Question on Convergent Series












-2












$begingroup$


This is the last question on my real analysis final exam. I didn't get to spend much time on it. Hopefully I remember the problem correctly.



${a_n}_{ngeq1}$ is a sequence of real numbers such that $sum_{ngeq1}a_n$ converges. Let $b_n$ be the cardinality of the set ${kin N:a_k>frac{1}{2^n}}$. Show that $limsup_{nrightarrowinfty}frac{1}{2^n}b_n=0$.



My thought is since $sum frac{1}{2^n}$ converges and $sum a_n$ converges, their terms need to be "comparable", but I don't know how to capture this.






real-analysis sequences-and-series







share|cite|improve this question











$endgroup$

















    -2












    $begingroup$


    This is the last question on my real analysis final exam. I didn't get to spend much time on it. Hopefully I remember the problem correctly.



    ${a_n}_{ngeq1}$ is a sequence of real numbers such that $sum_{ngeq1}a_n$ converges. Let $b_n$ be the cardinality of the set ${kin N:a_k>frac{1}{2^n}}$. Show that $limsup_{nrightarrowinfty}frac{1}{2^n}b_n=0$.



    My thought is since $sum frac{1}{2^n}$ converges and $sum a_n$ converges, their terms need to be "comparable", but I don't know how to capture this.






    real-analysis sequences-and-series







    share|cite|improve this question











    $endgroup$















      -2












      -2








      -2


      1



      $begingroup$


      This is the last question on my real analysis final exam. I didn't get to spend much time on it. Hopefully I remember the problem correctly.



      ${a_n}_{ngeq1}$ is a sequence of real numbers such that $sum_{ngeq1}a_n$ converges. Let $b_n$ be the cardinality of the set ${kin N:a_k>frac{1}{2^n}}$. Show that $limsup_{nrightarrowinfty}frac{1}{2^n}b_n=0$.



      My thought is since $sum frac{1}{2^n}$ converges and $sum a_n$ converges, their terms need to be "comparable", but I don't know how to capture this.






      real-analysis sequences-and-series







      share|cite|improve this question











      $endgroup$




      This is the last question on my real analysis final exam. I didn't get to spend much time on it. Hopefully I remember the problem correctly.



      ${a_n}_{ngeq1}$ is a sequence of real numbers such that $sum_{ngeq1}a_n$ converges. Let $b_n$ be the cardinality of the set ${kin N:a_k>frac{1}{2^n}}$. Show that $limsup_{nrightarrowinfty}frac{1}{2^n}b_n=0$.



      My thought is since $sum frac{1}{2^n}$ converges and $sum a_n$ converges, their terms need to be "comparable", but I don't know how to capture this.






      real-analysis sequences-and-series




      real-analysis sequences-and-series






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 19 at 3:58









      Martin Sleziak

      44.8k10119272




      44.8k10119272










      asked Dec 19 '18 at 11:11









      Fluffy SkyeFluffy Skye

      1749




      1749






















          1 Answer
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          $begingroup$

          Fix some positive sequence $(c_n)$ with limit $0$, and, for every $n$ and $m$, let $$A_m=sumlimits_{k=m}^infty a_kqquad B_n={kmid a_kgeqslant c_n}$$ Then, $$A_mgeqslantsum_{k=m}^inftymathbf 1_{kin B_n}a_kgeqslantsum_{k=m}^inftymathbf 1_{kin B_n}c_n=c_ncdot#B_ncap[m,infty)$$ hence $$c_ncdot#B_nleqslant c_ncdot m+c_ncdot#B_ncap[m,infty)leqslant c_ncdot m+A_m$$ In the limit $ntoinfty$, one gets, for every fixed $m$, $$limsup_{ntoinfty},c_ncdot#B_nleqslantlimsup_{ntoinfty},(c_ncdot m+A_m)=A_m$$ The series $sum a_k$ converges hence $inflimits_mA_m=0$, thus, $$limsup_{ntoinfty},c_ncdot#B_n=0$$ If $c_n=2^{-n}$ for every $n$, then $#B_n=b_n$ hence we are done.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            In the first chain of inequality, should it be $c_ncdot#B_ncap[m,infty)$?
            $endgroup$
            – Fluffy Skye
            Dec 19 '18 at 12:40












          • $begingroup$
            True! Well spotted, thanks.
            $endgroup$
            – Did
            Dec 19 '18 at 12:45










          • $begingroup$
            Emmm, I'm still a bit confused about the last part. You first take n to infinity and then take m to infinity? But then why $#B_m = b_n$ ?
            $endgroup$
            – Fluffy Skye
            Dec 19 '18 at 13:01












          • $begingroup$
            ?? $b_n=#B_n$ by definition.
            $endgroup$
            – Did
            Dec 19 '18 at 13:03










          • $begingroup$
            Yeah I know that. But the subscript is different? In the end you have c_n * #B_m not #B_n?
            $endgroup$
            – Fluffy Skye
            Dec 19 '18 at 13:10













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          $begingroup$

          Fix some positive sequence $(c_n)$ with limit $0$, and, for every $n$ and $m$, let $$A_m=sumlimits_{k=m}^infty a_kqquad B_n={kmid a_kgeqslant c_n}$$ Then, $$A_mgeqslantsum_{k=m}^inftymathbf 1_{kin B_n}a_kgeqslantsum_{k=m}^inftymathbf 1_{kin B_n}c_n=c_ncdot#B_ncap[m,infty)$$ hence $$c_ncdot#B_nleqslant c_ncdot m+c_ncdot#B_ncap[m,infty)leqslant c_ncdot m+A_m$$ In the limit $ntoinfty$, one gets, for every fixed $m$, $$limsup_{ntoinfty},c_ncdot#B_nleqslantlimsup_{ntoinfty},(c_ncdot m+A_m)=A_m$$ The series $sum a_k$ converges hence $inflimits_mA_m=0$, thus, $$limsup_{ntoinfty},c_ncdot#B_n=0$$ If $c_n=2^{-n}$ for every $n$, then $#B_n=b_n$ hence we are done.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            In the first chain of inequality, should it be $c_ncdot#B_ncap[m,infty)$?
            $endgroup$
            – Fluffy Skye
            Dec 19 '18 at 12:40












          • $begingroup$
            True! Well spotted, thanks.
            $endgroup$
            – Did
            Dec 19 '18 at 12:45










          • $begingroup$
            Emmm, I'm still a bit confused about the last part. You first take n to infinity and then take m to infinity? But then why $#B_m = b_n$ ?
            $endgroup$
            – Fluffy Skye
            Dec 19 '18 at 13:01












          • $begingroup$
            ?? $b_n=#B_n$ by definition.
            $endgroup$
            – Did
            Dec 19 '18 at 13:03










          • $begingroup$
            Yeah I know that. But the subscript is different? In the end you have c_n * #B_m not #B_n?
            $endgroup$
            – Fluffy Skye
            Dec 19 '18 at 13:10


















          0












          $begingroup$

          Fix some positive sequence $(c_n)$ with limit $0$, and, for every $n$ and $m$, let $$A_m=sumlimits_{k=m}^infty a_kqquad B_n={kmid a_kgeqslant c_n}$$ Then, $$A_mgeqslantsum_{k=m}^inftymathbf 1_{kin B_n}a_kgeqslantsum_{k=m}^inftymathbf 1_{kin B_n}c_n=c_ncdot#B_ncap[m,infty)$$ hence $$c_ncdot#B_nleqslant c_ncdot m+c_ncdot#B_ncap[m,infty)leqslant c_ncdot m+A_m$$ In the limit $ntoinfty$, one gets, for every fixed $m$, $$limsup_{ntoinfty},c_ncdot#B_nleqslantlimsup_{ntoinfty},(c_ncdot m+A_m)=A_m$$ The series $sum a_k$ converges hence $inflimits_mA_m=0$, thus, $$limsup_{ntoinfty},c_ncdot#B_n=0$$ If $c_n=2^{-n}$ for every $n$, then $#B_n=b_n$ hence we are done.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            In the first chain of inequality, should it be $c_ncdot#B_ncap[m,infty)$?
            $endgroup$
            – Fluffy Skye
            Dec 19 '18 at 12:40












          • $begingroup$
            True! Well spotted, thanks.
            $endgroup$
            – Did
            Dec 19 '18 at 12:45










          • $begingroup$
            Emmm, I'm still a bit confused about the last part. You first take n to infinity and then take m to infinity? But then why $#B_m = b_n$ ?
            $endgroup$
            – Fluffy Skye
            Dec 19 '18 at 13:01












          • $begingroup$
            ?? $b_n=#B_n$ by definition.
            $endgroup$
            – Did
            Dec 19 '18 at 13:03










          • $begingroup$
            Yeah I know that. But the subscript is different? In the end you have c_n * #B_m not #B_n?
            $endgroup$
            – Fluffy Skye
            Dec 19 '18 at 13:10
















          0












          0








          0





          $begingroup$

          Fix some positive sequence $(c_n)$ with limit $0$, and, for every $n$ and $m$, let $$A_m=sumlimits_{k=m}^infty a_kqquad B_n={kmid a_kgeqslant c_n}$$ Then, $$A_mgeqslantsum_{k=m}^inftymathbf 1_{kin B_n}a_kgeqslantsum_{k=m}^inftymathbf 1_{kin B_n}c_n=c_ncdot#B_ncap[m,infty)$$ hence $$c_ncdot#B_nleqslant c_ncdot m+c_ncdot#B_ncap[m,infty)leqslant c_ncdot m+A_m$$ In the limit $ntoinfty$, one gets, for every fixed $m$, $$limsup_{ntoinfty},c_ncdot#B_nleqslantlimsup_{ntoinfty},(c_ncdot m+A_m)=A_m$$ The series $sum a_k$ converges hence $inflimits_mA_m=0$, thus, $$limsup_{ntoinfty},c_ncdot#B_n=0$$ If $c_n=2^{-n}$ for every $n$, then $#B_n=b_n$ hence we are done.






          share|cite|improve this answer











          $endgroup$



          Fix some positive sequence $(c_n)$ with limit $0$, and, for every $n$ and $m$, let $$A_m=sumlimits_{k=m}^infty a_kqquad B_n={kmid a_kgeqslant c_n}$$ Then, $$A_mgeqslantsum_{k=m}^inftymathbf 1_{kin B_n}a_kgeqslantsum_{k=m}^inftymathbf 1_{kin B_n}c_n=c_ncdot#B_ncap[m,infty)$$ hence $$c_ncdot#B_nleqslant c_ncdot m+c_ncdot#B_ncap[m,infty)leqslant c_ncdot m+A_m$$ In the limit $ntoinfty$, one gets, for every fixed $m$, $$limsup_{ntoinfty},c_ncdot#B_nleqslantlimsup_{ntoinfty},(c_ncdot m+A_m)=A_m$$ The series $sum a_k$ converges hence $inflimits_mA_m=0$, thus, $$limsup_{ntoinfty},c_ncdot#B_n=0$$ If $c_n=2^{-n}$ for every $n$, then $#B_n=b_n$ hence we are done.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 19 '18 at 13:59

























          answered Dec 19 '18 at 12:07









          DidDid

          248k23224463




          248k23224463












          • $begingroup$
            In the first chain of inequality, should it be $c_ncdot#B_ncap[m,infty)$?
            $endgroup$
            – Fluffy Skye
            Dec 19 '18 at 12:40












          • $begingroup$
            True! Well spotted, thanks.
            $endgroup$
            – Did
            Dec 19 '18 at 12:45










          • $begingroup$
            Emmm, I'm still a bit confused about the last part. You first take n to infinity and then take m to infinity? But then why $#B_m = b_n$ ?
            $endgroup$
            – Fluffy Skye
            Dec 19 '18 at 13:01












          • $begingroup$
            ?? $b_n=#B_n$ by definition.
            $endgroup$
            – Did
            Dec 19 '18 at 13:03










          • $begingroup$
            Yeah I know that. But the subscript is different? In the end you have c_n * #B_m not #B_n?
            $endgroup$
            – Fluffy Skye
            Dec 19 '18 at 13:10




















          • $begingroup$
            In the first chain of inequality, should it be $c_ncdot#B_ncap[m,infty)$?
            $endgroup$
            – Fluffy Skye
            Dec 19 '18 at 12:40












          • $begingroup$
            True! Well spotted, thanks.
            $endgroup$
            – Did
            Dec 19 '18 at 12:45










          • $begingroup$
            Emmm, I'm still a bit confused about the last part. You first take n to infinity and then take m to infinity? But then why $#B_m = b_n$ ?
            $endgroup$
            – Fluffy Skye
            Dec 19 '18 at 13:01












          • $begingroup$
            ?? $b_n=#B_n$ by definition.
            $endgroup$
            – Did
            Dec 19 '18 at 13:03










          • $begingroup$
            Yeah I know that. But the subscript is different? In the end you have c_n * #B_m not #B_n?
            $endgroup$
            – Fluffy Skye
            Dec 19 '18 at 13:10


















          $begingroup$
          In the first chain of inequality, should it be $c_ncdot#B_ncap[m,infty)$?
          $endgroup$
          – Fluffy Skye
          Dec 19 '18 at 12:40






          $begingroup$
          In the first chain of inequality, should it be $c_ncdot#B_ncap[m,infty)$?
          $endgroup$
          – Fluffy Skye
          Dec 19 '18 at 12:40














          $begingroup$
          True! Well spotted, thanks.
          $endgroup$
          – Did
          Dec 19 '18 at 12:45




          $begingroup$
          True! Well spotted, thanks.
          $endgroup$
          – Did
          Dec 19 '18 at 12:45












          $begingroup$
          Emmm, I'm still a bit confused about the last part. You first take n to infinity and then take m to infinity? But then why $#B_m = b_n$ ?
          $endgroup$
          – Fluffy Skye
          Dec 19 '18 at 13:01






          $begingroup$
          Emmm, I'm still a bit confused about the last part. You first take n to infinity and then take m to infinity? But then why $#B_m = b_n$ ?
          $endgroup$
          – Fluffy Skye
          Dec 19 '18 at 13:01














          $begingroup$
          ?? $b_n=#B_n$ by definition.
          $endgroup$
          – Did
          Dec 19 '18 at 13:03




          $begingroup$
          ?? $b_n=#B_n$ by definition.
          $endgroup$
          – Did
          Dec 19 '18 at 13:03












          $begingroup$
          Yeah I know that. But the subscript is different? In the end you have c_n * #B_m not #B_n?
          $endgroup$
          – Fluffy Skye
          Dec 19 '18 at 13:10






          $begingroup$
          Yeah I know that. But the subscript is different? In the end you have c_n * #B_m not #B_n?
          $endgroup$
          – Fluffy Skye
          Dec 19 '18 at 13:10




















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