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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.

Martin Sleziak
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Rudy the Reindeer
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  • I deleted my original answer. – Rudy the Reindeer Mar 10 '11 at 18:03
  • Why -1 for this? – Aryabhata Mar 10 '11 at 18:04
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    "Pseudo" doesn't mean "wrong," exactly, but you're assuming some things about convergence of series that, at this level, you should be proving. (Also, you really should change the title of this question if you don't want it to also be closed as a duplicate. Nobody wants to talk about 0.999... anymore.) – Qiaochu Yuan Mar 10 '11 at 18:06
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    Looks like a valid proof to me... – Fabian Mar 10 '11 at 18:07
  • @Qiaochu: Sure, can do. What would you suggest as a title? – Rudy the Reindeer Mar 10 '11 at 18:09
  • @Moron: I think this question hits some people's sore spot and they get very emotional. (Judging from the responses to the old thread.) – Rudy the Reindeer Mar 10 '11 at 18:11
  • @Matt: Perhaps. Btw, there was another answer in the question you linked, perhaps the comment of "pseudo-proof" was regarding that? – Aryabhata Mar 10 '11 at 18:17
  • @Q: Should we rewrite the infinite series as a sequence of partial sums, or is there a different way to approach the more rigorous proof? – The Chaz 2.0 Mar 10 '11 at 18:31
  • @Matt: something like "validity of a proof about geometric series." Personally I have a strong averse reaction to the phrase $0.999... = 1$ and almost reflexively closed this question. – Qiaochu Yuan Mar 10 '11 at 18:58
  • @Moron: yes but there were 2 answers and the person wrote "Guys stop posting pseudo-proofs" which means both. I don't think I'm nit picking grammar in this case. – Rudy the Reindeer Mar 10 '11 at 21:33
  • This proof is correct, modulo the definition of $0.999\ldots$. (In your argument, it is implicitly understood to mean the limit of the finite decimals $0.9$, $0.99$, $0.999$, etc. This is certainly one reasonable definition.) I wouldn't take the pseudo-proof comment very seriously. – Matt E Mar 11 '11 at 08:47
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    Perhaps it might be worth mentioning that essentially the same proof is posted in this answer 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. – Martin Sleziak Apr 24 '17 at 04:54

2 Answers2

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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).

Qiaochu Yuan
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    How do you prove 1) ? – TROLLHUNTER Mar 10 '11 at 19:19
  • It's a straightforward consequence of the definition of convergence of an infinite series. All you have to verify is that if a sequence goes to $0$, then $c$ times that sequence also goes to $0$. – Qiaochu Yuan Mar 10 '11 at 19:25
  • @Qiaochu: I think I can prove 1) and 2), so your answer answers my question. Then my proof wasn't "pseudo", it was just incomplete. – Rudy the Reindeer Mar 10 '11 at 19:46
  • @Qiaochu: but both 1) and 2) are fairly easy to prove and both are among the simpler facts (proved almost immediately after the definition) about series. I see no problem in assuming both facts as known (understanding "known" as very easy to prove if need be). – Gonzalo Medina Mar 10 '11 at 19:50
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    @Gonzalo: yes, but $0.999... = 1$ is also fairly easy to prove and among the simpler facts about series. If someone asks you for a proof of this fact she is clearly confused about something and assuming that she'll understand all of the things you're omitting because you think they're obvious is not productive. Really, if someone asks you for a proof that $0.999... = 1$ the appropriate answer is not a proof: it is a careful explanation of the definition of convergence of infinite series, together with a link to some basic facts about the real numbers. – Qiaochu Yuan Mar 10 '11 at 19:54
  • @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. – Gonzalo Medina Mar 10 '11 at 19:58
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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.

Eivind
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