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
and
, 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.
![{\displaystyle 0.999\ldots =\sum _{i=1}^{\infty }{\frac {9}{10^{i}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dc48567f038a2125326263e3b4ce9dc6112a1555)
, and in particular,
such that
.
![{\displaystyle \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}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/56448e84d9915065908231a35407a397799526f8)
- Given a sequence
,
means (ie. is defined as)
such that ![{\displaystyle \forall n\in \mathbb {N} ,n\geq m\Rightarrow |L-a_{n}|<\epsilon }](https://wikimedia.org/api/rest_v1/media/math/render/svg/38e4a8e321ee08a3555079da657b0c12bc5ade81)
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
.
by point #3.
, where M is some finite number greater than
(which exists by point #2).
For any given
, set
. Then:
Therefore, we now have that:
, and since
, we then know that
.
By point #4,
. It has already been agreed that
, and therefore
. 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
, 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
is convergent. Then
is not undefined.
Assume that
and
, but that
. Then by the definition of the limit:
such that
, and similarly with a replaced by b.
Now,
by the Triangle Inequality. However, the fact that a and b are both limits of the
means that for
, say, there are values of n for which
, a clear contradiction. Therefore
.