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 $a=b$.