Deducing convergence from bounds on average












2














Assume I have a bound on the summatory function of the form
$$sum_{n leqslant x} |a_n|^2 leqslant x^alpha$$



Can I then deduce something about the convergence of
$$sum frac{a_n}{n^k} ?$$



I was expecting getting something using Cauchy-Schwarz inequality, but it is not enough since $x^alpha$ diverge and the sum of $n^{-2k}$ only converges, and does not compensate this $x^a$. Am I missing something?






analysis







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    2














    Assume I have a bound on the summatory function of the form
    $$sum_{n leqslant x} |a_n|^2 leqslant x^alpha$$



    Can I then deduce something about the convergence of
    $$sum frac{a_n}{n^k} ?$$



    I was expecting getting something using Cauchy-Schwarz inequality, but it is not enough since $x^alpha$ diverge and the sum of $n^{-2k}$ only converges, and does not compensate this $x^a$. Am I missing something?






    analysis







    share|cite|improve this question

























      2












      2








      2







      Assume I have a bound on the summatory function of the form
      $$sum_{n leqslant x} |a_n|^2 leqslant x^alpha$$



      Can I then deduce something about the convergence of
      $$sum frac{a_n}{n^k} ?$$



      I was expecting getting something using Cauchy-Schwarz inequality, but it is not enough since $x^alpha$ diverge and the sum of $n^{-2k}$ only converges, and does not compensate this $x^a$. Am I missing something?






      analysis







      share|cite|improve this question













      Assume I have a bound on the summatory function of the form
      $$sum_{n leqslant x} |a_n|^2 leqslant x^alpha$$



      Can I then deduce something about the convergence of
      $$sum frac{a_n}{n^k} ?$$



      I was expecting getting something using Cauchy-Schwarz inequality, but it is not enough since $x^alpha$ diverge and the sum of $n^{-2k}$ only converges, and does not compensate this $x^a$. Am I missing something?






      analysis




      analysis






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      share|cite|improve this question











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      asked 19 hours ago









      TheStudent

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          By the assumption, we have
          $$
          sum_{n= 2^j}^{2^{j+1}-1 }|a_n|^2 le C_1 2^{alpha j}
          $$           for all $jge 0$.
          Note that
          $$
          sum_{n= 2^j }^{2^{j+1}-1 }frac{|a_n|}{n^k}leq C_22^{-kj}sum_{n= 2^j }^{2^{j+1} -1}|a_n|leq C_32^{-(k-frac{1}{2})j}sqrt{sum_{n= 2^j }^{2^{j+1}-1 }|a_n|^2}le C_42^{-(k-frac{1}{2}-frac{alpha}{2})j}
          $$           for all $jge 0$. Here, $(C_p)_{ 1le ple 4}$ is a set of constants depending only on $alpha$ and $k$. Thus, if $k>frac{alpha+1}{2}$, then the sum
          $$
          sum_{j=0}^inftysum_{n= 2^j }^{2^{j+1}-1 }frac{|a_n|}{n^k}
          $$           converges absolutely. If $k le frac{alpha+1}{2}$, then $a_n = ccdot n^{frac{alpha-1}{2}}$ satisfies $sum_{jleq x} |a_j|^2 le x^alpha$ for sufficiently small $c>0$, but
          $$
          sum_n frac{a_n}{n^{k}} ge csum_n frac{1}{n}=infty.
          $$           Hence, the bound $k>frac{alpha+1}{2}$ cannot be improved.






          share|cite|improve this answer





















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            By the assumption, we have
            $$
            sum_{n= 2^j}^{2^{j+1}-1 }|a_n|^2 le C_1 2^{alpha j}
            $$             for all $jge 0$.
            Note that
            $$
            sum_{n= 2^j }^{2^{j+1}-1 }frac{|a_n|}{n^k}leq C_22^{-kj}sum_{n= 2^j }^{2^{j+1} -1}|a_n|leq C_32^{-(k-frac{1}{2})j}sqrt{sum_{n= 2^j }^{2^{j+1}-1 }|a_n|^2}le C_42^{-(k-frac{1}{2}-frac{alpha}{2})j}
            $$             for all $jge 0$. Here, $(C_p)_{ 1le ple 4}$ is a set of constants depending only on $alpha$ and $k$. Thus, if $k>frac{alpha+1}{2}$, then the sum
            $$
            sum_{j=0}^inftysum_{n= 2^j }^{2^{j+1}-1 }frac{|a_n|}{n^k}
            $$             converges absolutely. If $k le frac{alpha+1}{2}$, then $a_n = ccdot n^{frac{alpha-1}{2}}$ satisfies $sum_{jleq x} |a_j|^2 le x^alpha$ for sufficiently small $c>0$, but
            $$
            sum_n frac{a_n}{n^{k}} ge csum_n frac{1}{n}=infty.
            $$             Hence, the bound $k>frac{alpha+1}{2}$ cannot be improved.






            share|cite|improve this answer


























              1














              By the assumption, we have
              $$
              sum_{n= 2^j}^{2^{j+1}-1 }|a_n|^2 le C_1 2^{alpha j}
              $$               for all $jge 0$.
              Note that
              $$
              sum_{n= 2^j }^{2^{j+1}-1 }frac{|a_n|}{n^k}leq C_22^{-kj}sum_{n= 2^j }^{2^{j+1} -1}|a_n|leq C_32^{-(k-frac{1}{2})j}sqrt{sum_{n= 2^j }^{2^{j+1}-1 }|a_n|^2}le C_42^{-(k-frac{1}{2}-frac{alpha}{2})j}
              $$               for all $jge 0$. Here, $(C_p)_{ 1le ple 4}$ is a set of constants depending only on $alpha$ and $k$. Thus, if $k>frac{alpha+1}{2}$, then the sum
              $$
              sum_{j=0}^inftysum_{n= 2^j }^{2^{j+1}-1 }frac{|a_n|}{n^k}
              $$               converges absolutely. If $k le frac{alpha+1}{2}$, then $a_n = ccdot n^{frac{alpha-1}{2}}$ satisfies $sum_{jleq x} |a_j|^2 le x^alpha$ for sufficiently small $c>0$, but
              $$
              sum_n frac{a_n}{n^{k}} ge csum_n frac{1}{n}=infty.
              $$               Hence, the bound $k>frac{alpha+1}{2}$ cannot be improved.






              share|cite|improve this answer
























                1












                1








                1






                By the assumption, we have
                $$
                sum_{n= 2^j}^{2^{j+1}-1 }|a_n|^2 le C_1 2^{alpha j}
                $$                 for all $jge 0$.
                Note that
                $$
                sum_{n= 2^j }^{2^{j+1}-1 }frac{|a_n|}{n^k}leq C_22^{-kj}sum_{n= 2^j }^{2^{j+1} -1}|a_n|leq C_32^{-(k-frac{1}{2})j}sqrt{sum_{n= 2^j }^{2^{j+1}-1 }|a_n|^2}le C_42^{-(k-frac{1}{2}-frac{alpha}{2})j}
                $$                 for all $jge 0$. Here, $(C_p)_{ 1le ple 4}$ is a set of constants depending only on $alpha$ and $k$. Thus, if $k>frac{alpha+1}{2}$, then the sum
                $$
                sum_{j=0}^inftysum_{n= 2^j }^{2^{j+1}-1 }frac{|a_n|}{n^k}
                $$                 converges absolutely. If $k le frac{alpha+1}{2}$, then $a_n = ccdot n^{frac{alpha-1}{2}}$ satisfies $sum_{jleq x} |a_j|^2 le x^alpha$ for sufficiently small $c>0$, but
                $$
                sum_n frac{a_n}{n^{k}} ge csum_n frac{1}{n}=infty.
                $$                 Hence, the bound $k>frac{alpha+1}{2}$ cannot be improved.






                share|cite|improve this answer












                By the assumption, we have
                $$
                sum_{n= 2^j}^{2^{j+1}-1 }|a_n|^2 le C_1 2^{alpha j}
                $$                 for all $jge 0$.
                Note that
                $$
                sum_{n= 2^j }^{2^{j+1}-1 }frac{|a_n|}{n^k}leq C_22^{-kj}sum_{n= 2^j }^{2^{j+1} -1}|a_n|leq C_32^{-(k-frac{1}{2})j}sqrt{sum_{n= 2^j }^{2^{j+1}-1 }|a_n|^2}le C_42^{-(k-frac{1}{2}-frac{alpha}{2})j}
                $$                 for all $jge 0$. Here, $(C_p)_{ 1le ple 4}$ is a set of constants depending only on $alpha$ and $k$. Thus, if $k>frac{alpha+1}{2}$, then the sum
                $$
                sum_{j=0}^inftysum_{n= 2^j }^{2^{j+1}-1 }frac{|a_n|}{n^k}
                $$                 converges absolutely. If $k le frac{alpha+1}{2}$, then $a_n = ccdot n^{frac{alpha-1}{2}}$ satisfies $sum_{jleq x} |a_j|^2 le x^alpha$ for sufficiently small $c>0$, but
                $$
                sum_n frac{a_n}{n^{k}} ge csum_n frac{1}{n}=infty.
                $$                 Hence, the bound $k>frac{alpha+1}{2}$ cannot be improved.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 18 hours ago









                Song

                5,225318




                5,225318






























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