Existence of $a_k$ such that $sum_k a_kb_k<infty$ and $sum_k a_k=infty$ given $b_kto 0$












3












$begingroup$


I was working with a problem from functional analysis. I reduced the problem to the following problem:




Let $b_k>0$ be a decreasing sequence converging to $0$. Does there exist a non-negative sequence $a_k$ such that
$$sum_{kgeq 1} b_ka_k<infty$$
while at the same time
$$sum_{kgeq 1} a_k = infty$$




If I know what $b_k$ looks like, then I guess it is not hard to find out how to choose $a_k$. However we don't know what $b_k$ looks like and neither in which rate it decays. I really don't know what to do except that I have tried something like choosing for $n>m$
begin{align}
sum_{k=m}^n a_k = frac 1 {sqrt {b_mb_n^varepsilon}}
end{align}
for some $varepsilon>0$, so that the sequence is not Cauchy, that leads to the divergence of $sum_k a_k$ which is nice. At the same time I get
begin{align}
sum_{k=m}^n b_ka_k leq b_m sum_{k=m}^na_k =sqrt{frac{b_m}{b_n^varepsilon}}
end{align}
This tells us that, if there is $varepsilon$ such that
$$sqrt{frac{b_m}{b_n^varepsilon}}to 0 text{ as } n,mtoinfty$$
we are done. However, I don't think that will work, since $varepsilon$ is fixed...




Question. How can the problem be solved?







real-analysis sequences-and-series cauchy-sequences







share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    I was working with a problem from functional analysis. I reduced the problem to the following problem:




    Let $b_k>0$ be a decreasing sequence converging to $0$. Does there exist a non-negative sequence $a_k$ such that
    $$sum_{kgeq 1} b_ka_k<infty$$
    while at the same time
    $$sum_{kgeq 1} a_k = infty$$




    If I know what $b_k$ looks like, then I guess it is not hard to find out how to choose $a_k$. However we don't know what $b_k$ looks like and neither in which rate it decays. I really don't know what to do except that I have tried something like choosing for $n>m$
    begin{align}
    sum_{k=m}^n a_k = frac 1 {sqrt {b_mb_n^varepsilon}}
    end{align}
    for some $varepsilon>0$, so that the sequence is not Cauchy, that leads to the divergence of $sum_k a_k$ which is nice. At the same time I get
    begin{align}
    sum_{k=m}^n b_ka_k leq b_m sum_{k=m}^na_k =sqrt{frac{b_m}{b_n^varepsilon}}
    end{align}
    This tells us that, if there is $varepsilon$ such that
    $$sqrt{frac{b_m}{b_n^varepsilon}}to 0 text{ as } n,mtoinfty$$
    we are done. However, I don't think that will work, since $varepsilon$ is fixed...




    Question. How can the problem be solved?







    real-analysis sequences-and-series cauchy-sequences







    share|cite|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$


      I was working with a problem from functional analysis. I reduced the problem to the following problem:




      Let $b_k>0$ be a decreasing sequence converging to $0$. Does there exist a non-negative sequence $a_k$ such that
      $$sum_{kgeq 1} b_ka_k<infty$$
      while at the same time
      $$sum_{kgeq 1} a_k = infty$$




      If I know what $b_k$ looks like, then I guess it is not hard to find out how to choose $a_k$. However we don't know what $b_k$ looks like and neither in which rate it decays. I really don't know what to do except that I have tried something like choosing for $n>m$
      begin{align}
      sum_{k=m}^n a_k = frac 1 {sqrt {b_mb_n^varepsilon}}
      end{align}
      for some $varepsilon>0$, so that the sequence is not Cauchy, that leads to the divergence of $sum_k a_k$ which is nice. At the same time I get
      begin{align}
      sum_{k=m}^n b_ka_k leq b_m sum_{k=m}^na_k =sqrt{frac{b_m}{b_n^varepsilon}}
      end{align}
      This tells us that, if there is $varepsilon$ such that
      $$sqrt{frac{b_m}{b_n^varepsilon}}to 0 text{ as } n,mtoinfty$$
      we are done. However, I don't think that will work, since $varepsilon$ is fixed...




      Question. How can the problem be solved?







      real-analysis sequences-and-series cauchy-sequences







      share|cite|improve this question











      $endgroup$




      I was working with a problem from functional analysis. I reduced the problem to the following problem:




      Let $b_k>0$ be a decreasing sequence converging to $0$. Does there exist a non-negative sequence $a_k$ such that
      $$sum_{kgeq 1} b_ka_k<infty$$
      while at the same time
      $$sum_{kgeq 1} a_k = infty$$




      If I know what $b_k$ looks like, then I guess it is not hard to find out how to choose $a_k$. However we don't know what $b_k$ looks like and neither in which rate it decays. I really don't know what to do except that I have tried something like choosing for $n>m$
      begin{align}
      sum_{k=m}^n a_k = frac 1 {sqrt {b_mb_n^varepsilon}}
      end{align}
      for some $varepsilon>0$, so that the sequence is not Cauchy, that leads to the divergence of $sum_k a_k$ which is nice. At the same time I get
      begin{align}
      sum_{k=m}^n b_ka_k leq b_m sum_{k=m}^na_k =sqrt{frac{b_m}{b_n^varepsilon}}
      end{align}
      This tells us that, if there is $varepsilon$ such that
      $$sqrt{frac{b_m}{b_n^varepsilon}}to 0 text{ as } n,mtoinfty$$
      we are done. However, I don't think that will work, since $varepsilon$ is fixed...




      Question. How can the problem be solved?







      real-analysis sequences-and-series cauchy-sequences




      real-analysis sequences-and-series cauchy-sequences






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 10 at 19:50









      David Mitra

      63.4k6102164




      63.4k6102164










      asked Jan 10 at 19:22









      ShashiShashi

      7,2651628




      7,2651628






















          2 Answers
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          9












          $begingroup$

          Since $b_kto 0$, there is a strictly increasing subsequence $(k_n)_n$ such that
          $$|b_{k_n}|<frac{1}{2^{n}}.$$
          Now, for any $kin mathbb{N}$, let
          $$a_k=begin{cases}1 &text{if $k=k_n$}\ 0 &text{otherwise}end{cases}.$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            You didn't even use the fact that $b_n$ is decreasing.
            $endgroup$
            – N. S.
            Jan 10 at 20:39



















          0












          $begingroup$

          Define $$a_k=begin{cases}1&,quad b_k<{1over 2^k},notexists u<k,b_{u}<{1over 2^{k}}\0&,quad text{elsewhere}end{cases}$$for example if $b_n={1over n}$ then we have $$a_n=begin{cases}1&,quad n=1,2,4,8,16,cdots \0&,quad text{elsewhere}end{cases}$$






          share|cite|improve this answer











          $endgroup$













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            2 Answers
            2






            active

            oldest

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            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

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            active

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            9












            $begingroup$

            Since $b_kto 0$, there is a strictly increasing subsequence $(k_n)_n$ such that
            $$|b_{k_n}|<frac{1}{2^{n}}.$$
            Now, for any $kin mathbb{N}$, let
            $$a_k=begin{cases}1 &text{if $k=k_n$}\ 0 &text{otherwise}end{cases}.$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              You didn't even use the fact that $b_n$ is decreasing.
              $endgroup$
              – N. S.
              Jan 10 at 20:39
















            9












            $begingroup$

            Since $b_kto 0$, there is a strictly increasing subsequence $(k_n)_n$ such that
            $$|b_{k_n}|<frac{1}{2^{n}}.$$
            Now, for any $kin mathbb{N}$, let
            $$a_k=begin{cases}1 &text{if $k=k_n$}\ 0 &text{otherwise}end{cases}.$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              You didn't even use the fact that $b_n$ is decreasing.
              $endgroup$
              – N. S.
              Jan 10 at 20:39














            9












            9








            9





            $begingroup$

            Since $b_kto 0$, there is a strictly increasing subsequence $(k_n)_n$ such that
            $$|b_{k_n}|<frac{1}{2^{n}}.$$
            Now, for any $kin mathbb{N}$, let
            $$a_k=begin{cases}1 &text{if $k=k_n$}\ 0 &text{otherwise}end{cases}.$$






            share|cite|improve this answer











            $endgroup$



            Since $b_kto 0$, there is a strictly increasing subsequence $(k_n)_n$ such that
            $$|b_{k_n}|<frac{1}{2^{n}}.$$
            Now, for any $kin mathbb{N}$, let
            $$a_k=begin{cases}1 &text{if $k=k_n$}\ 0 &text{otherwise}end{cases}.$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 10 at 19:35

























            answered Jan 10 at 19:30









            Robert ZRobert Z

            99.4k1068140




            99.4k1068140












            • $begingroup$
              You didn't even use the fact that $b_n$ is decreasing.
              $endgroup$
              – N. S.
              Jan 10 at 20:39


















            • $begingroup$
              You didn't even use the fact that $b_n$ is decreasing.
              $endgroup$
              – N. S.
              Jan 10 at 20:39
















            $begingroup$
            You didn't even use the fact that $b_n$ is decreasing.
            $endgroup$
            – N. S.
            Jan 10 at 20:39




            $begingroup$
            You didn't even use the fact that $b_n$ is decreasing.
            $endgroup$
            – N. S.
            Jan 10 at 20:39











            0












            $begingroup$

            Define $$a_k=begin{cases}1&,quad b_k<{1over 2^k},notexists u<k,b_{u}<{1over 2^{k}}\0&,quad text{elsewhere}end{cases}$$for example if $b_n={1over n}$ then we have $$a_n=begin{cases}1&,quad n=1,2,4,8,16,cdots \0&,quad text{elsewhere}end{cases}$$






            share|cite|improve this answer











            $endgroup$


















              0












              $begingroup$

              Define $$a_k=begin{cases}1&,quad b_k<{1over 2^k},notexists u<k,b_{u}<{1over 2^{k}}\0&,quad text{elsewhere}end{cases}$$for example if $b_n={1over n}$ then we have $$a_n=begin{cases}1&,quad n=1,2,4,8,16,cdots \0&,quad text{elsewhere}end{cases}$$






              share|cite|improve this answer











              $endgroup$
















                0












                0








                0





                $begingroup$

                Define $$a_k=begin{cases}1&,quad b_k<{1over 2^k},notexists u<k,b_{u}<{1over 2^{k}}\0&,quad text{elsewhere}end{cases}$$for example if $b_n={1over n}$ then we have $$a_n=begin{cases}1&,quad n=1,2,4,8,16,cdots \0&,quad text{elsewhere}end{cases}$$






                share|cite|improve this answer











                $endgroup$



                Define $$a_k=begin{cases}1&,quad b_k<{1over 2^k},notexists u<k,b_{u}<{1over 2^{k}}\0&,quad text{elsewhere}end{cases}$$for example if $b_n={1over n}$ then we have $$a_n=begin{cases}1&,quad n=1,2,4,8,16,cdots \0&,quad text{elsewhere}end{cases}$$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 10 at 19:48

























                answered Jan 10 at 19:41









                Mostafa AyazMostafa Ayaz

                15.7k3939




                15.7k3939






























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