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.

The rearrangement theorem in real analysis tells us that given an absolutely convergent series we can permute its entries and the resulting series will have the same limit.

But what happens if my series contains only positive entries, but converges only improperly to $\infty$: Does every rearrangement of it also converges to $\infty$ ?

share|improve this question
    
Can you create an example of a positive, improperly convergent series? –  mousomer Dec 20 '12 at 16:31
    
@mousomer $\sum 1$ –  resu Dec 20 '12 at 16:38
    
I think convergent is used only for something which converges to a number. If the series goes to $\infty$ I think people say it diverges to infinity.... –  N. S. Dec 20 '12 at 16:41
    
OK. OK. This is a case of unclear notation. I haven't seen the term "improperly convergent" for a series in that sense before. It's not a widely used terminology. –  mousomer Dec 20 '12 at 17:31

3 Answers 3

up vote 4 down vote accepted

Yes. Is is easy to prove it with $\epsilon-N_\epsilon$, but here is a simpler proof:

Assume by contradiction that some rearangement doesn't diverge to $\infty$. Then the re-arangement is convergent (since positive) and hence absolutely convergent.

But then, any re-arangement of the re-arangement is convergent to the same real number... The original series is a re-arangement of any of his re-arangements....

P.S. Here is the $\epsilon-N_\epsilon$ proof.

Let $\sigma$ be any permutation of the indices. We will denote by $s_n$ respectively $t_n$ the partial sums of the original series and the permutation.

Let $\epsilon >0$ be arbitrary.

Since $\lim_n s_n =\infty$ then, there exists some $N_\epsilon$ so that for all $n > N_\epsilon$ we have

$$s_n > \epsilon \,.$$

Pick now some $M_\epsilon$ such that

$$\{1,2,3,.., N_\epsilon\} \subset \sigma( \{ 1,2,...., M_\epsilon \}) $$

Then

$$t_{M_\epsilon}=x_{\sigma(1)}+....+x_{\sigma(M_\epsilon)} \geq x_1+...+x_{N_\epsilon} =s_{N_\epsilon}$$

Thus, for all $n> M_\epsilon$ we have

$$T_n \geq t_{M_\epsilon} \geq s_{N_\epsilon} > \epsilon \,.$$

Since $\epsilon >0$ is arbitrary we are done.

share|improve this answer

For series with non-negative terms there is a nice way to think about sums: rather than worrying about reordering them, we may simply unorder them.

Namely, let $\sum_{n=1}^{\infty} a_n$ be a sequence of non-negative real numbers, with sum $A \in (-\infty,\infty]$. Then $A$ is the least upper bound of $\{\sum_{n \in S} a_n\}$ as $S$ ranges over all finite subsets of $\mathbb{Z}^+$. This description of $A$ is manifestly invariant under rearrangement, hence the sum ("improper" or otherwise) is invariant under rearrangement.

This idea can be generalized: for a(ny) set $\mathcal{S}$ and a(ny) function $a_{\bullet}: \mathcal{S} \rightarrow [0,\infty)$, we can define the sum $\sum_{s \in \mathcal{S}} a_s$ in terms of the supremum of sums over finite subsets of $\mathcal{S}$. This works in particular for things like $\sum_{n=-\infty}^{\infty} a_n$, $\sum_{m,n = 1}^{\infty} a_{m,n}$ and even sums over uncountable index sets (although the sum will be infinite unless all but countably many of the terms are zero).

More generally there is a theory of unordered summation of functions $a_{\bullet}: \mathcal{S} \rightarrow \mathbb{R}$ (or in any normed abelian group). See for instance $\S 9.4$ of these notes. However, it turns out to be equivalent to absolute convergence so does not give anything really new.

share|improve this answer

Yes, every rearrangement then "converges" to $\infty$. Let the orginal series be $\sum a_n$. For any $B$, there is an $N$ such that $\sum_1^N a_k \gt B$.

Let the rearranged series be $\sum b_n$. Then there is an $M(N)$ such that all the $a_i$ with index $\le N$ are $b_j$ for some $j \le M$. Therefore $\sum_1^M b_k \gt B$.

share|improve this answer

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.