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Prove that if every PROPER sub sequence of a sequence is convergent , then the sequence is convergent. ( By proper, i mean all the sub sequences excluding the main sequence )

$Attempt$ : Since all proper sub sequences are convergent

=> Each of them must be bounded And Each of them must have their unique limit points

Since EACH of the sub sequences are bounded, then, the main sequence must be also bounded

=> By Bolzanno Weirstrass Theorem, the bounded main sequence has a limit point. ......(1)

Now, If i prove that the bounded main sequence converges to a unique limit point, then , I can prove that the sequence is convergent.

To do this, If i prove that all the sub sequences converge to one limit point l only, then from (1) we can conclude that the unique limit of the sequence is l .

How do i do this? Any directions?

Thanks

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Suppose there are two subsequences that converge to different limits. Create a new subsequence by mixing the terms of the two subsequences. Is this new subsequence convergent?

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Just leave the first term out, that's a proper subsequence :D.

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  • $\begingroup$ It does answer the question: the sequence w/o first term converges to say $a$, then the whole sequence converges to $a$ since one may drop initial terms w/o affecting convergence or value of limit. $\endgroup$ – coffeemath Jan 24 '14 at 13:31
  • $\begingroup$ This is the answer I would give. It absolutely answers the question. $\endgroup$ – user223391 Oct 3 '15 at 20:07
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If every proper subsequence of $a_1,a_2,a_3,a_4,\ldots$ is convergent then $a_2,a_3,a_4,\ldots$ is convergent. But if $a_2,a_3,a_4,\ldots$ is convergent, then so is $a_1,a_2,a_3,a_4,\ldots$, and it has the same limit.

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