Convergent sequences out of bounded sequences Let us consider a bounded sequence  $\{a_n\}$ . Now as it is a bounded sequence it must contain a convergent sub-sequence, $\{b_n\}$. Now let us filter out $\{b_n\}$ out of $\{a_n\}$. As such we are still left with a bounded sequence. But , I want to know if this sequence can again contain infinite terms so as to have another convergent sub-sequence inside it . For a given sequence , does there exist anyway to determine how many times can we carry on this elimination of convergent sub-sequences each time ending up with a bounded infinite sequence to further have another convergent sub-sequence .
 A: It depends how we "filter out." If we remove all the $b_n$ we may end up with a sequence that is finite, or even empty. 
But if we remove say $b_1$ and $b_4$ and $b_9$ and $b_{16}$ and so on, then we are left with an infinite sequence and can repeat the process.
A: You can take as many subsequences of $\{a_n\}$ as you want. 
The subsequence $\{b_n\}$ is already convergent, so any further subsequence of $\{b_n\}$ will be convergent and will have the same limit as $\{b_n\}$.
If you have omitted only a finite number of terms from $\{a_n\}$ to get $\{b_n\}$ then by throwing away the terms of $\{b_n\}$ from $\{a_n\}$ you will end up with the same finite number of terms in $\{a_n\}$ which you have omitted first. Otherwise, if $\{a_n\}\setminus \{b_n\}$ is infinite, you can repeat the procedure.
A: If $(b_n)$ converges and you have only omitted finitely many terms from $(a_n)$ to get $(b_n)$, then the original sequence $(a_n)$ must have been convergent to begin with. Convergence only depends on the tail.
So if $(a_n)$ is not convergent, you can repeat the process infinitely many times on the "leftovers".
