) )end;
hence 2 to_power x <= z to_power x ; :: thesis: verum
end;
then 1 / (z to_power x) <= 1 / (2 to_power x) by P3, XREAL_1:118;
hence 1 / (z to_power x) < 1 / (x to_power k) by E1, XXREAL_0:2; :: thesis: verum
end;
hence for x being Nat st N <= x holds
1 / (z to_power x) < 1 / (x to_power k) ; :: thesis: verum