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My question concerns restrictions on the exercise of normal algebraic rules. The most well known restriction is the prohibition on division by zero (PDZ). This is justified by various 'proofs' of incorrect results, for example: $$a = b$$ $$ab = b^2$$ $$ab - a^2 = b^2 - a^2$$ $$a(b - a) = (b - a)(b + a)$$ $$a = b + a$$ $$a = 2a*$$ $$1 = 2$$ *From line 1. In this 'proof' the zero division is not immediately obvious. Another possible restriction would be a prohibition on multiplication by infinity (PMI), which is at least conceptually similar to PDZ. But is there any need for it? Consider the following series: $$x = 0.999...$$ $$10x = 9.999...$$ $$10x = 9 + x$$ $$9x = 9$$ $$x = 1$$ And: $$x = 1 - 1 + 1...$$ $$x = 1 - (1 - 1 + 1...$$ $$x = 1 - x$$ $$2x = 1$$ $$x = 1/2$$ And finally: $$x = 1 + 2 + 4...$$ $$x = 1 + 2(1 + 2 + 4...$$ $$x = 1 + 2x$$ $$-x = 1$$ $$x = -1$$ Debate about this issue has previously got stuck on whether the second series makes sense, but since the third series is obvious nonsense (the sum is increasing - it cannot be negative) the debate should be about how to disallow series such as the third using an uncontroversial restriction. The third series introduces a multiplication by infinity in line 2 (infinity can be defined as the sum of an additive sequence with equal or increasing terms) so disallowing such a step invalidates that series without affecting the other two. Is this a good way to treat the issue?

Edit: The above definition of infinity should include series with decreasing sequences which can be proven to be divergent because the last step in such proofs should explicitly show that they comply with it. Also note, nothing in this post has implications for conditionally convergent series.

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The usual statement of mathematical theorems includes positive generalities and negative counterexamples. For example:

(1) For all real numbers $a$ and $b$ and $x\neq0$, if $ax=bx$, then $a=b$.
(2) Note that $1\times0=2\times0$ but $1\neq 2$. So theorem (1) does not generalize to $x=0$.

It is not that division by zero is prohibited; there is simply no theorem that can be applied to the case $1\times0=2\times0$. It is not the case that algebraic manipulations are allowed until proven wrong. They are disallowed until proven right, in some particular setting.

Now consider the following theorems:

(3) If $\sum a_n$ is a convergent series of real numbers and $c$ is any real number, then $\sum ca_n=c\sum a_n$.
(4) If $\sum a_n$ and $\sum b_n$ are convergent series of real numbers, then $\sum(a_n+b_n)=\sum a_n + \sum b_n$.
(5) If $\sum a_n$ is a convergent series of real numbers, then $\sum a_n=a_0+\sum a_{n-1}$.

Theorems (3) and (4) state that summation of convergent series of real numbers is linear, and (5) states that it is stable. These theorems, together with the fact that $0.999\ldots$ converges, justify your proof that $0.999\ldots=1$.

Now for the next series, $1-1+1-1+\cdots$. This isn't a convergent series of real numbers, so theorems (3-5) don't apply to it. If we interpret the notation as meaning the sum of a convergent series, then the sum doesn't exist, so any equation involving it is absurd. It is not the "multiplication by infinity" is prohibited; we just don't have an interpretation of the notation in terms of a theory that would justify the manipulations.

Or do we? Let us interpret $1-1+1-1+\cdots$ as an Abel sum. We have theorems analogous to (3-5):

(6) Abel summation is linear.
(7) Abel summation is stable.

In light of theorems (6) and (7), your manipulations constitute a successful proof that if the Abel sum of $1-1+1-1+\cdots$ exists (and it does), then it is $\frac12$.

Now you say that "the third series is obvious nonsense". That is not so obvious. Consider the topological ring of $2$-adic integers $\mathbb Z_2$. This is not an ordered ring, so there is no way to evaluate whether a sequence or a series is increasing. Indeed, it is compact, so there is no way for a series to escape to a point at infinity. And it turns out that theorems (3-5) generalize to $\mathbb Z_2$. Together with the fact that $1+2+4+8+\cdots$ converges in $\mathbb Z_2$, your manipulations therefore prove that the sum is $-1$.

There is another way to interpret $1+2+4+8+\cdots$ in the real numbers using analytic continuation in the complex numbers, and similar considerations apply.

There is yet another interpretation in various compactifications of the real numbers, in which the sum of the third series is $x=\infty$ and $x=1+2x$ is a true statement. In this space, the series summation is fine, and the multiplication by infinity is fine, but the following step where you "cancel" $2x$ is wrong.

I wanted to spell out the above theorems to make two points. First, there is no need for a prohibition like "don't apply algebraic manipulations to divergent series". As long as you are considering classical summation, that prohibition would be redundant, as you haven't even defined what the sum of a divergent series would be.

Second, if you are using a summation method for divergent series, or if you are working in a space with a different notion of convergence, then there is even more reason to avoid such prohibitions: they may be wrong. You may even have theorems that state that the manipulations preserve generalized sums in exactly the way you want. Or the manipulations might fail, but at a different point from where you expect. It's impossible to say which manipulations succeed until you specify the interpretation.

Chris Culter
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  • That may be the best possible counterargument, but I can't be sure. –  Aug 24 '15 at 08:30