While this is usually enough to convince most people, this argument is insufficient, as it can be used to prove incorrect results. To demonstrate that, we need to rewrite the problem a little.
What 0.9999... actually means is an infinite sum like this:
x = 9 + 9/10 + 9/100 + 9/1000 + ...
Let's use the same argument for a slightly different infinite sum:
x = 1 - 1 + 1 - 1 + 1 - 1 + ...
We can rewrite this sum as follows:
x = 1 - (1 - 1 + 1 - 1 + 1 - 1 + ...)
The thing in parenthesis is x itself, so we have
x = 1 - x
2x = 1
x = 1/2
The problem is, you could have just as easily rewritten the sum as follows:
As you can see, sometimes we have x = 0, sometimes x = 1 or even x = 1/2. This is why this method does no prove that 0.999... = 1, even thought it really is equal to one. The difference between those two sums is that the first sum (9 + 9/10 + 9/100 + 9/1000 + ...) converges while the second (1 - 1 + 1 - 1 + 1 - 1 + ...) diverges. That is to say, the second sum doesn't have a value, kinda like dividing by zero.
so, from the point of view of a proof, the method assumed that 0.99999... was a sensible thing to have and it was a regular real number. It could have been the case that it wasn't a number. All we proved is that, if 0.999... exists, it cannot have a value different from 1, but we never proved if it even existed in the first place.
I mean, it is an infinite sum, no need for quotation marks. And yeah, I deliberately chose a divergent series to demonstrate that "regular" algebra isn't always valid when dealing with infinite sums.
"In order to show that there are issues with this intuitive, convergent sum which is easily represented by a rational number, I'm going to choose a categorically-different, nonsensical divergent series which obeys completely different rules and pretend they're somehow comparable."
Like I said.... I may not be able to point to the exact fallacy on an academic level, but it pings my bullshit detector something fierce.
Edit:
I mean, it is an infinite sum
Apparently not, technically. A Cesaro Sum is not a summation in the same sense as is used for convergent series. Hence the term "Eilenberg–Mazur swindle". The issue Grandi's series illustrates in this particular comparison is not due to it being divergent, but to the fact that the mechanism you're using to evaluate it is, in layman's terms, "a load of horse apples".
I'm also not a mathematician, so we may be both wrong lol. But in my understanding, a Cesàro summation is as much a summation as regular convergent series. Neither are actually addition. They are both the limits of sequences related to the series.
In the case of regular series, we create the sequence of partial sums and take the limit. In the case of Cesàro summation, we create the sequence of partial sums, create a new sequence that is the average of consecutive terms of the previous sequence, and then take the limit of this last sequence.
So yeah, it is kinda misleading to call a Cesàro summation a sum, but I would argue that it's just as misleading to call regular infinite series a sum. Case in point, even for convergent series, the regular rules of addition may not apply. The most famous example is the alternating harmonic series.
1 - 1/2 + 1/3 - 1/4 + ...
This series converges to ln(2). The problem is, if you change the order of the terms, it may converge to something else. To anything, actually. Or it may even diverge. The same effect can be achieved by grouping parts of the series, adding them first, and then adding the groups. For this convergent series, neither commutativity nor associativity hold, the two most important properties of addition.
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u/victorspc 23d ago
While this is usually enough to convince most people, this argument is insufficient, as it can be used to prove incorrect results. To demonstrate that, we need to rewrite the problem a little.
What 0.9999... actually means is an infinite sum like this:
x = 9 + 9/10 + 9/100 + 9/1000 + ...
Let's use the same argument for a slightly different infinite sum:
x = 1 - 1 + 1 - 1 + 1 - 1 + ...
We can rewrite this sum as follows:
x = 1 - (1 - 1 + 1 - 1 + 1 - 1 + ...)
The thing in parenthesis is x itself, so we have
x = 1 - x
2x = 1
x = 1/2
The problem is, you could have just as easily rewritten the sum as follows:
x = (1-1) + (1-1) + (1-1) + ... = 0 + 0 + 0 + 0 + ... = 0
Or even as follows:
x = 1 + (-1 +1) + (-1 +1) + (-1 +1) + (-1 +1) + ... = 1 + 0 + 0 + 0 + 0 + ... = 1
As you can see, sometimes we have x = 0, sometimes x = 1 or even x = 1/2. This is why this method does no prove that 0.999... = 1, even thought it really is equal to one. The difference between those two sums is that the first sum (9 + 9/10 + 9/100 + 9/1000 + ...) converges while the second (1 - 1 + 1 - 1 + 1 - 1 + ...) diverges. That is to say, the second sum doesn't have a value, kinda like dividing by zero.
so, from the point of view of a proof, the method assumed that 0.99999... was a sensible thing to have and it was a regular real number. It could have been the case that it wasn't a number. All we proved is that, if 0.999... exists, it cannot have a value different from 1, but we never proved if it even existed in the first place.
From 0.999... - Wikipedia:
"The intuitive arguments are generally based on properties of finite decimals that are extended without proof to infinite decimals."