One divided by Infinity? Okay, I'm not much of a mathematician (I'm an 8th grader in Algebra I), but I have a question about something that's been bugging me.
I know that $0.999 \cdots$ (repeating) = $1$. So wouldn't $1 - \frac{1}{\infty} = 1$ as well? Because $\frac{1}{\infty} $ would be infinitely close to $0$, perhaps as $1^{-\infty}$?
So $1 - 1^{-\infty}$, or $\frac{1}{\infty}$ would be equivalent to $0.999 \cdots$? Or am I missing something? Is infinity something that can even be used in this sort of mathematics?

There is one issue that has not been raised in the fine answers given earlier.  The issue is implicit in the OP's phrasing and it is worth making it explicit.  Namely, the OP is assuming that, just as $0.9$ or $0.99$ or $0.999$ denote terminating decimals with a finite number of 9s, so also $0.999\ldots$ denotes a terminating decimal with an infinite number of 9s, the said infinite number being denoted $\infty$. Changing the notation from $\infty$ to $H$ for this infinite number so as to avoid a clash with traditional notation, we get that indeed that 0.999etc. with an infinite number $H$ of 9s falls infinitesimally short of $1$.  
More specifically, it falls short of $1$ by the infinitesimal $\frac{1}{10^H}$, and there is no paradox. Here one does not especially need the hyperreal number system.  It is sufficient to use the field of fractions of Skolem's nonstandard integers whose construction is completely constructive (namely does not use the axiom of choice or any of its weaker forms). As the OP points out, the infinitesimal $\frac{1}{H}$ (or more precisely $\frac{1}{10^H}$) is infinitely close to $0$ without being $0$ itself.
