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Someone asked this question about how many ways there are to prove $0.999\dots = 1$ and I posted this:

$ 0.99999 \dots = \sum_{k = 1}^\infty \frac{9}{10^k} = 9 \sum_{k = 1}^\infty \frac{1}{10^k} = 9 \Big ( \frac{1}{1 - \frac{1}{10}} - 1\Big ) = \frac{9}{9} = 1$

The question was a duplicate so in the end it was closed but before that someone wrote in a comment to the question: "Guys, please stop posting pseudo-proofs on an exact duplicate!" and I got down votes, so I assume this proof is wrong.

Now I would like to know, of course, why this proof is wrong. I have thought about it but somehow I can't seem to find the mistake.

Many thanks for your help. The original can be found here.

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    Perhaps it might be worth mentioning that essentially the same proof is posted [in this answer](https://math.stackexchange.com/questions/11/is-it-true-that-0-999999999-ldots-1/116#116) which is at score over 100. Perhaps the fact that the poster explicitly stated that he assumes formula for sum of geometric series is the only relevant difference I see.2017-04-24

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The problem is that you are assuming 1) that multiplication by constants distributes over infinite sums, and 2) the validity of the geometric series formula. Most of the content of the result is in 2), so it doesn't make much sense to me to assume it in order to prove the result. Instead you should prove 2), and if you really want to be precise you should also prove 1).

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    @Qiaochu: in that we agree. If someone has problems understanding that 0.9999...=1 then the problem comes from not understanding the notions of convergence of a series and basic facts about real numbers.2011-03-10
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I got this as an assignment in my first university-level calculus course: To prove $1 = 0.999...$. My proof was essentially the same as yours, and it was good enough for them. The only "mistake" I can think of must be the one Qiaochu is pointing out, that you assume to much about convergence of series.