User:ConMan/Proof that 0.999... equals 1 (Limit proof)

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The following proof was originally posted in Talk:Proof that 0.999... equals 1/Archive05, and I am keeping it here for my, and others', future reference.

Contents

[edit] Motivation

A number of anonymous posters on the Talk page claimed that while 0.999\ldots = \sum_{i=1}^{\infty}\frac{9}{10^i} and \lim_{n\rightarrow\infty}\sum_{i=1}^n\frac{9}{10^i} = 1, the two were not equal because "the infinite sum is not equal to the limit of the partial sums". I seemed to recall something about the infinite sum being defined as the limit of partial sums, because nothing else makes sense, but I wondered if it was in fact provable - and in this case at least it was.

[edit] Accepted definitions and statements

These were agreed upon by people claiming both that 0.999... equals and does not equal 1.

  1. 0.999\ldots = \sum_{i=1}^{\infty}\frac{9}{10^i}
  2. 0.999\ldots < \infty, and in particular, \exists M\in\mathbb{R} such that 0.999\ldots < M.
  3. \sum_{i=1}^{\infty}\frac{9}{10^i} - \sum_{i=1}^{n}\frac{9}{10^i} = \sum_{i=n+1}^{\infty}\frac{9}{10^i} = \frac{1}{10^n}\sum_{i=n+1}^{\infty}\frac{9}{10^{i-n}} = \frac{1}{10^n}\sum_{i=1}^{\infty}\frac{9}{10^i}
  4. Given a sequence (a_n) = (a_1,a_2,a_3,\ldots,a_n,\ldots), \lim_{n\rightarrow\infty}{a_n} = L means (ie. is defined as) \forall\epsilon > 0, \exists m\in\mathbb{N} such that \forall n\in\mathbb{N}, n \geq m \Rightarrow |L-a_n| < \epsilon

[edit] The proof

Let a_n = \sum_{i=1}^n\frac{9}{10^i}.

0.999\ldots - a_n = \frac{1}{10^n}\sum_{i=1}^{\infty}\frac{9}{10^i} = \frac{1}{10^n}0.999\ldots by point #3.

0.999\ldots - a_n = \frac{1}{10^n}0.999\ldots < \frac{1}{10^n}M, where M is some finite number greater than 0.999\ldots (which exists by point #2).

For any given ε > 0, set m = \lceil log_{10}\frac{M}{\epsilon} \rceil + 1. Then:

m > log_{10}\frac{M}{\epsilon}

10^m > \frac{M}{\epsilon}

10^{-m} < \frac{\epsilon}{M}

\frac{M}{10^m} < \epsilon

Therefore, we now have that:

|0.999\ldots - a_m| < \frac{M}{10^m} < \epsilon, and since |0.999\ldots - a_n| \leq |0.999\ldots - a_m| \quad \forall n \geq m, we then know that |0.999\ldots - a_n| < \epsilon.

By point #4, 0.999\ldots = \lim_{n\rightarrow\infty}{\sum_{i=1}^n\frac{9}{10^i}}. It has already been agreed that \lim_{n\rightarrow\infty}{\sum_{i=1}^n\frac{9}{10^i}} = 1, and therefore 0.999\ldots = 1. In other words, I have shown that, in fact, that "the infinite sum is equal to the limit of the partial sums" is not a definition, but a provable statement.

[edit] Holes

I admitted that the proof as stated above was not 100% rigorous, so here is a list of some of the spots where the rigor is lacking, and an attempt to correct that.

[edit] Two limits?

As pointed out by User:Rasmus Faber, the proof assumes that if a sequence has a limit, that limit is unique. In the real numbers, this can be shown by the following Lemma:

Lemma 1: If a sequence of real numbers converges, it has a unique limit.

Proof:

Suppose that the sequence (a_n) = (a_1, a_2, a_3, ...), a_n \in\mathbb{R}\ \forall n is convergent. Then \lim_{n\rightarrow\infty}a_n is not undefined.

Assume that \lim_{n\rightarrow\infty}a_n = a and \lim_{n\rightarrow\infty}a_n = b, but that a \neq b. Then by the definition of the limit:

\forall\epsilon > 0, \exists m\in\mathbb{N} such that \forall n\in\mathbb{N}, n \geq m \Rightarrow |a-a_n| < \epsilon, and similarly with a replaced by b.

Now, \forall n \in\mathbb{N}, |b - a| = |(b - a_n) - (a - a_n)| \leq |b - a_n| + |a - a_n| by the Triangle Inequality. However, the fact that a and b are both limits of the an means that for \epsilon = |b - a|/3\,, say, there are values of n for whice |b - a| \leq |b - a_n| + |a - a_n| < \epsilon + \epsilon = \frac{2}{3}(b - a), a clear contradiction. Therefore a = b.