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I'm a PreCalculus student trying to find a rigorous proof that $\displaystyle\frac{1}{3} = 0.333\ldots$, but I couldn't find it. I think (just think) that this proof would start by proving that

$\displaystyle\sum_{i=1}^{\infty}3\cdot10^{-i} = \frac{1}{3}$. My guesses (assuming that proving that $\displaystyle\sum_{i=1}^{\infty}\left(\frac{1}{10}\right)^i$ converges is trivial):

$\displaystyle\sum_{i=1}^{\infty}3\cdot10^{-i} = 3\cdot\sum_{i = 1}^{\infty}10^{-i} = 3\cdot\sum_{i=1}^{\infty}\left(\frac{1}{10}\right)^i = 3\cdot\left(\frac{1}{1 - \frac{1}{10}}-1\right) = 3\cdot\left(\frac{10}{9}-1\right) = \frac{1}{3}$.

Questions: is this completely rigorous? Which flaws could be found in this proof? How can I improve it?

PS. I'm not sure how to tag this. Feel free to edit, if necessary.

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2 Answers 2

up vote 11 down vote accepted

Since $\sum_{i=1}^\infty 3\cdot 10^{-i}$ is what the notation "$0.333...$" means, your argument is perfectly good. It's not just the "start of a proof", it is all there is to it.

Okay, perhaps it is not really trivial to prove that the geometric series converges, but straightforward it is. Just plug in the definition of the sum of a series and crank the handle, using standard tricks to rewrite each of the partial sums in turn.

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Sorry, but I didn't understand what is wrong in using the fact of $10x - x = 3,333\ldots - 0,333\ldots = 3$. It seems like abuse of notation, but I'm not sure. Could you clarify, please? –  Ian Mateus Jun 13 '12 at 22:24
    
@Ian: The problem with it is that if you want to go that way, you first need to prove that arithmetic on infinite decimals actually works the way you're implicitly using it here. In order to prove that you first need a definition of what those things you're adding and multiplying are in the first place, and there the infinite series (or minor variations on that theme) is essentially the only idea that works. It's doable, but doing it rigorously involves much more work than just showing $0.333...=\frac13$ directly. –  Henning Makholm Jun 13 '12 at 22:31
1  
$3.333\ldots = 3\cdot\sum_{i=0}^{\infty}10^{-i} = 3 + 3\cdot\sum_{i=1}^{\infty}10^{-i}$ and $0.333\ldots = 3\cdot\sum_{i=1}^{\infty}10^{-i}$. Subtracting gives us the term $\left(3\cdot\sum_{i=1}^{\infty}10^{-i} - 3\cdot\sum_{i=1}^{\infty}10^{-i}\right)$ which would be $\infty - \infty$ (undefined) if $\sum_{i=1}^{\infty}10^{-i}$ diverges. Is it a "reason", nonsense or just irrelevant? –  Ian Mateus Jun 13 '12 at 22:44
    
@Ian: Hmmm... you make a good case. Do you have a result available that allows you to subtract series term by term here? –  Henning Makholm Jun 13 '12 at 22:50
    
If I got it, defining $3.333\ldots = 3 + 3\cdot\sum_{i=1}^{\infty}10^{-i}$, $0.333\ldots = 3\cdot\sum_{i=1}^{\infty}10^{-i}$, and proving convergence of $3\cdot\sum_{i=1}^{\infty}10^{-i}$ makes it rigorous, right? –  Ian Mateus Jun 13 '12 at 22:56

Your proof is correct. Essentially you proved it was the convergence of the geometric series.

Alternatively if you want to avoid using the geometric series formula and just use a very simple limit argument, note that , $(3)(.333...) = .999... = 1$ $\text{ }$ To see this, note that $.999... = \sum_{i = 1}^\infty \frac{9}{10^i}$ converges to $1$. Take a look at partial sums. Thus $.333... = \frac{1}{3}$.

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