Show that $limsup_{k to infty} 2^{-k} N_k = 0$ where $N_k$ is the number of $a_n geq 2^{-k}$.












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Let ${a_n}_{n geq 1}$ be a non-negative sequence of reals such that $sum_{n geq 1} a_n$ converges to $s$. Define $N_k = |{n in mathbb{N} : a_n geq 2^{-k}}|$. Show that
begin{equation} limsup_{k to infty} 2^{-k} N_k = 0 end{equation}
The idea I had was to bound the terms $v_N :=sup{2^{-k} N_k : k geq N}$ by $s c_n$ where $c_n$ is some sequence converging to 0, though this approach didn't seem to go anywhere.



Some other facts I observed were that for all $k in mathbb{N}$ there exists $n_k in mathbb{N}$ such that $a_n < 2^{-k}$ for all $n geq n_k$ (just the definition of convergence), and I was thinking there would be a way to split up the partial sums and write in the form of $2^{-k} N_k + text{stuff}$, though I couldn't work it out. Any suggestions would be appreciated.






real-analysis analysis convergence summation limsup-and-liminf







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    1












    $begingroup$


    Let ${a_n}_{n geq 1}$ be a non-negative sequence of reals such that $sum_{n geq 1} a_n$ converges to $s$. Define $N_k = |{n in mathbb{N} : a_n geq 2^{-k}}|$. Show that
    begin{equation} limsup_{k to infty} 2^{-k} N_k = 0 end{equation}
    The idea I had was to bound the terms $v_N :=sup{2^{-k} N_k : k geq N}$ by $s c_n$ where $c_n$ is some sequence converging to 0, though this approach didn't seem to go anywhere.



    Some other facts I observed were that for all $k in mathbb{N}$ there exists $n_k in mathbb{N}$ such that $a_n < 2^{-k}$ for all $n geq n_k$ (just the definition of convergence), and I was thinking there would be a way to split up the partial sums and write in the form of $2^{-k} N_k + text{stuff}$, though I couldn't work it out. Any suggestions would be appreciated.






    real-analysis analysis convergence summation limsup-and-liminf







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Let ${a_n}_{n geq 1}$ be a non-negative sequence of reals such that $sum_{n geq 1} a_n$ converges to $s$. Define $N_k = |{n in mathbb{N} : a_n geq 2^{-k}}|$. Show that
      begin{equation} limsup_{k to infty} 2^{-k} N_k = 0 end{equation}
      The idea I had was to bound the terms $v_N :=sup{2^{-k} N_k : k geq N}$ by $s c_n$ where $c_n$ is some sequence converging to 0, though this approach didn't seem to go anywhere.



      Some other facts I observed were that for all $k in mathbb{N}$ there exists $n_k in mathbb{N}$ such that $a_n < 2^{-k}$ for all $n geq n_k$ (just the definition of convergence), and I was thinking there would be a way to split up the partial sums and write in the form of $2^{-k} N_k + text{stuff}$, though I couldn't work it out. Any suggestions would be appreciated.






      real-analysis analysis convergence summation limsup-and-liminf







      share|cite|improve this question









      $endgroup$




      Let ${a_n}_{n geq 1}$ be a non-negative sequence of reals such that $sum_{n geq 1} a_n$ converges to $s$. Define $N_k = |{n in mathbb{N} : a_n geq 2^{-k}}|$. Show that
      begin{equation} limsup_{k to infty} 2^{-k} N_k = 0 end{equation}
      The idea I had was to bound the terms $v_N :=sup{2^{-k} N_k : k geq N}$ by $s c_n$ where $c_n$ is some sequence converging to 0, though this approach didn't seem to go anywhere.



      Some other facts I observed were that for all $k in mathbb{N}$ there exists $n_k in mathbb{N}$ such that $a_n < 2^{-k}$ for all $n geq n_k$ (just the definition of convergence), and I was thinking there would be a way to split up the partial sums and write in the form of $2^{-k} N_k + text{stuff}$, though I couldn't work it out. Any suggestions would be appreciated.






      real-analysis analysis convergence summation limsup-and-liminf




      real-analysis analysis convergence summation limsup-and-liminf






      share|cite|improve this question













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      asked Dec 10 '18 at 21:46









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          Let $E_k={n: 2^{-(k+1)} leq a_n < 2^{-k}}$. Then $sum_k sum_{n in E_k} a_n leq sum_n a_n$ because $E_k$ 's are disjoint. Since $a_n geq 2^{-(k+1)}$ for $n in E_k$ we get $sum_k sum_{n in E_k} 2^{-(k+1)} leq sum_n a_n$. This gives $sum_k 2^{-(k+1)} (N_{k+1} -N_k) <infty$. Use partial summation to conclude that $sum N_k b_k <infty$ where $b_k =sum_{j=1}^{k} 2^{-(j+1)}$. Conclude the proof by computing the geometric sum $sum_{j=1}^{k} 2^{-(j+1)}$.






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            $begingroup$

            Let $E_k={n: 2^{-(k+1)} leq a_n < 2^{-k}}$. Then $sum_k sum_{n in E_k} a_n leq sum_n a_n$ because $E_k$ 's are disjoint. Since $a_n geq 2^{-(k+1)}$ for $n in E_k$ we get $sum_k sum_{n in E_k} 2^{-(k+1)} leq sum_n a_n$. This gives $sum_k 2^{-(k+1)} (N_{k+1} -N_k) <infty$. Use partial summation to conclude that $sum N_k b_k <infty$ where $b_k =sum_{j=1}^{k} 2^{-(j+1)}$. Conclude the proof by computing the geometric sum $sum_{j=1}^{k} 2^{-(j+1)}$.






            share|cite|improve this answer









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              $begingroup$

              Let $E_k={n: 2^{-(k+1)} leq a_n < 2^{-k}}$. Then $sum_k sum_{n in E_k} a_n leq sum_n a_n$ because $E_k$ 's are disjoint. Since $a_n geq 2^{-(k+1)}$ for $n in E_k$ we get $sum_k sum_{n in E_k} 2^{-(k+1)} leq sum_n a_n$. This gives $sum_k 2^{-(k+1)} (N_{k+1} -N_k) <infty$. Use partial summation to conclude that $sum N_k b_k <infty$ where $b_k =sum_{j=1}^{k} 2^{-(j+1)}$. Conclude the proof by computing the geometric sum $sum_{j=1}^{k} 2^{-(j+1)}$.






              share|cite|improve this answer









              $endgroup$
















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                $begingroup$

                Let $E_k={n: 2^{-(k+1)} leq a_n < 2^{-k}}$. Then $sum_k sum_{n in E_k} a_n leq sum_n a_n$ because $E_k$ 's are disjoint. Since $a_n geq 2^{-(k+1)}$ for $n in E_k$ we get $sum_k sum_{n in E_k} 2^{-(k+1)} leq sum_n a_n$. This gives $sum_k 2^{-(k+1)} (N_{k+1} -N_k) <infty$. Use partial summation to conclude that $sum N_k b_k <infty$ where $b_k =sum_{j=1}^{k} 2^{-(j+1)}$. Conclude the proof by computing the geometric sum $sum_{j=1}^{k} 2^{-(j+1)}$.






                share|cite|improve this answer









                $endgroup$



                Let $E_k={n: 2^{-(k+1)} leq a_n < 2^{-k}}$. Then $sum_k sum_{n in E_k} a_n leq sum_n a_n$ because $E_k$ 's are disjoint. Since $a_n geq 2^{-(k+1)}$ for $n in E_k$ we get $sum_k sum_{n in E_k} 2^{-(k+1)} leq sum_n a_n$. This gives $sum_k 2^{-(k+1)} (N_{k+1} -N_k) <infty$. Use partial summation to conclude that $sum N_k b_k <infty$ where $b_k =sum_{j=1}^{k} 2^{-(j+1)}$. Conclude the proof by computing the geometric sum $sum_{j=1}^{k} 2^{-(j+1)}$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 10 '18 at 23:35









                Kavi Rama MurthyKavi Rama Murthy

                57.6k42160




                57.6k42160






























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