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 am trying to come up with example of sequence whose countably many subsequences converge to single point but the original sequences does not.

I came up with a series, which seems to converge but created a little confusion. Consider a nested sequence $$\left(\left(\frac{k}{n}\right)_{n\ge 1}\right)_{k\ge 1}, n,k\in\mathbb{N}$$.

For finitely many $k$, it sure does converge, what about countable but infinite k. There should be no problem that it converges but $\infty$ leaves some dilemma.

Edit Could you rather suggest how I can achieve such sequence that does not converge but whose countably many subsequences do?

Thank You

share|improve this question
    
You need to explain how you "intertwine" the sequences. Just taking a union of the sets doesn't guarantee you even have the same limit as the original sequences had. –  Asaf Karagila Jan 24 '13 at 9:19
    
@AsafKaragila, I wish to intertwine this way: $1,1/2,1/3,...2,2/2,2/3,2/4,...$. This seems to converge to $0$. But I have some questions, Is this still a sequence, because there are infinities in two direction? Further, How can I (if possible) arrange so that this does not converge. And, If it cannot be made so can you suggest some other examples? –  007resu Jan 24 '13 at 9:26
    
"This way" doesn't cut it. You have to be very precise when talking about infinite sets. –  Asaf Karagila Jan 24 '13 at 9:27
    
@AsafKaragila I mean: $\{(1/n)_{n\ge 1},(2/n)_{n\ge 1},(3/n)_{n\ge 1},(4/n)_{n\ge 1},...\}$ I am not sure how to clarify it. $\{1,1/2,1/3...,0,2,2/2,2/3,2/4,2/5,...0,3,3/2,3/3,3/4,...,0,\}$ Is it clearer? –  007resu Jan 24 '13 at 9:31
2  
But this has order type of $\omega^2$ whereas sequences are usually taken to be with order type of $\omega$. You may want to use some pairing function $f(n,m)=k$ so $a_k$ is the $m$-th member of the $n$-th sequence. One example is Cantor's pairing function $f(n,m)=\frac{(n+m)(n+m+1)}{2}+n$. –  Asaf Karagila Jan 24 '13 at 9:35
show 1 more comment

1 Answer

First note that you cannot mean that all subsequences converge to the same limit, yet the series does not converge, since the entire series is a subseries of itself.

It isn't clear what you want in your sequence, but here is an example of a series having a countable collection of subseries $A_1,A_2,...$, with no two terms equal in the entire series, but for which each subseries $A_k$ converges to $0$.

First we need the notion of a squarefree number, which may be defined as any number whose prime factorization has each prime to the power 1. For example since $15=3 \cdot 5$, $15$ is squarefree, while since $12=2\cdot 2 \cdot 3$ has the repeated prime $2$, $12$ is not squarefree.

Now define the sequence $x_n$ to be $n$ if $n$ is not squarefree, and $1/n$ if $n$ is squarefree. Further for each $k \ge 1$ define $A_k$ to be the set of squarefree numbers having exactly $k$ prime factors. For example $A_1$ is the sequence of primes, $A_2$ is the sequence of numbers (put into increasing order) of the form $pq$ where $p,q$ are distinct primes, and so on.

Then each subsequence $A_k$ converges to $0$, and there are countably many such subsequences $A_k$. The series itself does not converge, since it has the subsequence $2^2,3^2,4^2,...$ which diverges to $+\infty.$

share|improve this answer
add comment

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.