Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Warmup (you've probably seen this before)

Suppose $\sum_{n\ge 1} a_n$ is a conditionally convergent series of real numbers, then by rearranging the terms, you can make "the same series" converge to any real number $x$. To do this, let $P=\{n\ge 1|a_n\ge 0\}$ and $N=\{n\ge 1|a_n<0\}$. Since $\sum_{n\ge 1} a_n$ converges conditionally, each of $\sum_{n\in P}a_n$ and $\sum_{n\in N}a_n$ diverge and $\lim a_n=0$.

Starting with the empty sum (namely zero), build the rearrangement inductively. Suppose $\sum_{i=1}^m a_{n_i}=x_m$ is the (inductively constructed) $m$-th partial sum of the rearrangement. If $x_m\le x$, take $n_{m+1}$ to be the smallest element of $P$ which hasn't already been used. If $x_m> x$, take $n_{m+1}$ to be the smallest element of $N$ which hasn't already been used.

Since $\sum_{n\in P}a_n$ diverges, there will be infinitely many $m$ for which $x_m\ge x$, so $n_{m+1}$ will be in $N$ infinitely often. Similarly, $n_{m+1}$ will be in $P$ infinitely often, so we've really constructed a rearrangement of the original series. Note that $|x-x_m|\le \max\{|a_n|\bigm| n\not\in\{n_1,\dots, n_m\}\}$, so $\lim x_m=x$ because $\lim a_n=0$.


Suppose $\sum_{n\ge 1}v_n$ is a conditionally convergent series with $v_n\in \mathbb R^k$. Can the sum be rearranged to converge to any given $w\in \mathbb R^k$?

Obviously not! If $\lambda$ is a linear functional on $\mathbb R^k$ such that $\sum \lambda(v_n)$ converges absolutely, then $\lambda$ applied to any rearrangement will be equal to $\sum \lambda(v_n)$. So let's also suppose that $\sum \lambda(v_n)$ is conditionally convergent for every non-zero linear functional $\lambda$. Under this additional hypothesis, I'm pretty sure the answer should be "yes".

share|improve this question

4 Answers 4

up vote 31 down vote accepted

The Levy--Steinitz theorem says the set of all convergent rearrangements of a series of vectors, if nonempty, is an affine subspace of ${\mathbf R}^k$. There is an article on this by Peter Rosenthal in the Amer. Math. Monthly from 1987, called "The Remarkable Theorem of Levy and Steinitz". Also see Remmert's Theory of Complex Functions, pp. 30--31.

As an example, taking $k = 2$, suppose $v_n = ((-1)^{n-1}/n,(-1)^{n-1}/n)$. Then the convergent rearrangments fill up the line $y = x$. The linear function $\lambda(x,y) = x-y$ of course kills the series, which makes Anton's observation explicit in this instance.

The Rosenthal article, at the end, discusses Anton's question. Indeed if there is no absolute convergence in any direction then the set of all rearranged series is all of ${\mathbf R}^k$. Note by the above example that this condition is stronger than saying the series in each standard coordinate is conditionally convergent. Rosenthal said this stronger form of the Levy-Steinitz theorem was in the papers by Levy (1905) and Steinitz (1913). He also refers to I. Halperin, Sums of a Series Permitting Rearrangements, C. R. Math Rep. Acad. Sci. Canada VIII (1986), 87--102.

share|improve this answer

I discuss (with references, but without proof) the Levy-Steinitz Theorem in Section 2 of the following document:

http://www.math.uga.edu/~pete/UGAVIGRE08.pdf

In particular, the version I give describes precisely the set of limits of convergent rearrangements in terms of the subspace of directions of absolute convergence of the series. As a special case, if no one-dimensional projection is absolutely convergent, then indeed one can rearrange the series to converge to any vector in $\mathbb{R}^n$.

share|improve this answer

To a conditionally convergent series $\sum_{n\geq 1}v_n$ in $\mathbb{R}^d$ one can attach so called convergence functionals $f$, which are linear functionals $f:\mathbb{R}^d\to\mathbb{R}$ with the property $\sum_{n=1}^{\infty}|f(v_n)|<\infty$. Let $\Gamma ((v_n))$ be the set of all these functionals. Then the set of values of the possible rearrangements of the series $\sum_{n=0}^{\infty}v_n$ is exactly the affine space $\sum_{n=0}^{\infty}v_n + \Gamma ((v_n))_0$, where $\Gamma ((v_n))_0$ denotes the annihilator of $\Gamma ((v_n))$, i.e. $\bigcap_{f\in\Gamma ((v_n))}\mathrm{ker}(f)$. This is precisely the Steinitz` Theorem mentioned by KConrad. Let me just add that this result does not hold in general for infinite-dimensional spaces. However, a generalization of Steinitz theorem seems very approachable for locally convex spaces -> see e.g. "The Steinitz theorem on rearrangement of series for nuclear spaces" by W. Banaszczyk (1990), in Journal für die reine und angewandte Mathematik 403, 187-200.

EDIT: added the condition on $v_n$ per KConrad´s comment.

share|improve this answer
    
You should also add the condition that the specific series you wrote down a_1 + a_2 + a_3 + ... is convergent. It wouldn't make sense to take the translate of that annihilator by a divergent ordering of the terms. –  KConrad Jun 24 '10 at 21:51
    
I meant the same sequence/series $a_n$ as in the OP. But you are right, for the sake of completeness I should add it. –  efq Jun 24 '10 at 23:16
    
You don't want the series to be in R, since there nothing of much interest is happening (linear functionals on the real line?). Your series and linear functionals belong in R^k, as you had edited it once before. Maybe change a_n to v_n? –  KConrad Jun 25 '10 at 0:25
    
Ok, notation also fixed to be in correspondence with the original post. –  efq Jun 25 '10 at 0:36
    
The exponents in the R^d were enclosed in the \mathbb, so they weren't showing up (at least on my screen). I took them outside of \mathbb and now it looks okay. –  KConrad Jun 25 '10 at 2:10

One way to think of this is to have a vector whose $x$-component is conditionally convergent and whose $y$-component is absolutely convergent. Then the answer is seen to be obviously "no".

Next, change to a different basis and do the same thing, and go to higher dimensions, and you then quickly see that you can have an affine space to some point of which every rearrangement converges, but to any point of which some rearrangement converges.

share|improve this answer
1  
Not precisely worded correctly. Some rearrangements do not converge at all. Thus "to some point of which every rearrangement converges" is wrong. –  Gerald Edgar Jun 25 '10 at 3:02
    
OK: "....to some point of which every convergent rearrangement converges". –  Michael Hardy Jun 25 '10 at 16:13
    
Your example doesn't satisfy the hypothesis that the series remains conditionally convergent after applying any linear functional. Specifically, it becomes absolutely convergent after applying the linear functional $x$. The example demonstrates that it's possible to have a 1-dimensional affine space of limit points, but the question is essentially whether it's always possible to have a higher-dimensional space of possible limit points. –  Anton Geraschenko Jun 26 '10 at 0:54

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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