Consider Z(s)=Sum(1/N^s) For n=1 to infinity. Let s=(xi+1/2). The divisor is then: N^(xi+1/2) This is equivalent to (N^xi)(N^(1/2)) As n increases, the n^1/2 term will more greatly slow the increase of the divisor and accelerate z(s) away from zero. This means that zeroes will occur less often for larger n. The 1/2 term is necessary so that the zeta function z(s) will encounter fewer primes as it iterates. Is this a valid understanding of z(s)' behavior?