What's wrong with using algebra on infinite series?

Question

I've recently found an article (referred somewhere on this site) criticizing the use of common rules of algebra on infinite series. To be honest, the video referred is one of the videos of Numberphile I liked the most. I mean, informally, to say a rule doesn't hold, I think one should find an example (in modern logic, a $\forall$ statement is true by default, and a $\exists$ false); say, associativity of addition for infinite series:

$$ S_1=(1-1)+(1-1)+(1-1)+(1-1)\cdots=0\\ S_2=1+(-1+1)+(-1+1)+(-1+1)+\cdots=1\\ \therefore S_1\neq S_2 $$

But what inconsistency does:

$$ \begin{align} S=1&-1+1-1+\cdots\\ S+S=1&-1+1-1+\cdots+\\ &+1-1+1-1+\cdots=1\iff\\ \iff2S=1&\iff S=\frac12 \end{align} $$

create? I'm not even getting into Cesàro summation. Why does the limit of a sum have to equal the sum itself?

Why can't we have

$$ \frac12=\sum_{n=0}^\infty\ (-1)^n\neq \lim_{x\to\infty}\sum_{n=0}^x\ (-1)^n= \text{st}\sum_{n=0}^H\ (-1)^n= \text{undefined} $$

After all, in here $x$ is an arbitrarily big real number, $H$ is a positive infinite hyperinteger number and $\infty$ is NaN, not a number. Where is the inconsistency?

Edit:

There are already a lot of comments, and I feel I haven't made myself clear. Maybe the question is more philosophical than I thought. Here is an attempt to make my still developing points clearer:

1. $\cdots$ means the continuation to infinity of a series that continues the most simple pattern.

   • Example: $\displaystyle\sum_{n=0}^\infty\ (-1)^n$ means that for whatever number you have taken the partial sum, you are as far from the result as you were in the beginning. As by $6.$, such non-converging sum cannot be computed directly.

2. In the first example, it is proven that associativity does not hold for all infinite series, at least for divergent series, the same way $\sqrt a \sqrt b=\sqrt{ab}$ does not hold in $\mathbb C$. However, non-contradicting laws for associativity can be found:

   • Associativity may not work infinitely for numbers within the same series, as $S\neq S_1\neq S_2\neq S$ shows. However, it works pairwise between infinite series. $$ \begin{align} S+S=1&-1+1-1+\cdots+\\ &+1-1+1-1+\cdots \end{align} $$ is the same as $$ S+S=1+(-1+1)+(+1-1)+(-1+1)+\cdots $$ which is $1$, as we have seen. This rule is consistent. Other pairwise associations for this will either give the same, or $2-2+2-2+\cdots$, which is also $1$.
   • In other words, for any numbers, associativity works. So it works pairwise (2 series, 2 numbers per application), an infinite number of times. But it doesn't work within the same series, as each set of numbers would be a finite series, infinity is not a number, so it can't have not-a-number of times per application. That is, $$ S_1=(1-1)+(1-1)+(1-1)+\cdots $$ is the same as $$ \begin{align} S_1=1+&1+1+\cdots\\ -1-&1-1+\cdots \end{align} $$ and not the same as $$ \begin{align} S=1-&1+\\ +1-&1+\\ +1-&1+\\ +\cdots \end{align} $$

3. The limit of a sum equals the sum when the sum converges. Example: $$ \sum_{n=0}^\infty 2^{-n}=\lim_{x\to\infty}\sum_{n=0}^x 2^{-n}=2 $$

4. The limit of a non-converging sum does not exist or is infinity, not a number. All sums have a value, even thought their limits might not have one, or the value of the sum is infinity.

   • If that is the case, infinity, as not a number, cannot be directly summed with another sum (eliminating problems as $\infty-\infty$ by reason of lack of information).
   • If according to non-contradictory rules, a value can be assigned to a sum, that is the value of the sum. See the example above for $1-1+1-1+\cdots$

5. To distinguish between the values of two non-convergent sums, they first must be computed according to non-contradictory rules. Then their values can be compared, by transitivity of equality.

6. A divergent sum cannot be computed directly (the reason why $S\neq S_1\neq S_2\neq S$), as by definition of infinity, one cannot reach it. Again, use non-contradictory rules, making a finite number of changes that maintain the value of the divergent sum (see the examples' consistency).

Thank you for reading,

Answer 1

What's wrong is that it's gibberish. You haven't actually made any concrete claim at all, you've just written some symbols down.

$$S=1-1+1-1...$$

What the hell does that mean? You haven't told me what the symbols on the right hand side represent, so I have no idea what's being claimed when you say that $S$ "equals" the right hand side.

$$S+S=...$$

And now you want to take this meaningless, undefined object, and add it to itself?

When I say it's gibberish, I really mean it. You may as well have written down:

$$S=SDG\mathcal{RREG}20358!!!?++()))($$

When you write down an expression using symbols, you need to specify what those symbols mean. That sounds obvious, but sometimes we write down meaningless things without realizing it. You simply never gave a definition for what object the symbolic expression $1-1+1-1...$ refers to.

Now, you could provide such a definition. And if, after doing that, you could prove that, under that definition, we are indeed justified in adding $S+S$ and rearranging the terms in the way that you did, you would be fine. But you have to prove that.

Answer 2

I was asked to develop my comment into an answer. I haven't taken the time to read the somewhat lengthy discussion surrounding this post, but here is my take.

For a given sequence $\{a_n\}_{n \geq 1}$ of numbers one can associate many possible values. In the case that $\{a_n \}_{n \geq 1}$ is summable, a common value to associate to it is $$ S = \lim_{n \to \infty} \sum_{i=1}^n a_i $$ which the notations $$ \sum_{i=1}^\infty a_i \quad \text{ and} \quad a_1 + a_2 + \ldots $$ are used to denote. It happens that for convergent series some nice properties hold, such as $\sum_{i=1}^\infty a_i + \sum_{i=1}^\infty b_i = \sum_{i=1}^\infty (a_i+b_i)$.

Of course, there may be other value association rules one would wish to employ. One you mentioned is Cesaro summation, which it happens specifies a value for a strictly larger set of sequences than does the convergent series rule, as I'll term it, and which yields the same value for members of both sets. The series you mention is a prominent example of something for which the Cesaro rule produces a value, but the convergent series rule does not.

Now, a few points:

1. I am not sure of any well studied rules which assign values to all possible sequences (though trivial rules like "the value is the magnitude of the $10^{th}$ term" could be chosen).
2. Since the notation $a_1 + a_2 + \ldots$ is, as far as I am aware, only standard for sequences $\{a_n\}_{n \geq 1}$ yielding convergent series, I'd be careful with notation, or at least specify clearly what your notation means a priori.
3. Your ultimate question about whether something like associativity of the terms holds will depend on your rule. Without precisely specifying an exact rule, no one can say whether a given property will hold.

I suspect that the sensation of the question is that many people, possibly including yourself, are unsure exactly which rule you are using, and this confusion is compounded by the fact that the notation you are using is standard notation for a rule which you clearly cannot be using (namely the convergent series rule).

I hope that was helpful!

Answer 3

You ask what inconsistency is there in \begin{align} S=1&-1+1-1+\cdots\\ S+S=1&-1+1-1+\cdots+\\ &+1-1+1-1+\cdots=1\iff\\ \iff2S=1&\iff S=\frac12 \end{align}

You said before that $$S_1=(1-1)+(1-1)+(1-1)+(1-1)\cdots=0\\ S_2=1+(-1+1)+(-1+1)+(-1+1)+\cdots=1\\ $$ In the reasoning I copied first, you decided to use the "$S_2$" convention. But unless you believe that what you wrote for $S_1$ is wrong, you could have used that one too. So $$ 2S=0. $$ And you have proven that $1=0$.