I disagree with this. For any integer number n this is true, indeed (actually, there's a small error in there, it should be "i=1, n". For infinity this no longer holds.

That's because "sum(i=1, n) 9(10^-i)" is in itself a limit for n is infinity. It's an implicit limit, but a limit nonetheless. And you can't say that:

Code:

1 - sum(i=1, n) 9(10^-i) = 1/10^n

for n is infinity here, as the implicit limit is disappearing on the right hand side of the equation. Also, "1/10^n" is never actually 0, so that would actually disproof 1 = 0.999... (edit: well, not actually disproof, but it would assume 0 = infinitesimal, which Mario wrongfully accused me of doing).

So it should rather be:

Code:

1 - sum(i=1, n) 9(10^-i) = lim(x->n) 1/10^x

Here, the right hand side isn't "a very small number", but it actually is zero, proving that 1 = 0.9999...