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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,

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    $\begingroup$ Nothing wrong in algebra with converging infinite series... $\endgroup$ – DonAntonio May 3 '14 at 15:36
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    $\begingroup$ Yes @JMCF125, that's exactly what it means, by definition. $\endgroup$ – DonAntonio May 3 '14 at 15:38
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    $\begingroup$ As far as you don't give a mathematical, consistent definition of "value of infinite series" other than the usual, standard one, I can't understand what we're talking about here, @JMCF . $\endgroup$ – DonAntonio May 3 '14 at 15:52
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    $\begingroup$ @JMCF125 You can, but that doesn't mean it will be meaningful. You have to define how you assign those values, and then it will only be meaningful in that context. $\endgroup$ – Eff May 3 '14 at 16:09
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    $\begingroup$ @JMCF125, it seems like you've some sense of an idea, but either you can't express it in a mathematically coherent fashion...or else I'm unable to understand you: too many things left flying in the air... $\endgroup$ – DonAntonio May 3 '14 at 16:22
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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.

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  • $\begingroup$ It means to take the sum all the way to infinity, not to see the limit of greater and greater partial sums. What else could it mean? $\endgroup$ – JMCF125 May 3 '14 at 15:52
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    $\begingroup$ What does "taking the sum all the way to infinity" mean mathematically ? $\endgroup$ – DonAntonio May 3 '14 at 15:52
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    $\begingroup$ @JMCF125 Okay, so let $(a_n)$ be a sequence of real numbers. What is your definition for what number $\sum a_n$ equals? Note that "You contine to sum up the $a_n$" doesn't refer to number, it refers to a process. $\endgroup$ – Jack M May 3 '14 at 16:04
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    $\begingroup$ @JMCF125 No, the sum is the result. Additions of addends is the process. $\endgroup$ – Eff May 3 '14 at 16:17
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    $\begingroup$ @JMCF125, how long do I have to sum up numbers? What if you get tired before I do? Will our results be different? I can't talk for you, but I've stuff to do: I cannot sum all the time... $\endgroup$ – DonAntonio May 3 '14 at 16:20
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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!

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  • $\begingroup$ +1, indeed I'm not sure what that associativity (referred in my question edit point $2.$) means, but there does seem to be some coherence. While searching similar questions, I have found the values to any sum I'm pointing out are probably Ramanujan summations. $\endgroup$ – JMCF125 May 3 '14 at 21:47
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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 $1=0$.

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  • $\begingroup$ As I said, one can prove that way associativity doesn't hold for infinite series. That means $S\neq S_1\neq S_2\neq S$. I'm not adding an infinite number of parentheses in $S$, as I did to show the incorrect associativity in $S_1$ and $S_2$. $\endgroup$ – JMCF125 May 3 '14 at 15:42
  • $\begingroup$ And what is the "correct associativity"? $\endgroup$ – Martin Argerami May 3 '14 at 15:55
  • $\begingroup$ maybe infinite series just don't have that property, just as as complex numbers don't have $\sqrt a \sqrt b=\sqrt{ab}$. $\endgroup$ – JMCF125 May 3 '14 at 15:57
  • $\begingroup$ Of course they don't (unless they converge absolutely). But then how do you know that $1-1+1-1+\cdots+1-1+1-1+\cdots=1$? $\endgroup$ – Martin Argerami May 3 '14 at 17:47
  • $\begingroup$ I don't know that $1-1+1-1+\cdots=1$, and I can't, because that is false. But $1+(-1+1)+(-1+1)+\cdots=1$. That is the point: "infinite associativity" doesn't work, at least for divergent series. I know that $S+S=1$ because of commutativity. $\endgroup$ – JMCF125 May 3 '14 at 18:30

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