Help with a sequence proof












1














How can I show that if $a_n$ is a sequence of real numbers, then the following statements are equivalent:





  • $a_n$ has a subsequence, $a_{n_k}$, such that the summation of the subsequence converges,i.e:


$$ sum_{k=0}^infty a_{n_k} $$ converges



and



$$ liminf |a_n| = 0$$



This makes some intuitive sense to me, because if a_n has a convergent subsequence, it must be bounded somehow (above or below or both), and if liminf of the absolute value sequence is zero, then it again means that the sequence must be in some way bounded, because if it didnt it would diverge to infinity.



However I have no idea where to start on this in a formal proof sense. Does anyone have any tips for how to form this connection? I don't really know anything concrete that I can conclude from the second statement (the liminf one) I feel like maybe bolzano-weierstrauss can give me a hand somehow??






real-analysis proof-writing







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    1














    How can I show that if $a_n$ is a sequence of real numbers, then the following statements are equivalent:





    • $a_n$ has a subsequence, $a_{n_k}$, such that the summation of the subsequence converges,i.e:


    $$ sum_{k=0}^infty a_{n_k} $$ converges



    and



    $$ liminf |a_n| = 0$$



    This makes some intuitive sense to me, because if a_n has a convergent subsequence, it must be bounded somehow (above or below or both), and if liminf of the absolute value sequence is zero, then it again means that the sequence must be in some way bounded, because if it didnt it would diverge to infinity.



    However I have no idea where to start on this in a formal proof sense. Does anyone have any tips for how to form this connection? I don't really know anything concrete that I can conclude from the second statement (the liminf one) I feel like maybe bolzano-weierstrauss can give me a hand somehow??






    real-analysis proof-writing







    share|cite|improve this question



























      1












      1








      1







      How can I show that if $a_n$ is a sequence of real numbers, then the following statements are equivalent:





      • $a_n$ has a subsequence, $a_{n_k}$, such that the summation of the subsequence converges,i.e:


      $$ sum_{k=0}^infty a_{n_k} $$ converges



      and



      $$ liminf |a_n| = 0$$



      This makes some intuitive sense to me, because if a_n has a convergent subsequence, it must be bounded somehow (above or below or both), and if liminf of the absolute value sequence is zero, then it again means that the sequence must be in some way bounded, because if it didnt it would diverge to infinity.



      However I have no idea where to start on this in a formal proof sense. Does anyone have any tips for how to form this connection? I don't really know anything concrete that I can conclude from the second statement (the liminf one) I feel like maybe bolzano-weierstrauss can give me a hand somehow??






      real-analysis proof-writing







      share|cite|improve this question















      How can I show that if $a_n$ is a sequence of real numbers, then the following statements are equivalent:





      • $a_n$ has a subsequence, $a_{n_k}$, such that the summation of the subsequence converges,i.e:


      $$ sum_{k=0}^infty a_{n_k} $$ converges



      and



      $$ liminf |a_n| = 0$$



      This makes some intuitive sense to me, because if a_n has a convergent subsequence, it must be bounded somehow (above or below or both), and if liminf of the absolute value sequence is zero, then it again means that the sequence must be in some way bounded, because if it didnt it would diverge to infinity.



      However I have no idea where to start on this in a formal proof sense. Does anyone have any tips for how to form this connection? I don't really know anything concrete that I can conclude from the second statement (the liminf one) I feel like maybe bolzano-weierstrauss can give me a hand somehow??






      real-analysis proof-writing




      real-analysis proof-writing






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      share|cite|improve this question













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      edited Nov 20 '18 at 21:53

























      asked Nov 20 '18 at 21:16









      ktuggle

      184




      184






















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          Check first that $liminf |a_n|=0$ if and only if some subsequence of the $a_n$ converges to $0$.



          Assuming that $liminf|a_n|=0$, use the equivalence above to show that, indeed, there is a subsequence $(a_{n_k})_{kge0}$ that, not only converges to 0, but does it very quickly, say $|a_{n_k}|<1/2^k$ for all $k$. Check that the sum of this subsequence converges.



          For the converse, if there is a subsequence whose series converges, this subsequence converges to 0 (right?). Use the equivalence mentioned in the first paragraph to conclude.






          share|cite|improve this answer























          • How do I go about showing the "very quickly" part. Given that i'm working with a very general sequence, how can i prove its always smaller than $1/2^k$
            – ktuggle
            Nov 21 '18 at 3:32






          • 1




            Use the definition of "the sequence converges to 0" to show that from some point on all terms are in absolute value less than 1, and from some point on they are all in absolute value less than $1/2$, and so on, and use this to extract the required subsequence.
            – Andrés E. Caicedo
            Nov 21 '18 at 4:02











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          1 Answer
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          active

          oldest

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          1














          Check first that $liminf |a_n|=0$ if and only if some subsequence of the $a_n$ converges to $0$.



          Assuming that $liminf|a_n|=0$, use the equivalence above to show that, indeed, there is a subsequence $(a_{n_k})_{kge0}$ that, not only converges to 0, but does it very quickly, say $|a_{n_k}|<1/2^k$ for all $k$. Check that the sum of this subsequence converges.



          For the converse, if there is a subsequence whose series converges, this subsequence converges to 0 (right?). Use the equivalence mentioned in the first paragraph to conclude.






          share|cite|improve this answer























          • How do I go about showing the "very quickly" part. Given that i'm working with a very general sequence, how can i prove its always smaller than $1/2^k$
            – ktuggle
            Nov 21 '18 at 3:32






          • 1




            Use the definition of "the sequence converges to 0" to show that from some point on all terms are in absolute value less than 1, and from some point on they are all in absolute value less than $1/2$, and so on, and use this to extract the required subsequence.
            – Andrés E. Caicedo
            Nov 21 '18 at 4:02
















          1














          Check first that $liminf |a_n|=0$ if and only if some subsequence of the $a_n$ converges to $0$.



          Assuming that $liminf|a_n|=0$, use the equivalence above to show that, indeed, there is a subsequence $(a_{n_k})_{kge0}$ that, not only converges to 0, but does it very quickly, say $|a_{n_k}|<1/2^k$ for all $k$. Check that the sum of this subsequence converges.



          For the converse, if there is a subsequence whose series converges, this subsequence converges to 0 (right?). Use the equivalence mentioned in the first paragraph to conclude.






          share|cite|improve this answer























          • How do I go about showing the "very quickly" part. Given that i'm working with a very general sequence, how can i prove its always smaller than $1/2^k$
            – ktuggle
            Nov 21 '18 at 3:32






          • 1




            Use the definition of "the sequence converges to 0" to show that from some point on all terms are in absolute value less than 1, and from some point on they are all in absolute value less than $1/2$, and so on, and use this to extract the required subsequence.
            – Andrés E. Caicedo
            Nov 21 '18 at 4:02














          1












          1








          1






          Check first that $liminf |a_n|=0$ if and only if some subsequence of the $a_n$ converges to $0$.



          Assuming that $liminf|a_n|=0$, use the equivalence above to show that, indeed, there is a subsequence $(a_{n_k})_{kge0}$ that, not only converges to 0, but does it very quickly, say $|a_{n_k}|<1/2^k$ for all $k$. Check that the sum of this subsequence converges.



          For the converse, if there is a subsequence whose series converges, this subsequence converges to 0 (right?). Use the equivalence mentioned in the first paragraph to conclude.






          share|cite|improve this answer














          Check first that $liminf |a_n|=0$ if and only if some subsequence of the $a_n$ converges to $0$.



          Assuming that $liminf|a_n|=0$, use the equivalence above to show that, indeed, there is a subsequence $(a_{n_k})_{kge0}$ that, not only converges to 0, but does it very quickly, say $|a_{n_k}|<1/2^k$ for all $k$. Check that the sum of this subsequence converges.



          For the converse, if there is a subsequence whose series converges, this subsequence converges to 0 (right?). Use the equivalence mentioned in the first paragraph to conclude.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 20 '18 at 23:37

























          answered Nov 20 '18 at 21:57









          Andrés E. Caicedo

          64.8k8158246




          64.8k8158246












          • How do I go about showing the "very quickly" part. Given that i'm working with a very general sequence, how can i prove its always smaller than $1/2^k$
            – ktuggle
            Nov 21 '18 at 3:32






          • 1




            Use the definition of "the sequence converges to 0" to show that from some point on all terms are in absolute value less than 1, and from some point on they are all in absolute value less than $1/2$, and so on, and use this to extract the required subsequence.
            – Andrés E. Caicedo
            Nov 21 '18 at 4:02


















          • How do I go about showing the "very quickly" part. Given that i'm working with a very general sequence, how can i prove its always smaller than $1/2^k$
            – ktuggle
            Nov 21 '18 at 3:32






          • 1




            Use the definition of "the sequence converges to 0" to show that from some point on all terms are in absolute value less than 1, and from some point on they are all in absolute value less than $1/2$, and so on, and use this to extract the required subsequence.
            – Andrés E. Caicedo
            Nov 21 '18 at 4:02
















          How do I go about showing the "very quickly" part. Given that i'm working with a very general sequence, how can i prove its always smaller than $1/2^k$
          – ktuggle
          Nov 21 '18 at 3:32




          How do I go about showing the "very quickly" part. Given that i'm working with a very general sequence, how can i prove its always smaller than $1/2^k$
          – ktuggle
          Nov 21 '18 at 3:32




          1




          1




          Use the definition of "the sequence converges to 0" to show that from some point on all terms are in absolute value less than 1, and from some point on they are all in absolute value less than $1/2$, and so on, and use this to extract the required subsequence.
          – Andrés E. Caicedo
          Nov 21 '18 at 4:02




          Use the definition of "the sequence converges to 0" to show that from some point on all terms are in absolute value less than 1, and from some point on they are all in absolute value less than $1/2$, and so on, and use this to extract the required subsequence.
          – Andrés E. Caicedo
          Nov 21 '18 at 4:02


















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