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.

I'm asked to find an example of a diverging sequence $\sum_{n\in\mathbb{N}}a_n$ such that $\lim_{n\to\infty} a_n = 0$ but there exists parenthesization such that $\sum_{n\in\mathbb{N}}a'_n<\infty$ ($a'_n$ is the parenthesised sequence).

Clearly, we need to find a "forbidden" parenthesization that will change the convergence status of the sequence.

We learned that parenthesization is allowed in 2 cases:

  1. If the general term of the sequence $\to_{n\to\infty}{0}$ and the number of terms in each parentheses is bounded.
  2. All the terms in each parentheses are of the same sign.

Couldn't find one so far... Will appreciate your help.

Thanks.

share|improve this question

1 Answer 1

up vote 3 down vote accepted

Let $s_n = \sum_{i=1}^n$ be the sequence of partial sums of your series. "Parenthesizing" amounts to picking a subsequence of $s_n$. However, if $s_n$ diverges, say $s_n \rightarrow \infty$, then no subsequence can be convergent.

Perhaps the original series isn't meant to be divergent but rather non-convergent? In this case, you can take $1,-1,1/2,1/2,-1/2,-1/2,1/3,1/3,1/3,-1/3,-1/3,-1/3,\ldots$.

share|improve this answer
2  
I've seen a few others define "diverging" to mean "non-converging". But +1 either way. –  Jason DeVito May 29 '12 at 13:46
    
Thanks. Why do we need $2n$ terms in each parentheses? –  Amihai Zivan May 29 '12 at 13:58
    
OK, I understood: otherwise it would converge, right? –  Amihai Zivan May 29 '12 at 14:18
1  
The reason is that you want $a_n\rightarrow 0$. Otherwise you could take $1,-1,1,-1,\ldots$. –  Yuval Filmus May 29 '12 at 21:57

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.