If $sum a_n^{3/2}$ is convergent then $sum frac{a_n}n$ is convergent












0












$begingroup$



If $sum a_n^{3/2}$ is convergent then $sum frac{a_n}n$ is convergent.




To prove this result: How to approach ?



Actually, when I first encountered this problem, I searched for an example that would counter this claim.



But I only have the usual example of harmonic series.



Hint only, please.






real-analysis sequences-and-series







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Is $(a_n)$ assumed positive?
    $endgroup$
    – Math_QED
    Jan 4 at 17:12










  • $begingroup$
    Yes a_n >0 .Because other wise by alternating test directly we can argue.
    $endgroup$
    – MathLover
    Jan 4 at 17:13










  • $begingroup$
    Young's inequality $xyle x^p/p+y^q/q$, valid for positive $x,y,p,q$ where $1/p+1/q=1$, is useful here.
    $endgroup$
    – Mike Earnest
    Jan 4 at 17:19
















0












$begingroup$



If $sum a_n^{3/2}$ is convergent then $sum frac{a_n}n$ is convergent.




To prove this result: How to approach ?



Actually, when I first encountered this problem, I searched for an example that would counter this claim.



But I only have the usual example of harmonic series.



Hint only, please.






real-analysis sequences-and-series







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Is $(a_n)$ assumed positive?
    $endgroup$
    – Math_QED
    Jan 4 at 17:12










  • $begingroup$
    Yes a_n >0 .Because other wise by alternating test directly we can argue.
    $endgroup$
    – MathLover
    Jan 4 at 17:13










  • $begingroup$
    Young's inequality $xyle x^p/p+y^q/q$, valid for positive $x,y,p,q$ where $1/p+1/q=1$, is useful here.
    $endgroup$
    – Mike Earnest
    Jan 4 at 17:19














0












0








0


0



$begingroup$



If $sum a_n^{3/2}$ is convergent then $sum frac{a_n}n$ is convergent.




To prove this result: How to approach ?



Actually, when I first encountered this problem, I searched for an example that would counter this claim.



But I only have the usual example of harmonic series.



Hint only, please.






real-analysis sequences-and-series







share|cite|improve this question











$endgroup$





If $sum a_n^{3/2}$ is convergent then $sum frac{a_n}n$ is convergent.




To prove this result: How to approach ?



Actually, when I first encountered this problem, I searched for an example that would counter this claim.



But I only have the usual example of harmonic series.



Hint only, please.






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 4 at 17:36









Did

247k23223460




247k23223460










asked Jan 4 at 17:08









MathLoverMathLover

51410




51410








  • 1




    $begingroup$
    Is $(a_n)$ assumed positive?
    $endgroup$
    – Math_QED
    Jan 4 at 17:12










  • $begingroup$
    Yes a_n >0 .Because other wise by alternating test directly we can argue.
    $endgroup$
    – MathLover
    Jan 4 at 17:13










  • $begingroup$
    Young's inequality $xyle x^p/p+y^q/q$, valid for positive $x,y,p,q$ where $1/p+1/q=1$, is useful here.
    $endgroup$
    – Mike Earnest
    Jan 4 at 17:19














  • 1




    $begingroup$
    Is $(a_n)$ assumed positive?
    $endgroup$
    – Math_QED
    Jan 4 at 17:12










  • $begingroup$
    Yes a_n >0 .Because other wise by alternating test directly we can argue.
    $endgroup$
    – MathLover
    Jan 4 at 17:13










  • $begingroup$
    Young's inequality $xyle x^p/p+y^q/q$, valid for positive $x,y,p,q$ where $1/p+1/q=1$, is useful here.
    $endgroup$
    – Mike Earnest
    Jan 4 at 17:19








1




1




$begingroup$
Is $(a_n)$ assumed positive?
$endgroup$
– Math_QED
Jan 4 at 17:12




$begingroup$
Is $(a_n)$ assumed positive?
$endgroup$
– Math_QED
Jan 4 at 17:12












$begingroup$
Yes a_n >0 .Because other wise by alternating test directly we can argue.
$endgroup$
– MathLover
Jan 4 at 17:13




$begingroup$
Yes a_n >0 .Because other wise by alternating test directly we can argue.
$endgroup$
– MathLover
Jan 4 at 17:13












$begingroup$
Young's inequality $xyle x^p/p+y^q/q$, valid for positive $x,y,p,q$ where $1/p+1/q=1$, is useful here.
$endgroup$
– Mike Earnest
Jan 4 at 17:19




$begingroup$
Young's inequality $xyle x^p/p+y^q/q$, valid for positive $x,y,p,q$ where $1/p+1/q=1$, is useful here.
$endgroup$
– Mike Earnest
Jan 4 at 17:19










4 Answers
4






active

oldest

votes


















2












$begingroup$

Use Holder's inequality to deduce that
$$
sum_{n} frac{a_n}{n}leq left(sum_n a_n^{3/2}right)^{2/3}left(sum_n frac{1}{n^3}right)^{1/3}
$$






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    One does not need to use Holder, Young, or anything else that advanced. The part of the series in the RHS corresponding to those $n$ with $a_n<frac1{n^2}$ converges since $a_n<frac1{n^2}$ yields $frac{a_n}n<frac1{n^3}$, the part corresponding to those $n$ with $a_ngefrac1{n^2}$ converges since in this case $frac{a_n}nle a_n^{3/2}$.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      Your thought of the harmonic series is a good one. If $a_n$ is nice enough, you can say $a_n^{3/2} lt frac 1n$, so $frac {a_n}n lt n^{-5/3}$. Of course, you were not given that $a_n$ is nice, so there is still work to do.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        This seems at most a comment.
        $endgroup$
        – Did
        Jan 4 at 17:36



















      0












      $begingroup$

      Note first that, for $a_1,ldots,a_nge 0$, according to Hölder's inequality
      $$
      frac{a_1}{1}+frac{a_2}{2}+cdots+frac{a_n}{n}le (a_1^p+cdots+a_n^p)^{1/p}
      left(frac{1}{1^q}+cdots+frac{1}{n^q}right)^{1/q}
      $$
      whenever $displaystyle frac{1}{p}+frac{1}{q}=1$. In particular, for $p=3/2$ and $q=3$, we have
      $$
      frac{a_1}{1}+frac{a_2}{2}+cdots+frac{a_n}{n}le (a_1^{3/2}+cdots+a_n^{3/2})^{2/3}
      left(frac{1}{1^3}+cdots+frac{1}{n^3}right)^{1/3}le left(sum_{n=1}^infty a_n^{3/2}right)^{2/3}left(sum_{n=1}^inftyfrac{1}{n^3}right)^{1/3}=M<infty.
      $$
      Hence, $sum_{n=1}^inftyfrac{a_n}{n}$ converges.






      share|cite|improve this answer









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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        2












        $begingroup$

        Use Holder's inequality to deduce that
        $$
        sum_{n} frac{a_n}{n}leq left(sum_n a_n^{3/2}right)^{2/3}left(sum_n frac{1}{n^3}right)^{1/3}
        $$






        share|cite|improve this answer









        $endgroup$


















          2












          $begingroup$

          Use Holder's inequality to deduce that
          $$
          sum_{n} frac{a_n}{n}leq left(sum_n a_n^{3/2}right)^{2/3}left(sum_n frac{1}{n^3}right)^{1/3}
          $$






          share|cite|improve this answer









          $endgroup$
















            2












            2








            2





            $begingroup$

            Use Holder's inequality to deduce that
            $$
            sum_{n} frac{a_n}{n}leq left(sum_n a_n^{3/2}right)^{2/3}left(sum_n frac{1}{n^3}right)^{1/3}
            $$






            share|cite|improve this answer









            $endgroup$



            Use Holder's inequality to deduce that
            $$
            sum_{n} frac{a_n}{n}leq left(sum_n a_n^{3/2}right)^{2/3}left(sum_n frac{1}{n^3}right)^{1/3}
            $$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 4 at 17:27









            Foobaz JohnFoobaz John

            22k41352




            22k41352























                2












                $begingroup$

                One does not need to use Holder, Young, or anything else that advanced. The part of the series in the RHS corresponding to those $n$ with $a_n<frac1{n^2}$ converges since $a_n<frac1{n^2}$ yields $frac{a_n}n<frac1{n^3}$, the part corresponding to those $n$ with $a_ngefrac1{n^2}$ converges since in this case $frac{a_n}nle a_n^{3/2}$.






                share|cite|improve this answer









                $endgroup$


















                  2












                  $begingroup$

                  One does not need to use Holder, Young, or anything else that advanced. The part of the series in the RHS corresponding to those $n$ with $a_n<frac1{n^2}$ converges since $a_n<frac1{n^2}$ yields $frac{a_n}n<frac1{n^3}$, the part corresponding to those $n$ with $a_ngefrac1{n^2}$ converges since in this case $frac{a_n}nle a_n^{3/2}$.






                  share|cite|improve this answer









                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    One does not need to use Holder, Young, or anything else that advanced. The part of the series in the RHS corresponding to those $n$ with $a_n<frac1{n^2}$ converges since $a_n<frac1{n^2}$ yields $frac{a_n}n<frac1{n^3}$, the part corresponding to those $n$ with $a_ngefrac1{n^2}$ converges since in this case $frac{a_n}nle a_n^{3/2}$.






                    share|cite|improve this answer









                    $endgroup$



                    One does not need to use Holder, Young, or anything else that advanced. The part of the series in the RHS corresponding to those $n$ with $a_n<frac1{n^2}$ converges since $a_n<frac1{n^2}$ yields $frac{a_n}n<frac1{n^3}$, the part corresponding to those $n$ with $a_ngefrac1{n^2}$ converges since in this case $frac{a_n}nle a_n^{3/2}$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 4 at 17:42









                    W-t-PW-t-P

                    1,015610




                    1,015610























                        0












                        $begingroup$

                        Your thought of the harmonic series is a good one. If $a_n$ is nice enough, you can say $a_n^{3/2} lt frac 1n$, so $frac {a_n}n lt n^{-5/3}$. Of course, you were not given that $a_n$ is nice, so there is still work to do.






                        share|cite|improve this answer









                        $endgroup$













                        • $begingroup$
                          This seems at most a comment.
                          $endgroup$
                          – Did
                          Jan 4 at 17:36
















                        0












                        $begingroup$

                        Your thought of the harmonic series is a good one. If $a_n$ is nice enough, you can say $a_n^{3/2} lt frac 1n$, so $frac {a_n}n lt n^{-5/3}$. Of course, you were not given that $a_n$ is nice, so there is still work to do.






                        share|cite|improve this answer









                        $endgroup$













                        • $begingroup$
                          This seems at most a comment.
                          $endgroup$
                          – Did
                          Jan 4 at 17:36














                        0












                        0








                        0





                        $begingroup$

                        Your thought of the harmonic series is a good one. If $a_n$ is nice enough, you can say $a_n^{3/2} lt frac 1n$, so $frac {a_n}n lt n^{-5/3}$. Of course, you were not given that $a_n$ is nice, so there is still work to do.






                        share|cite|improve this answer









                        $endgroup$



                        Your thought of the harmonic series is a good one. If $a_n$ is nice enough, you can say $a_n^{3/2} lt frac 1n$, so $frac {a_n}n lt n^{-5/3}$. Of course, you were not given that $a_n$ is nice, so there is still work to do.







                        share|cite|improve this answer












                        share|cite|improve this answer



                        share|cite|improve this answer










                        answered Jan 4 at 17:14









                        Ross MillikanRoss Millikan

                        295k23198371




                        295k23198371












                        • $begingroup$
                          This seems at most a comment.
                          $endgroup$
                          – Did
                          Jan 4 at 17:36


















                        • $begingroup$
                          This seems at most a comment.
                          $endgroup$
                          – Did
                          Jan 4 at 17:36
















                        $begingroup$
                        This seems at most a comment.
                        $endgroup$
                        – Did
                        Jan 4 at 17:36




                        $begingroup$
                        This seems at most a comment.
                        $endgroup$
                        – Did
                        Jan 4 at 17:36











                        0












                        $begingroup$

                        Note first that, for $a_1,ldots,a_nge 0$, according to Hölder's inequality
                        $$
                        frac{a_1}{1}+frac{a_2}{2}+cdots+frac{a_n}{n}le (a_1^p+cdots+a_n^p)^{1/p}
                        left(frac{1}{1^q}+cdots+frac{1}{n^q}right)^{1/q}
                        $$
                        whenever $displaystyle frac{1}{p}+frac{1}{q}=1$. In particular, for $p=3/2$ and $q=3$, we have
                        $$
                        frac{a_1}{1}+frac{a_2}{2}+cdots+frac{a_n}{n}le (a_1^{3/2}+cdots+a_n^{3/2})^{2/3}
                        left(frac{1}{1^3}+cdots+frac{1}{n^3}right)^{1/3}le left(sum_{n=1}^infty a_n^{3/2}right)^{2/3}left(sum_{n=1}^inftyfrac{1}{n^3}right)^{1/3}=M<infty.
                        $$
                        Hence, $sum_{n=1}^inftyfrac{a_n}{n}$ converges.






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          Note first that, for $a_1,ldots,a_nge 0$, according to Hölder's inequality
                          $$
                          frac{a_1}{1}+frac{a_2}{2}+cdots+frac{a_n}{n}le (a_1^p+cdots+a_n^p)^{1/p}
                          left(frac{1}{1^q}+cdots+frac{1}{n^q}right)^{1/q}
                          $$
                          whenever $displaystyle frac{1}{p}+frac{1}{q}=1$. In particular, for $p=3/2$ and $q=3$, we have
                          $$
                          frac{a_1}{1}+frac{a_2}{2}+cdots+frac{a_n}{n}le (a_1^{3/2}+cdots+a_n^{3/2})^{2/3}
                          left(frac{1}{1^3}+cdots+frac{1}{n^3}right)^{1/3}le left(sum_{n=1}^infty a_n^{3/2}right)^{2/3}left(sum_{n=1}^inftyfrac{1}{n^3}right)^{1/3}=M<infty.
                          $$
                          Hence, $sum_{n=1}^inftyfrac{a_n}{n}$ converges.






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            Note first that, for $a_1,ldots,a_nge 0$, according to Hölder's inequality
                            $$
                            frac{a_1}{1}+frac{a_2}{2}+cdots+frac{a_n}{n}le (a_1^p+cdots+a_n^p)^{1/p}
                            left(frac{1}{1^q}+cdots+frac{1}{n^q}right)^{1/q}
                            $$
                            whenever $displaystyle frac{1}{p}+frac{1}{q}=1$. In particular, for $p=3/2$ and $q=3$, we have
                            $$
                            frac{a_1}{1}+frac{a_2}{2}+cdots+frac{a_n}{n}le (a_1^{3/2}+cdots+a_n^{3/2})^{2/3}
                            left(frac{1}{1^3}+cdots+frac{1}{n^3}right)^{1/3}le left(sum_{n=1}^infty a_n^{3/2}right)^{2/3}left(sum_{n=1}^inftyfrac{1}{n^3}right)^{1/3}=M<infty.
                            $$
                            Hence, $sum_{n=1}^inftyfrac{a_n}{n}$ converges.






                            share|cite|improve this answer









                            $endgroup$



                            Note first that, for $a_1,ldots,a_nge 0$, according to Hölder's inequality
                            $$
                            frac{a_1}{1}+frac{a_2}{2}+cdots+frac{a_n}{n}le (a_1^p+cdots+a_n^p)^{1/p}
                            left(frac{1}{1^q}+cdots+frac{1}{n^q}right)^{1/q}
                            $$
                            whenever $displaystyle frac{1}{p}+frac{1}{q}=1$. In particular, for $p=3/2$ and $q=3$, we have
                            $$
                            frac{a_1}{1}+frac{a_2}{2}+cdots+frac{a_n}{n}le (a_1^{3/2}+cdots+a_n^{3/2})^{2/3}
                            left(frac{1}{1^3}+cdots+frac{1}{n^3}right)^{1/3}le left(sum_{n=1}^infty a_n^{3/2}right)^{2/3}left(sum_{n=1}^inftyfrac{1}{n^3}right)^{1/3}=M<infty.
                            $$
                            Hence, $sum_{n=1}^inftyfrac{a_n}{n}$ converges.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 4 at 17:27









                            Yiorgos S. SmyrlisYiorgos S. Smyrlis

                            63.3k1384163




                            63.3k1384163






























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