I'm usually careful to not use exclamation points in math equations because they could represent a factorial and not just the normal excitement inherent in any discussion of mathematics but in this case it is ok since \(1! = 1\).
First off, the most popular "proof" that 0.9999... = 1 goes like this. Suppose that
\[
n = 0.9999\ldots
\]
then multiple both sides by 10, which perseveres the equality:
\[
10n = 9.9999\ldots.
\]
Then you can subtract the first equation from the second and get:
\[
9n = 9.0000\ldots
\]
which gives that n = 1. The problem with this is that it's a bit of mathematical slight of hand and hides what's really going on. The problem is what do we even mean by 0.9999...?
Wherefore art thou 0.9999...?
\[
9/10 + 9/100 + 9/1000 + \cdots = \frac{0}{10^0} + \frac{9}{10^1} + \frac{9}{10^2} + \frac{9}{10^3} + \cdots.
\]
This can be succinctly written as:
\[
\sum_{i = 1}^{\infty} \frac{9}{10^i}.
\]
So 0.9999... isn't number. It's a never ending sum. There is no point where the process stops and spits out the final number. Since it's not a number, what does it mean to ask what number does 0.9999... equal? How can an infinite process equal a number, one is a process and the other is a number? There is a mismatching of concepts.
There is a way (several in fact) where one can assign a value to the above infinite sum. First we start by looking at the partial sum, i.e. the sum where we stop after n terms:
\[
S(n) = \sum_{i = 1}^{n} \frac{9}{10^i}
\]
Two observations:
- for a given value of n, the sum S(n) is a finite and has a definite value.
- for larger and larger values of n, S(n) does get closer and closer to 1.
\[
\forall \epsilon >0 \ \ \exists n \ \ \ \text{s. t.}\ \ \ \forall m > n \left| S(m) - 1\right| < \epsilon.
\]
I think that morass of symbols obscures more than it revivals. The ideas are the thing. In this case, the difference between S(n) and 1 has a simple closed form (left to the reader to derive) and it's clear that we can pick n such that difference is less than \(\epsilon\) for any positive \(\epsilon\). When this is the case, we say that S(n) converges to 1 as n goes to infinity. It's important to note that n never reaches infinity (whatever that might mean).
Thus \(0.9999... = \sum_{i = 1}^{\infty} \frac{9}{10^i} = 1\) is really shorthand for \(\sum_{i = 1}^{n} \frac{9}{10^i}\) converges to 1.
There are lots of infinite sums that converge to 1. Here's another famous one:
\[\sum_{i = 1}^{\infty} \frac{1}{2^i}.\]
Once you wrap your head around that one, what about this:
\[\sum_{i = 1}^{\infty} (-1)^n.\]
Does that converge? If so, to what?
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