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Suppose the sequence $(a_n)$ is bounded and sequence$(b_n)$ converges to non-zero limit. Does the sequence $(a_nb_n)$ converge ? My claim is that it converges. I plug in few examples to see and it does converge. But I don know how to prove it. I stuck at $|a_nb_n-c|<\epsilon$ where $c$ is the value which $(a_nb_n)$ converges. Any idea on how to prove this ?

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No. Suppose $$a_n=(-1)^n$$ and $$b_n=1.$$

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  • $\begingroup$ thx. I keep on thinking of reciprocal. no wonder I can't get the right conterexample $\endgroup$ – Idonknow Feb 3 '13 at 14:28
  • $\begingroup$ @Idonknow: No problem! Hopefully it's pretty clear :) $\endgroup$ – Clayton Feb 3 '13 at 14:33
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no,let's use counterexample

i have sequence $(a_n)$ with $(a_n)=(1,-1,1,-1,...)$ and $(b_n)=(1,1,1,1,...)$

we get $(a_n)$ is bounded because $\forall n\in\mathbb{N},|a_n|\le 1$ and $(b_n)$ converges to $1$.

and then we get $(a_nb_n)=(a_n)$ which is divergen

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  • $\begingroup$ i'm sorry,i'm late when i'm typing :) $\endgroup$ – A Ricko Maulidar Feb 3 '13 at 14:38
  • $\begingroup$ I can relate; that's happened to me, even now that I'm more accustomed to formatting. You'll get quicker. At any rate, no worries: "duplicate answers" happen all the time! $\endgroup$ – Namaste Feb 3 '13 at 14:44
  • $\begingroup$ thankyou for your advise :) $\endgroup$ – A Ricko Maulidar Feb 3 '13 at 14:47

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