4
$\begingroup$

Before I really ask my question, I want to give my train of reasoning. Suppose we have some method of summation (as I understand, assigning a number to a series) that satisfies some or all of regularity, linearity, or stability, as defined in Hagen von Eitzen's answer to this question.

Suppose this method also assigns $1-1+1-1+\dots=1/2$---say it's Cesaro summation, or whatever. Evidently this method cannot admit rebracketing, because $(1-1)+(1-1)+\dots=0$ while $1+(-1+1)+(-1+1)\dots = 1$.

However, we can rebracket series that converge in the usual sense, even though they do not give the same results after rearrangement. Right? I think the limits of the partial sums $S_N = \sum^N_{k=1} a_k$ and $S_M = \sum_{k=1}^M(a_{2k-1}+a_{2k})$ must be the same---suppose that $S_N$ converges, and then evaluating for finite values of $N$ and $M$ shows that $S_M$ is a subsequence of $S_N$, so converges to the same limit. But $S_M$ is a rebracketing of $S_N$.

Therefore if a series converges, we enjoy the power to rebracket, which we do not have with Cesaro summation. However, in both cases, the axioms show we can still sum term by term, etc, so Cesaro summation keeps some nice things. What things doesn't it keep? Rebracketing appears to be an example.

So: What does a convergent series grant us that a series that is simply Cesaro summable (or otherwise) does not?

===============

I had some additional questions, which I found answers to here: Can we show that $1+2+3+\dotsb=-\frac{1}{12}$ using only stability or linearity, not both, and without regularizing or specifying a summation method?, but I'm leaving this paragraph anyway because comments refer to it.

As I was reading this link: http://www.nottingham.ac.uk/~ppzap4/response.html, where a physicist explains why some steps in a simple derivation he made were behind-the-scenes justified, one thing he says is that his manipulations do not contradict the three axioms, given in Hardy's Divergent Series. They seem to encapsulate linearity and stability, but not regularity.

  1. if $\sum a_n = A$ then $\sum ka_n = kA$
  2. if $\sum a_n = A$ and $\sum b_n = B$ then $\sum (a_n + b_n) = A+B$
  3. $\sum\limits_{n=0}a_n = A$ if and only if $\sum\limits_{n=1} a_n = A - a_0$

I'm sure that from axiom 3 (stability) we can obtain that if $\sum a_n = A$, then $\sum c_n = A$, where $\{c_n\}$ is just $\{a_n\}$, but with maybe infinitely many $0$'s thrown between the $a$'s. For instance $c_1 = a_1$, $c_2 = a_2$, but $c_i=0$ for $3 \leq i \leq 7$, and then $c_8 = a_3$, and so on.

Assume $S_1 = 1-2+3-4+\dots = 1/4$ in whatever sense we consider. Let $Z = 1+2+3+4+\dots$. It is not summable by any linear/stable method. But he writes

$$S_1 + 4Z = (1+0) + (-2 + 4\cdot1) + (3+0) + (-4 + 4\cdot2)=1+2+3+4+\dots=Z$$ $$-3Z = S \Rightarrow Z = -1/12$$

which requires stability. I suppose obtaining a "correct" value was a coincidence, then. He claims in the link they do not contradict the axioms, but the axioms don't hold in the first place.

$\endgroup$
  • $\begingroup$ As has been explained ad nauseam on the site and elsewhere, it is quite wrong to believe that $Z=\sum\limits_{n=1}^\infty n$ is $Z=-\frac1{12}$. Instead, $Z=+\infty$ and some (meromorphic complex) function, called $\zeta$, which happens to have the value $\zeta(s)=\sum\limits_{n=1}^\infty n^{-s}$ at every $\Re s>1$ can be defined on the whole complex plane except at $1$, in such a way that, for very good reasons, $\zeta(-1)=-\frac1{12}$. Hence the argument in the last paragraph of your post is, sorry to say, crap. $\endgroup$ – Did Aug 2 '16 at 11:18
  • 3
    $\begingroup$ When I use the equals sign there, I don't actually mean converges. I mean "equals" under whatever summation method we are considering. (Additionally, I learned some extra things about properties of summation methods, so let me edit the question a bit...) $\endgroup$ – Drew N Aug 2 '16 at 11:20
  • 1
    $\begingroup$ Additionally one might note that the "behind-the-scene" page you refer to is (notwithstanding its quite I-am-o-so-satisfied-with-myself tone) actually a damage control operation by one of the authors of the infamous Numberphile video on the subject which led to some (quite justified) stern reactions. $\endgroup$ – Did Aug 2 '16 at 11:31
  • 1
    $\begingroup$ Was the damage control not effective? I was pretty steamed watching the video, too, but if it's true that his manipulations are justified, then he has a shorthand way to sum whatever series he wants, doesn't he? $\endgroup$ – Drew N Aug 2 '16 at 11:35
  • 3
    $\begingroup$ @Drew: But the manipulations are not justified. No linear stable sumation method gives $-\frac1{12}$, and the Numberphile video relies on both linearity and stability. $\endgroup$ – Deusovi Aug 2 '16 at 14:07
-3
$\begingroup$

The answer to your question has to do with topology. The notion of limit coming from topology as applied to the standard topology of the real line coincides with the more elementary $\epsilon$-$\delta$ definition. Hence all of the theory applicable to the notion of limit of a sequence in a topological space is applicable to the familiar notion of limit defined by $\epsilon$-$\delta$ definition of limit.

You may find learning about topology and limits in topological spaces interesting, and even more so, combining that with knowledge from order theory about orderings of a countably infinite set.

Edit: To expand upon this, a topology for a set $X$ is a collection of subsets of that set, $\tau$, which are defined to be open such that:

1) $\emptyset,X\in\tau$, or in other words, the empty set and the entire set are in the topology.

2) Given any collection of open sets, $U_i\in\tau,i\in I$, the union $\bigcup_{i\in I}U_i\in\tau$. (Closed under arbitrary unions)

3) Given any finite collection of open sets, $U_n\in\tau$, $n\in\mathbb{N}_N$, the intersection, $\bigcap_{n\in \mathbb{N}_N}U_n\in\tau$. (Closed under finite intersections)

The pair $(X,\tau)$ is called a topological space, $\tau$ is the topology of the space, and $X$ the point set.

Because most topologies or interest are infinite and usually uncountably infinite, it is important to have ways of describing a topology in simpler terms. This leads one to consider a basis for the topology, as well as the related idea of a subbasis.

A basis for a topology, $\tau$ is a subset of the topology, $\mathcal{B}\subset\tau$, such that:

$\forall V\in\tau,\exists U_i\in\mathcal{B}$ such that $\bigcup_{i\in I}U_i=V$.

The point is: every open set in the topology $\tau$ can be represented as a union of open sets from the basis $\mathcal{B}$. The notion of subbasis takes this idea further:

A subbasis for a topology $\tau$ is a subset of $\mathcal{S}\subset\tau$ such that:

$\forall V\in\tau,\exists U_{(i,n_i)}\in\mathcal{S}$ such that $\bigcup_{i\in I}\bigcap_{n_i\in\mathbb{N}_{N_i}}U_{(i,n_i)}=V$

The point being that every open set of in the topology $\tau$ can be represented as a union of finite intersections of subbasis elements. The collection created from a subbasis, $\mathcal{S}$ by taking intersections only is the corresponding basis for that topology $\mathcal{B}$ (note: more than one basis may exist for a given topology, and likewise, more than one subbasis may exist for a given basis).

To give a familiar concrete example of a topology via a basis. We consider the point set $X=\mathbb{R}$, the real numbers. The standard topology for the real numbers has basis:

$\mathcal{B}=\{(a,b)|a<b;a,b\in\mathbb{R}\}\cup\{(-\infty,b)|b\in\mathbb{R}\}\cup\{(a,\infty)|a\in\mathbb{R}\}$

Aka, the set of all open intervals is the basis for the standard topology of the reals.

The familiar definition of limit of a sequence says that a sequence $(a_n)_{n\in\mathbb{N}}$ converges to a point $x$ if and only if $\forall\epsilon>0$, $\exists N\in\mathbb{N}$ such that $\forall n>N$, $|x-a_n|<\epsilon$.

Rephrasing the above in terms of open intervals containing $x$, aka open neighborhoods of $x$. The above definition says that the sequence $(a_n)_{n\in\mathbb{N}}$ converges to the point $x$ if and only if the sequence is eventually entirely within that neighborhood. Explicitly, $(a_n)_{n\in\mathbb{N}}$ converges to the point $x$ if and only if

$\forall(a,b)\in\mathcal{B}$ such that $x\in(a,b)$, $\exists N\in\mathbb{N}$, such that, $\forall n>N$ $a_n\in(a,b)$.

The topological definition of limit of a sequence generalizes this to contexts where there is no analog of intervals, and in a way that is independent of the basis:

In a topological space, $(X,\tau)$ a sequence of points, $(x_n)_{n\in\mathbb{N}}$, $x_n\in X$, is said to converge to a point $x\in X$ if:

For every open set $U$ containing $x$, there exists $N\in\mathbb{N}$ such that for every $n>N$, $x_n\in U$.

This is actually equivalent to the above description involving intervals in the reals, but is applicable in any topological space. Hence, with this definition, the notion of limit of a sequence is topological. Or in other words, depends solely on the topology of the space in question.

What does all of this have to do with the convergence or divergence of an infinite series (and its value)? The answer is in the definition of infinite series:

The infinite series, $\Sigma_{i=1}^{\infty}a_i$, is said to converge to the value $S$ of the limit of the sequence of partial sums, $\lim_{n\rightarrow\infty}S_n=\lim_{n\rightarrow\infty}\Sigma_{i=1}^na_i$, provided this limit exists. Hence the converge of an infinite series depends on the existence of the limit of the sequence of partial sums, which in turn depends on the topology of the space in question ($\mathbb{R}$ in the present case).

$\endgroup$
  • $\begingroup$ Can you explain in a little more detail? I admit I don't know much topology yet, just the Real Analysis point set stuff. I certainly can't see what it has to do with bracketing series and different summation methods. $\endgroup$ – Drew N Aug 2 '16 at 20:08
  • $\begingroup$ @DrewN Edited in an intro to topology and what it has to do w/ limits of sequences. (and its relevance to converge of infinite series) $\endgroup$ – Justin Benfield Aug 3 '16 at 2:53
  • 1
    $\begingroup$ So why would we consider different topologies? Do different topologies correspond to different summation regimes? And what does it have to do with rebracketing? $\endgroup$ – Drew N Aug 3 '16 at 5:43
  • $\begingroup$ @DrewN: Different topologies have a correspondingly different notion of limit (in some spaces, convergent sequences can have more than one limit point!). They don't correspond directly to different summation regimes, rather, what those do, is replace the standard definition of sum of an infinite series (limit of partial sums) with something else (the result is that the meaning of $\Sigma$ has changed. $\endgroup$ – Justin Benfield Aug 3 '16 at 6:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.