Help with a sequence proof











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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??










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    up vote
    1
    down vote

    favorite












    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??










    share|cite|improve this question


























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      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??










      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






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      edited Nov 20 at 21:53

























      asked Nov 20 at 21:16









      ktuggle

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      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 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 at 4:02











          Your Answer





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          1 Answer
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          active

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          1 Answer
          1






          active

          oldest

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          active

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          active

          oldest

          votes








          up vote
          1
          down vote



          accepted










          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 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 at 4:02















          up vote
          1
          down vote



          accepted










          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 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 at 4:02













          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          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 at 23:37

























          answered Nov 20 at 21:57









          Andrés E. Caicedo

          64.6k8158246




          64.6k8158246












          • 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 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 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 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 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 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 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 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 at 4:02


















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