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?
You could say that $\frac{1}{\infty} = 0$, so $1-\frac{1}{\infty} = 1$. But then, you're stretching the definition of division past breaking point - division as you know it isn't defined for infinity, so the answer is undefined. Otherwise, you can quickly get yourself into a pickle and end up saying 1=2.
Arithmetic operators - add, subtract, divide, multiply, raise to the power of - are defined on a particular set of numbers: such as real numbers, or complex numbers.
The set you use for definition, will determine what you can and can't say meaningfully. Typically (but not always), infinity is excluded from that set.
If we take the set of real numbers, and look at "raise to the power of", then $1^x$ is equal to 1 for any x, as x -> infinity. So in that case, you could have a convention of saying that $1^\infty = 1$. But $\frac{1}{1} = 1$, so $1^{-\infty}$ would also equal 1. However, when you go about defining these new conventions, you have to be extremely careful - sometimes, a convention will seem obvious, but if you run with it, you end up seeming to prove 1=2, which means that your convention wasn't that helpful.
Let's compare with raising to the power 0.5, i.e. taking the square root. $-1^{0.5}$ is undefined when we are working on the reals - so, just as dividing by infinity, you can't include it in your arithmetic. Only when you expand to the complex numbers and extend your definition of the arithmetic operators to cope, can you say something meaningful about $(-1)^{0.5}$
Similarly, the reals and the complex numbers each exclude infinity, so arithmetic isn't defined for it.
You can extend those sets to include infinity - but then you have to extend the definition of the arithmetic operators, to cope with that extended set. And then, you need to start thinking about arithmetic differently. If you want to learn more about that, then there are lots of friendly places on the web to get into the work of Cantor on the different types of infinity (of which there are an infinite number of different infinities).
Whether one defines $1/\infty$ may be a matter of convention, but one can say that $1/x$ approaches $0$ as $x$ approaches $\infty$, and what that means is that $1/x$ can be made as close as desired to $0$ by making $x$ big enough. How big is big enough depends on how close you want to make $x$ to $0$. If that's what you mean by $1/\infty = 0$, then that statement seems unobjectionable. More specifically, if you want the distance between $1/x$ and $0$ to be less than a small positive number $\varepsilon$, then that is so whenever $x > 1/\varepsilon$.
As pointed out in the other answers, in the real number system there is no item "$\infty$". Nor is there in the complex number system. There are some other number systems that DO have such an item. One is called the "Riemann sphere" ... consisting of the complex numbers with an extra point $\infty$. Legitimate caluclations defined on the Riemann sphere do, indeed, include the equation $1/\infty = 0$.
I believe it should be pointed out that there is a very significant and important similarity between $0.\dot 9$ and $\frac1\infty$. It may seem easy to accept that $0.\dot9 = 1$ because calculations like $0.\dot9-0.0\dot9$ seem obvious, but you are none the less manipulating two expressions, each of which represents a number with a countably infinite number of non zero digits. That difference involves cancelling, an infinite number of times, two digits equal to $9$.
The dot over the $9$ makes it easy to forget this and makes it feel like you can hold it in your hand and manipulate it like any other finite quantity like $1$. Infinite decimals are introduced very loosely in secondary education and the subtleties are not always fully grasped until arriving at university.
By the way, there is a group of very strict Mathematicians who find it very difficult to accept the manipulation of infinite quantities in any way. I'm not one of their number, if I joined them, my number would have to be $\infty$, which is a contradiction ;-).
What we actually have here is $0.\dot9 = 9\sum_{n=1}^\infty \frac1{10^n}$. What we actually show is, in the limit, as $n\to\infty$, $0.\dot9 \to 1$. Using the same process, as $n\to\infty$, $1 - \frac1n \to 1$.
Note: I'm not saying that $\frac1\infty=\lim_{n\to\infty}\frac1n$ in exactly the same way that $0.\dot9=1$ is defined in mathematics, but just pointing out that $0.\dot9$ is a decimal representation with an infinite number of digits in it and has to be handled no less carefully than any other quantity that deals with infinity in some way or another. As has been pointed out elsewhere, even if you define $\frac1\infty$ as $\lim_{n\to\infty}\frac1n$ you can get into trouble combining it with limits that don't otherwise exist.
So, an important difference between defining $0.\dot9$ as $1$ and $\frac1\infty$ as $0$ is that in the latter case it's tempting to use $0$ to hide what's really going on and you can't do that with $1$.
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.
