An example of a sequence which does not have any subsequence with a finite limit.











up vote
2
down vote

favorite













Find and example of a sequence ${x_n}$ which does not contain any subsequences having a finite limit.




I've been thinking of the following seqeunce:
$$
x_n = sin(n)cdotsin(sqrt{3}n)
$$



But is that true? Also what are other examples of such sequences?






calculus limits examples-counterexamples







share|cite|improve this question


















  • 8




    Natural numbers $1,2,3,...$? No subsequence is Cauchy, hence convergent?
    – астон вілла олоф мэллбэрг
    5 hours ago












  • @астонвіллаолофмэллбэрг oh my, shame on me
    – roman
    5 hours ago






  • 3




    Remember, every bounded sequence of real numbers has a convergent subsequence. Therefore, your sequence being bounded does not have a convergent subsequence. The best bet is to take an unbounded sequence where the distance between any two terms is greater than some fixed number.
    – астон вілла олоф мэллбэрг
    5 hours ago










  • @астонвіллаолофмэллбэрг thank you! it's crystal clear now
    – roman
    5 hours ago










  • Great. You can now answer this question yourself and close it.
    – астон вілла олоф мэллбэрг
    5 hours ago

















up vote
2
down vote

favorite













Find and example of a sequence ${x_n}$ which does not contain any subsequences having a finite limit.




I've been thinking of the following seqeunce:
$$
x_n = sin(n)cdotsin(sqrt{3}n)
$$



But is that true? Also what are other examples of such sequences?






calculus limits examples-counterexamples







share|cite|improve this question


















  • 8




    Natural numbers $1,2,3,...$? No subsequence is Cauchy, hence convergent?
    – астон вілла олоф мэллбэрг
    5 hours ago












  • @астонвіллаолофмэллбэрг oh my, shame on me
    – roman
    5 hours ago






  • 3




    Remember, every bounded sequence of real numbers has a convergent subsequence. Therefore, your sequence being bounded does not have a convergent subsequence. The best bet is to take an unbounded sequence where the distance between any two terms is greater than some fixed number.
    – астон вілла олоф мэллбэрг
    5 hours ago










  • @астонвіллаолофмэллбэрг thank you! it's crystal clear now
    – roman
    5 hours ago










  • Great. You can now answer this question yourself and close it.
    – астон вілла олоф мэллбэрг
    5 hours ago















up vote
2
down vote

favorite









up vote
2
down vote

favorite












Find and example of a sequence ${x_n}$ which does not contain any subsequences having a finite limit.




I've been thinking of the following seqeunce:
$$
x_n = sin(n)cdotsin(sqrt{3}n)
$$



But is that true? Also what are other examples of such sequences?






calculus limits examples-counterexamples







share|cite|improve this question














Find and example of a sequence ${x_n}$ which does not contain any subsequences having a finite limit.




I've been thinking of the following seqeunce:
$$
x_n = sin(n)cdotsin(sqrt{3}n)
$$



But is that true? Also what are other examples of such sequences?






calculus limits examples-counterexamples




calculus limits examples-counterexamples






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 5 hours ago









roman

1,23711017




1,23711017








  • 8




    Natural numbers $1,2,3,...$? No subsequence is Cauchy, hence convergent?
    – астон вілла олоф мэллбэрг
    5 hours ago












  • @астонвіллаолофмэллбэрг oh my, shame on me
    – roman
    5 hours ago






  • 3




    Remember, every bounded sequence of real numbers has a convergent subsequence. Therefore, your sequence being bounded does not have a convergent subsequence. The best bet is to take an unbounded sequence where the distance between any two terms is greater than some fixed number.
    – астон вілла олоф мэллбэрг
    5 hours ago










  • @астонвіллаолофмэллбэрг thank you! it's crystal clear now
    – roman
    5 hours ago










  • Great. You can now answer this question yourself and close it.
    – астон вілла олоф мэллбэрг
    5 hours ago
















  • 8




    Natural numbers $1,2,3,...$? No subsequence is Cauchy, hence convergent?
    – астон вілла олоф мэллбэрг
    5 hours ago












  • @астонвіллаолофмэллбэрг oh my, shame on me
    – roman
    5 hours ago






  • 3




    Remember, every bounded sequence of real numbers has a convergent subsequence. Therefore, your sequence being bounded does not have a convergent subsequence. The best bet is to take an unbounded sequence where the distance between any two terms is greater than some fixed number.
    – астон вілла олоф мэллбэрг
    5 hours ago










  • @астонвіллаолофмэллбэрг thank you! it's crystal clear now
    – roman
    5 hours ago










  • Great. You can now answer this question yourself and close it.
    – астон вілла олоф мэллбэрг
    5 hours ago










8




8




Natural numbers $1,2,3,...$? No subsequence is Cauchy, hence convergent?
– астон вілла олоф мэллбэрг
5 hours ago






Natural numbers $1,2,3,...$? No subsequence is Cauchy, hence convergent?
– астон вілла олоф мэллбэрг
5 hours ago














@астонвіллаолофмэллбэрг oh my, shame on me
– roman
5 hours ago




@астонвіллаолофмэллбэрг oh my, shame on me
– roman
5 hours ago




3




3




Remember, every bounded sequence of real numbers has a convergent subsequence. Therefore, your sequence being bounded does not have a convergent subsequence. The best bet is to take an unbounded sequence where the distance between any two terms is greater than some fixed number.
– астон вілла олоф мэллбэрг
5 hours ago




Remember, every bounded sequence of real numbers has a convergent subsequence. Therefore, your sequence being bounded does not have a convergent subsequence. The best bet is to take an unbounded sequence where the distance between any two terms is greater than some fixed number.
– астон вілла олоф мэллбэрг
5 hours ago












@астонвіллаолофмэллбэрг thank you! it's crystal clear now
– roman
5 hours ago




@астонвіллаолофмэллбэрг thank you! it's crystal clear now
– roman
5 hours ago












Great. You can now answer this question yourself and close it.
– астон вілла олоф мэллбэрг
5 hours ago






Great. You can now answer this question yourself and close it.
– астон вілла олоф мэллбэрг
5 hours ago












2 Answers
2






active

oldest

votes

















up vote
5
down vote



accepted










Any sequence ${x_n}$ with $|x_n| to infty$ will do.



By Bolzano-Weierstrass Theorem others wont: if $|x_n| not to infty$, there is $M$, such that you can find a subsequence of ${x_n}$ that is bounded by $M$. Then you can find a converging subsequence of that subsequence (and hence of the original sequence).






share|cite|improve this answer






























    up vote
    6
    down vote













    The example in the OP is wrong. This can be shown by Bolzano-Weierstrass theorem which states that every bounded sequence contains a convergent subsequence.



    Now if we consider:
    $$
    x_n = sin(n)sin(sqrt{3}n)
    $$
    its clear that the sequence is bounded by $[-1, 1]$. Hence it contains a convergent subsequence.






    share|cite|improve this answer























    • Upvoted. Good job.
      – астон вілла олоф мэллбэрг
      5 hours ago











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    5
    down vote



    accepted










    Any sequence ${x_n}$ with $|x_n| to infty$ will do.



    By Bolzano-Weierstrass Theorem others wont: if $|x_n| not to infty$, there is $M$, such that you can find a subsequence of ${x_n}$ that is bounded by $M$. Then you can find a converging subsequence of that subsequence (and hence of the original sequence).






    share|cite|improve this answer



























      up vote
      5
      down vote



      accepted










      Any sequence ${x_n}$ with $|x_n| to infty$ will do.



      By Bolzano-Weierstrass Theorem others wont: if $|x_n| not to infty$, there is $M$, such that you can find a subsequence of ${x_n}$ that is bounded by $M$. Then you can find a converging subsequence of that subsequence (and hence of the original sequence).






      share|cite|improve this answer

























        up vote
        5
        down vote



        accepted







        up vote
        5
        down vote



        accepted






        Any sequence ${x_n}$ with $|x_n| to infty$ will do.



        By Bolzano-Weierstrass Theorem others wont: if $|x_n| not to infty$, there is $M$, such that you can find a subsequence of ${x_n}$ that is bounded by $M$. Then you can find a converging subsequence of that subsequence (and hence of the original sequence).






        share|cite|improve this answer














        Any sequence ${x_n}$ with $|x_n| to infty$ will do.



        By Bolzano-Weierstrass Theorem others wont: if $|x_n| not to infty$, there is $M$, such that you can find a subsequence of ${x_n}$ that is bounded by $M$. Then you can find a converging subsequence of that subsequence (and hence of the original sequence).







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 49 mins ago

























        answered 5 hours ago









        ploosu2

        4,5001023




        4,5001023






















            up vote
            6
            down vote













            The example in the OP is wrong. This can be shown by Bolzano-Weierstrass theorem which states that every bounded sequence contains a convergent subsequence.



            Now if we consider:
            $$
            x_n = sin(n)sin(sqrt{3}n)
            $$
            its clear that the sequence is bounded by $[-1, 1]$. Hence it contains a convergent subsequence.






            share|cite|improve this answer























            • Upvoted. Good job.
              – астон вілла олоф мэллбэрг
              5 hours ago















            up vote
            6
            down vote













            The example in the OP is wrong. This can be shown by Bolzano-Weierstrass theorem which states that every bounded sequence contains a convergent subsequence.



            Now if we consider:
            $$
            x_n = sin(n)sin(sqrt{3}n)
            $$
            its clear that the sequence is bounded by $[-1, 1]$. Hence it contains a convergent subsequence.






            share|cite|improve this answer























            • Upvoted. Good job.
              – астон вілла олоф мэллбэрг
              5 hours ago













            up vote
            6
            down vote










            up vote
            6
            down vote









            The example in the OP is wrong. This can be shown by Bolzano-Weierstrass theorem which states that every bounded sequence contains a convergent subsequence.



            Now if we consider:
            $$
            x_n = sin(n)sin(sqrt{3}n)
            $$
            its clear that the sequence is bounded by $[-1, 1]$. Hence it contains a convergent subsequence.






            share|cite|improve this answer














            The example in the OP is wrong. This can be shown by Bolzano-Weierstrass theorem which states that every bounded sequence contains a convergent subsequence.



            Now if we consider:
            $$
            x_n = sin(n)sin(sqrt{3}n)
            $$
            its clear that the sequence is bounded by $[-1, 1]$. Hence it contains a convergent subsequence.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 5 hours ago

























            answered 5 hours ago









            roman

            1,23711017




            1,23711017












            • Upvoted. Good job.
              – астон вілла олоф мэллбэрг
              5 hours ago


















            • Upvoted. Good job.
              – астон вілла олоф мэллбэрг
              5 hours ago
















            Upvoted. Good job.
            – астон вілла олоф мэллбэрг
            5 hours ago




            Upvoted. Good job.
            – астон вілла олоф мэллбэрг
            5 hours ago


















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