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How can i actually use the fact of the subsequences being proper. Are $x_{2n+1}$ and $x_{2n}$ proper subsequences? because being so solves this immediately.

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  • $\begingroup$ How does it solve it immediately? It would do so only if you know both of those sequences converge. $\endgroup$ – Michael Hardy Sep 30 '17 at 21:58
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    $\begingroup$ Hint: you do need that the original sequence is bounded. $\endgroup$ – Mark Bennet Sep 30 '17 at 22:00
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A bounded sequence of real numbers must have at least one convergent subsequence. All of those converge to $\ell,$ and there is at least one of those; call it $(x_{n_k})_{k=1}^\infty$. So for every $\varepsilon>0,$ all except finitely many terms of $(x_{n_k})_{k=1}^\infty$ are in the interval with endpoints $\ell\pm\varepsilon.$ If infinitely many terms of the original sequence lie outside that small interval, then that is a bounded sequence, which therefore has at least one convergent subsequence. That subsequence converges to some point outside that interval small interval. But that contradicts the hypothesis.

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  • $\begingroup$ @Mariam : Note that you can "accept" the answer. $\endgroup$ – Michael Hardy Sep 30 '17 at 22:48
  • $\begingroup$ @MichaelHardy I was wondering if it would be possible for you to take a look at math.stackexchange.com/questions/2453786/… I think you could be able to help me a great deal! Thank you. $\endgroup$ – ALannister Oct 2 '17 at 3:04
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Assume the contrary.

Then there exists $\epsilon >0$ and a subsequence $(y_{n})$ of $(x_{n})$ such that $|y_{n}-l| \geq \epsilon$. But $(y_{n})$ being bounded, you can extract a convergent subsequence $(z_{n})$ from it (using Bolzano-Weierstrass theorem), which must then converge to $l$ by hypothesis: a contradiction.

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  • $\begingroup$ so the thing is : (yn) according to assumption is actually divergent, but the Bolzano Weierstrass conveys that a bounded sequence admits infinitely many convergent subsequences and that's why it's a contradiction of the theorem? so is (yn) a representation of any subsequence of (xn)? $\endgroup$ – Mariam Sep 30 '17 at 22:27
  • $\begingroup$ According to the assumption (yn) does not converge to l, but the Bolzano Weierstrass conveys that a bounded sequence (in this case yn) admits at least one convergent subsequence, which in this case should be converging to l by hypothesis. However this is not the case. (yn) is the construction of a subsequence of (xn) that does not converge to l (which can be constructed by "assuming the contrary of your result"). By the way, the proofs provided by Michael & jjj below are equivalent to mine. $\endgroup$ – Julien Sep 30 '17 at 22:39
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suppose the sequence does not convergent tto 1 then there exist a nbd of 1 s.t. it does not contain infinitely many term of the sequence.now,we take that terms which is outside of the nbd,and a it is a sunsequence of the sequence and as the sequence is bounded therefore the subsequence has a monotone convergent subsequence. but the subsequence does not converge to 1.its a contradiction.

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  • $\begingroup$ by 1 i guess u mean l, and thank you $\endgroup$ – Mariam Sep 30 '17 at 22:45

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