I hate to feed into another one of these questions (they come up regularly here it seems), but this time a few people seem quite convinced of the non-equality, so I'll try to pitch in. I doubt that I'll be able to convince anyone, but it's worth a shot.
Shawty: 1/3 is indeed equal to .3333repeating. If you simply do the long division you may be able to convince yourself that dividing 1 by 3 results in infinitely many 3's after the decimal point. Otherwise, your problem is a common one--misunderstanding the mathematical concept of infinity. I'll address this further down. Your rebuke to kyle's answer doesn't hold, as 9.999repeating-x IS possible: just remember what x was defined as (.999repeating), and it's easy to subtract.
Bucky: Your refutation of converging limits is admirable, but the thing to remember here is that when we say .999 repeats forever, we really do mean infinitely; we do not mean that .999999 has a number of digits that approaches infinity, but rather than it has infinitely many digits (all of which are 9's).
As mentioned before, when you consider 3N, a 7 does not appear at the "end" of the digit string, because there is no end: the nines just keep on coming. But suppose for a moment that there are infinitely many 9's, followed by a seven. Then the rest of your work goes to show that .99999repeating = .99999repeating. Okay, that's fine, but we've also shown that .99999repeating=1, using valid mathematical operations. Your last comment (before edits) seems to indicate you think that any manipulation of .999repeating should end up with it becoming 1. This is not the case. Think of y=x*(x+1). Using a bit of manipulation, we can see that y=x^2+x. But doing some more manipulations, we can likewise see that y=(x+1)*x. There's no problem here though, they're simply different ways of expressing the same thing.
Can you agree that 1/3=.333repeating? If so, then do your final manipulation backwards: Let N=1, so N/3=.333repeating, so N=.9999repeating.
Basically I think your roadblock here is that you consider .9999repeating to be a series of 9's that grows without limit. Technically though, the 9's are there, and they're infinite. They don't really expand outward, and that's why a 7 is never found at the end of 3N, as mentioned above. In your definition of convergence, notice the "as the number of terms increases"; the number of 9's in N is not increasing, they're there already.
rathan d: yes, 1-1=0, and 1-.99999repeating=0 as well.
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I rather like Nisovin's (the question asker) approach in the second add. details. On the real number line (and hopefully we can agree that N is a real number), there are several properties. One is that for any non-equivalent numbers a and b, there is a number between them. In fact, there should be infinitely many numbers between them. Also, every number has a decimal representation. So if there is a difference between .999repeating and 1, someone should be able to give me a decimal expansion that is between them.
And if anyone didn't read the links given in geezah's answer, please do at least peruse them. They might provide a new way to look at this problem.