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

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.

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.

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 $$\forall n\in\mathbb{N}, n \geq m \Rightarrow |L-a_n| < \epsilon$$

Not true. I was the other debater in this argument with Rasmus. His assertions are false and what you have in the archive is no proof at all. Here are a few recent articles I wrote on this subject:

There is a problem defining the sum of an infinite sum as a limit: http://thenewcalculus.weebly.com/uploads/5/6/7/4/5674177/magnitude_and_number.pdf

And, 0.999... is not really a number: http://thenewcalculus.weebly.com/uploads/5/6/7/4/5674177/proof_that_0.999_not_equal_1.pdf

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 $$\epsilon > 0$$, 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.


 * No. It has not been agreed anywhere that $$\lim_{n\rightarrow\infty}{\sum_{i=1}^n\frac{9}{10^i}} = 1$$, this is what you are trying to prove.

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.

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 $$a_n$$ means that for $$\epsilon = |b - a|/3\,$$, say, there are values of n for which $$|b - a| \leq |b - a_n| + |a - a_n| < \epsilon + \epsilon = \frac{2}{3}(b - a)$$, a clear contradiction. Therefore $$a = b$$.