## Convergent sub sequence. [closed]

Let us consider a convergent sequence {a1, a2, a3...........}. Now let me choose few terms out of it { b1.b2,b3....bn}. Now let us consider the sequence , { 1,2,3,4, b1,b2,b3 .....bn, 5,6,7,8,9........}. Is this sequence a divergent sequence? If yes then how can it have a convergent subsequence { b1, b2,b3.....bn}?

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Please consult the FAQ for the scope of this website. Your question is better suited for math.stackexchange.com – Willie Wong Aug 25 2011 at 16:19
Hello Willie, math.stackexchange says this question does not meet their qulity standars. I hope you will be glad to reply to this question. – primeczar Aug 25 2011 at 16:34
primeczar, I am well informed on sequences in general, however, I have great difficulty understanding your question, and I assume many people have the same problem. The main problem I see is that the definition of your sequence is unclear. The way you write I assume there are just a view terms b1,...,bn inserted into the sequence of naturam numbers. This b1,...,bn is not a subsequence in the usual sense as it is finite. Also, with the typoical def of divergence it is not problem to have a convergent subsequence. Perhaps you mean what is also called convergence to infinity. cont – quid Aug 25 2011 at 17:21
Then a convergent subsequence would be a problem. Yet you do not have one, in the way I understand your sequence. In short, your precise question is impossible to understand for me. Perhaps what I wrote is helpful to make your question easier to understand. But, as Willie said, please try to ask a possibly more precise question along these lines on stackexchange not here. – quid Aug 25 2011 at 17:25
Importantly, you need to be clear what you mean by "divergent". Do you simply mean "not convergent", or do you mean "converging to infinity"? In either case, your question seems to belong better on mathstackexchange.com, as quid has suggested – Yemon Choi Aug 25 2011 at 18:56