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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

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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
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

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 $+\infty.$

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