The ellipsis is critical here. 0.999... ≠ 0.999, because the former decimal is nonterminating.

Look again at the one-third proof. I've numbered the lines this time.

[1] 1/3 = 10/30
[2]     = 9/30 + 1/30
[3]     = 9/3 * 1/10 + 1/3 * 1/10
[4]     = 3 * 1/10 + 1/3 * 1/10
[5]     = 3 * 1/10 + 3 * 1/10 * 1/10 + 1/3 * 1/10 * 1/10
[6]     = 3 * 1/10 + 3 * 1/102 + 3 * 1/103 + ...
[7]     = .3 + .03 + .003 + ...
[8]     = .333...

What you are pointing out is essentially what is stated on [4]: that if you terminate the decimal at any point, you need a 1/3 * 1/10^-n to represent the remaining part of the fraction. However, the notation does not represent the termination of the decimal at any point - rather, the decimal never terminates. The roundoff error you point out never comes. Instead, you get the infinite series (copied from Sam Hughe's page (http://qntm.org/?pointnine)):

0.9999... = 0.9 + 0.09 + 0.009 + 0.0009 + ...

            n=∞
          =  Σ 0.9 * 0.1n
            n=0

            n=∞
          =  Σ a * rn where a=0.9 and r=0.1
            n=0

          = a / (1-r)
          = 0.9 / (1-0.1)
          = 0.9 / 0.9
          = 1

If you require that all decimals terminate, 0.999... is not a decimal.