08/30/2010, 11:40 PM
the post contains some mistakes ...
will correct later.
meanwhile for abs(z) < 1 and <> 0 sum (n, -oo , oo) x^2^n
seems a solution.
will correct later.
meanwhile for abs(z) < 1 and <> 0 sum (n, -oo , oo) x^2^n
seems a solution.

