Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose that the complex infinite sum $ \sum_{n=1}^{\infty}(-1)^{n}Z_n$ converges.

Define $A \subset \mathbb{C}$ by

$A=\{{z\in\mathbb{C}\mid\exists f:\mathbb{N}\underset{on}{\overset{1-1}{\rightarrow}}\mathbb{N},z= \sum_{n=1}^{\infty}Z_{f(n)}}\}$

Prove that $A=\mathbb{C}$ or that $\exists a, b \in \mathbb{C}$ such that $A = \{a+tb \mid t\in \mathbb{R}\}$.

--edit-- in bounty message I mean solution of the problem.

share|improve this question
6  
Please don't just dump undigested problems here with no indication that you have put any thought into them. Why are you interested in this problem? What thoughts do you have about it? What pertinent facts do you already know? Where are you getting stuck? Engage with us, so we can write more useful answers. –  Gerry Myerson Jan 6 '13 at 15:05
1  
This is a very uncommon task, I completelly have no idea how to start. I need hints how to start thinking on it. –  Steve Jan 6 '13 at 15:08
3  
When you have no idea how to start on something, it's often helpful to construct some examples, and play with them. Know any such convergent series? Pick one, and see what happens when you apply such an $f$ to it. Do a few examples, see if you can work out what's going on. Also, see if there's anything like this in whatever text or notes you're learning from. –  Gerry Myerson Jan 6 '13 at 15:13
3  
I don't understand why there are four close votes for this question. Two are for the reason "not constructive" and two for "not a real question". As annoying as carelessly typed questions with imperatives and no work shown are, closing them seems like a significant overreaction to me. It's perfectly clear (at least to me) what this question is asking, and it's a perfectly legitimate mathematical question. No-one has to answer it in its present sloppy form if they don't want to, but that's not a reason for closing it. –  joriki Jan 9 '13 at 18:44
2  
@did: The result is stated in Wikipedia. Your counterexample isn't one because the case of an unconditionally convergent series is covered by $b=0$. –  joriki Jan 9 '13 at 19:44

1 Answer 1

up vote 4 down vote accepted
+50

Let $\displaystyle \sum_{n \geq 1} X_n$ be a conditionally convergent series in the $d$-dimensional real Euclidean space $\mathbb{R}^{d}$. Then the set $Y$ of all possible sums of rearrangments of this series is an affine subspace of $\mathbb{R}^{d}$. When $d=1$, this is the well know Riemann rearrangement theorem and the proof is not very difficult.

Addendum: A good reference on this is the following: The Remarkable Theorem of Levy and Steinitz, Peter Rosenthal, The American Mathematical Monthly, Vol. 94, No. 4 (Apr., 1987), pp. 342-351

I also realized that you do not need conditional convergence because of the way you stated your conclusion (b = 0 covers absolutely convergent case). So I removed that trivial remark.

share|improve this answer
1  
The cited article can be found on JSTOR. –  robjohn Jan 14 '13 at 2:28

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.