Actually, it will. 2^-25 is represented exactly, since it is really represented as 0b 1.00000000000000000000000 * 2^(0b100011001 - 128). The exponent is represented exactly, and the mantissa (the 1.xxxx term) is representable exactly since it is exactly 1.
If you repeatedly add 2^-25 to a single accumulator register, then you'll hit a wall (1-2^-25 is not representable). If you do something less efficient like pairwise addition or Kahan summation, then you can avoid that problem.
You are completely correct, and I was wrong. My sincere apologies for filling the net with misinformation. Thanks for pointing out the correction, and I now better see what you are saying.
