Find a sequence ${a_n}$ such that both $sum {a_n}$ and $sum{frac{1}{n^2 a_n}}$ converge.












2












$begingroup$



Find a sequence ${a_n}$ such that both $sum_1^infty {a_n}$ and
$sum_1^infty {frac{1}{n^2 a_n}}$ converge. If no such sequence
exists, prove that.




Actually the question was for $sum_1^infty {n{a_n}}$ and $sum_1^infty {frac{1}{n^2 a_n}}$. For this problem; If we suppose that both are convergent, then their multiply $sum_1^infty 1/n$ must be convergent and this is a contradiction.



But for $sum_1^infty {a_n}$ and $sum_1^infty {frac{1}{n^2 a_n}}$, if we multiply we get $sum_1^infty 1/n^2$ which is convergent. I tried to find a sequence such that both the series converge but I couldn't find any but also I couldn't prove that there is no such sequence exists.



Edit. $a_n >0$






sequences-and-series convergence







share|cite|improve this question











$endgroup$












  • $begingroup$
    I think that $a_n>0$ should be a condition to make the question really interesting
    $endgroup$
    – ajotatxe
    Jan 7 at 21:08










  • $begingroup$
    @ajotatxe yes. The book I found the first question, said at the first of the series chapter, that for all questions, consider $a_n>0$ and I forgot to mention that.
    $endgroup$
    – amir na
    Jan 7 at 21:23












  • $begingroup$
    With the condition $a_n > 0$ it is a duplicate of math.stackexchange.com/q/1933001/42969.
    $endgroup$
    – Martin R
    Jan 7 at 21:31










  • $begingroup$
    If $a_n>0$, by Cauchy-Schwarz $$sum_{n=1}^{N}a_nsum_{n=1}^{N}frac{1}{n^2 a_n}geq H_N^2geq log^2(N).$$
    $endgroup$
    – Jack D'Aurizio
    Jan 7 at 21:34


















2












$begingroup$



Find a sequence ${a_n}$ such that both $sum_1^infty {a_n}$ and
$sum_1^infty {frac{1}{n^2 a_n}}$ converge. If no such sequence
exists, prove that.




Actually the question was for $sum_1^infty {n{a_n}}$ and $sum_1^infty {frac{1}{n^2 a_n}}$. For this problem; If we suppose that both are convergent, then their multiply $sum_1^infty 1/n$ must be convergent and this is a contradiction.



But for $sum_1^infty {a_n}$ and $sum_1^infty {frac{1}{n^2 a_n}}$, if we multiply we get $sum_1^infty 1/n^2$ which is convergent. I tried to find a sequence such that both the series converge but I couldn't find any but also I couldn't prove that there is no such sequence exists.



Edit. $a_n >0$






sequences-and-series convergence







share|cite|improve this question











$endgroup$












  • $begingroup$
    I think that $a_n>0$ should be a condition to make the question really interesting
    $endgroup$
    – ajotatxe
    Jan 7 at 21:08










  • $begingroup$
    @ajotatxe yes. The book I found the first question, said at the first of the series chapter, that for all questions, consider $a_n>0$ and I forgot to mention that.
    $endgroup$
    – amir na
    Jan 7 at 21:23












  • $begingroup$
    With the condition $a_n > 0$ it is a duplicate of math.stackexchange.com/q/1933001/42969.
    $endgroup$
    – Martin R
    Jan 7 at 21:31










  • $begingroup$
    If $a_n>0$, by Cauchy-Schwarz $$sum_{n=1}^{N}a_nsum_{n=1}^{N}frac{1}{n^2 a_n}geq H_N^2geq log^2(N).$$
    $endgroup$
    – Jack D'Aurizio
    Jan 7 at 21:34
















2












2








2





$begingroup$



Find a sequence ${a_n}$ such that both $sum_1^infty {a_n}$ and
$sum_1^infty {frac{1}{n^2 a_n}}$ converge. If no such sequence
exists, prove that.




Actually the question was for $sum_1^infty {n{a_n}}$ and $sum_1^infty {frac{1}{n^2 a_n}}$. For this problem; If we suppose that both are convergent, then their multiply $sum_1^infty 1/n$ must be convergent and this is a contradiction.



But for $sum_1^infty {a_n}$ and $sum_1^infty {frac{1}{n^2 a_n}}$, if we multiply we get $sum_1^infty 1/n^2$ which is convergent. I tried to find a sequence such that both the series converge but I couldn't find any but also I couldn't prove that there is no such sequence exists.



Edit. $a_n >0$






sequences-and-series convergence







share|cite|improve this question











$endgroup$





Find a sequence ${a_n}$ such that both $sum_1^infty {a_n}$ and
$sum_1^infty {frac{1}{n^2 a_n}}$ converge. If no such sequence
exists, prove that.




Actually the question was for $sum_1^infty {n{a_n}}$ and $sum_1^infty {frac{1}{n^2 a_n}}$. For this problem; If we suppose that both are convergent, then their multiply $sum_1^infty 1/n$ must be convergent and this is a contradiction.



But for $sum_1^infty {a_n}$ and $sum_1^infty {frac{1}{n^2 a_n}}$, if we multiply we get $sum_1^infty 1/n^2$ which is convergent. I tried to find a sequence such that both the series converge but I couldn't find any but also I couldn't prove that there is no such sequence exists.



Edit. $a_n >0$






sequences-and-series convergence




sequences-and-series convergence






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 7 at 21:23







amir na

















asked Jan 7 at 21:01









amir naamir na

696




696












  • $begingroup$
    I think that $a_n>0$ should be a condition to make the question really interesting
    $endgroup$
    – ajotatxe
    Jan 7 at 21:08










  • $begingroup$
    @ajotatxe yes. The book I found the first question, said at the first of the series chapter, that for all questions, consider $a_n>0$ and I forgot to mention that.
    $endgroup$
    – amir na
    Jan 7 at 21:23












  • $begingroup$
    With the condition $a_n > 0$ it is a duplicate of math.stackexchange.com/q/1933001/42969.
    $endgroup$
    – Martin R
    Jan 7 at 21:31










  • $begingroup$
    If $a_n>0$, by Cauchy-Schwarz $$sum_{n=1}^{N}a_nsum_{n=1}^{N}frac{1}{n^2 a_n}geq H_N^2geq log^2(N).$$
    $endgroup$
    – Jack D'Aurizio
    Jan 7 at 21:34




















  • $begingroup$
    I think that $a_n>0$ should be a condition to make the question really interesting
    $endgroup$
    – ajotatxe
    Jan 7 at 21:08










  • $begingroup$
    @ajotatxe yes. The book I found the first question, said at the first of the series chapter, that for all questions, consider $a_n>0$ and I forgot to mention that.
    $endgroup$
    – amir na
    Jan 7 at 21:23












  • $begingroup$
    With the condition $a_n > 0$ it is a duplicate of math.stackexchange.com/q/1933001/42969.
    $endgroup$
    – Martin R
    Jan 7 at 21:31










  • $begingroup$
    If $a_n>0$, by Cauchy-Schwarz $$sum_{n=1}^{N}a_nsum_{n=1}^{N}frac{1}{n^2 a_n}geq H_N^2geq log^2(N).$$
    $endgroup$
    – Jack D'Aurizio
    Jan 7 at 21:34


















$begingroup$
I think that $a_n>0$ should be a condition to make the question really interesting
$endgroup$
– ajotatxe
Jan 7 at 21:08




$begingroup$
I think that $a_n>0$ should be a condition to make the question really interesting
$endgroup$
– ajotatxe
Jan 7 at 21:08












$begingroup$
@ajotatxe yes. The book I found the first question, said at the first of the series chapter, that for all questions, consider $a_n>0$ and I forgot to mention that.
$endgroup$
– amir na
Jan 7 at 21:23






$begingroup$
@ajotatxe yes. The book I found the first question, said at the first of the series chapter, that for all questions, consider $a_n>0$ and I forgot to mention that.
$endgroup$
– amir na
Jan 7 at 21:23














$begingroup$
With the condition $a_n > 0$ it is a duplicate of math.stackexchange.com/q/1933001/42969.
$endgroup$
– Martin R
Jan 7 at 21:31




$begingroup$
With the condition $a_n > 0$ it is a duplicate of math.stackexchange.com/q/1933001/42969.
$endgroup$
– Martin R
Jan 7 at 21:31












$begingroup$
If $a_n>0$, by Cauchy-Schwarz $$sum_{n=1}^{N}a_nsum_{n=1}^{N}frac{1}{n^2 a_n}geq H_N^2geq log^2(N).$$
$endgroup$
– Jack D'Aurizio
Jan 7 at 21:34






$begingroup$
If $a_n>0$, by Cauchy-Schwarz $$sum_{n=1}^{N}a_nsum_{n=1}^{N}frac{1}{n^2 a_n}geq H_N^2geq log^2(N).$$
$endgroup$
– Jack D'Aurizio
Jan 7 at 21:34












3 Answers
3






active

oldest

votes


















4












$begingroup$

The answer is yes, an example is
$$a_n = frac{(-1)^n}{n}$$
However, there is no such a sequence with positive terms.
Indeed, if you suppsose that both $sum a_n$ and $sumfrac{1}{n^2a_n}$ are convergent and with positive terms, then you have by the AM-GM inequality
$$sum frac{1}{n}=sum sqrt{a_n cdot frac{1}{n^2a_n}} le sum frac{1}{2} left(a_n + frac{1}{n^2a_n} right) < +infty$$
which implies that the harmonic series is convergent: a contradiction.






share|cite|improve this answer









$endgroup$





















    5












    $begingroup$

    Perhaps a bit too trivial?



    $a_n=(-1)^n(1/n)$






    share|cite|improve this answer









    $endgroup$





















      5












      $begingroup$

      The above answer shows that if the $a_n$s are allowed to change sign, then the sum may converge. However, if the $a_n$s are all positive, then by the Cauchy-Schwarz inequality,



      begin{align*}
      sum_{n=1}^kfrac{1}{n}&=sum_{n= 1}^kfrac{sqrt{a_n}}{sqrt{n^2a_n}}\
      &leq left(sum_{n =1}^ka_nright)^{1/2}left(sum_{n= 1}^kfrac{1}{n^2a_n}right)^{1/2}
      end{align*}



      Since $sum_{n=1}^kfrac{1}{n}toinfty$ as $ktoinfty$, one of $sum a_n$ and $sum frac{1}{n^2a_n}$ must diverge.






      share|cite|improve this answer









      $endgroup$













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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        4












        $begingroup$

        The answer is yes, an example is
        $$a_n = frac{(-1)^n}{n}$$
        However, there is no such a sequence with positive terms.
        Indeed, if you suppsose that both $sum a_n$ and $sumfrac{1}{n^2a_n}$ are convergent and with positive terms, then you have by the AM-GM inequality
        $$sum frac{1}{n}=sum sqrt{a_n cdot frac{1}{n^2a_n}} le sum frac{1}{2} left(a_n + frac{1}{n^2a_n} right) < +infty$$
        which implies that the harmonic series is convergent: a contradiction.






        share|cite|improve this answer









        $endgroup$


















          4












          $begingroup$

          The answer is yes, an example is
          $$a_n = frac{(-1)^n}{n}$$
          However, there is no such a sequence with positive terms.
          Indeed, if you suppsose that both $sum a_n$ and $sumfrac{1}{n^2a_n}$ are convergent and with positive terms, then you have by the AM-GM inequality
          $$sum frac{1}{n}=sum sqrt{a_n cdot frac{1}{n^2a_n}} le sum frac{1}{2} left(a_n + frac{1}{n^2a_n} right) < +infty$$
          which implies that the harmonic series is convergent: a contradiction.






          share|cite|improve this answer









          $endgroup$
















            4












            4








            4





            $begingroup$

            The answer is yes, an example is
            $$a_n = frac{(-1)^n}{n}$$
            However, there is no such a sequence with positive terms.
            Indeed, if you suppsose that both $sum a_n$ and $sumfrac{1}{n^2a_n}$ are convergent and with positive terms, then you have by the AM-GM inequality
            $$sum frac{1}{n}=sum sqrt{a_n cdot frac{1}{n^2a_n}} le sum frac{1}{2} left(a_n + frac{1}{n^2a_n} right) < +infty$$
            which implies that the harmonic series is convergent: a contradiction.






            share|cite|improve this answer









            $endgroup$



            The answer is yes, an example is
            $$a_n = frac{(-1)^n}{n}$$
            However, there is no such a sequence with positive terms.
            Indeed, if you suppsose that both $sum a_n$ and $sumfrac{1}{n^2a_n}$ are convergent and with positive terms, then you have by the AM-GM inequality
            $$sum frac{1}{n}=sum sqrt{a_n cdot frac{1}{n^2a_n}} le sum frac{1}{2} left(a_n + frac{1}{n^2a_n} right) < +infty$$
            which implies that the harmonic series is convergent: a contradiction.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 7 at 21:15









            CrostulCrostul

            27.9k22352




            27.9k22352























                5












                $begingroup$

                Perhaps a bit too trivial?



                $a_n=(-1)^n(1/n)$






                share|cite|improve this answer









                $endgroup$


















                  5












                  $begingroup$

                  Perhaps a bit too trivial?



                  $a_n=(-1)^n(1/n)$






                  share|cite|improve this answer









                  $endgroup$
















                    5












                    5








                    5





                    $begingroup$

                    Perhaps a bit too trivial?



                    $a_n=(-1)^n(1/n)$






                    share|cite|improve this answer









                    $endgroup$



                    Perhaps a bit too trivial?



                    $a_n=(-1)^n(1/n)$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 7 at 21:06









                    Peter SzilasPeter Szilas

                    11.3k2822




                    11.3k2822























                        5












                        $begingroup$

                        The above answer shows that if the $a_n$s are allowed to change sign, then the sum may converge. However, if the $a_n$s are all positive, then by the Cauchy-Schwarz inequality,



                        begin{align*}
                        sum_{n=1}^kfrac{1}{n}&=sum_{n= 1}^kfrac{sqrt{a_n}}{sqrt{n^2a_n}}\
                        &leq left(sum_{n =1}^ka_nright)^{1/2}left(sum_{n= 1}^kfrac{1}{n^2a_n}right)^{1/2}
                        end{align*}



                        Since $sum_{n=1}^kfrac{1}{n}toinfty$ as $ktoinfty$, one of $sum a_n$ and $sum frac{1}{n^2a_n}$ must diverge.






                        share|cite|improve this answer









                        $endgroup$


















                          5












                          $begingroup$

                          The above answer shows that if the $a_n$s are allowed to change sign, then the sum may converge. However, if the $a_n$s are all positive, then by the Cauchy-Schwarz inequality,



                          begin{align*}
                          sum_{n=1}^kfrac{1}{n}&=sum_{n= 1}^kfrac{sqrt{a_n}}{sqrt{n^2a_n}}\
                          &leq left(sum_{n =1}^ka_nright)^{1/2}left(sum_{n= 1}^kfrac{1}{n^2a_n}right)^{1/2}
                          end{align*}



                          Since $sum_{n=1}^kfrac{1}{n}toinfty$ as $ktoinfty$, one of $sum a_n$ and $sum frac{1}{n^2a_n}$ must diverge.






                          share|cite|improve this answer









                          $endgroup$
















                            5












                            5








                            5





                            $begingroup$

                            The above answer shows that if the $a_n$s are allowed to change sign, then the sum may converge. However, if the $a_n$s are all positive, then by the Cauchy-Schwarz inequality,



                            begin{align*}
                            sum_{n=1}^kfrac{1}{n}&=sum_{n= 1}^kfrac{sqrt{a_n}}{sqrt{n^2a_n}}\
                            &leq left(sum_{n =1}^ka_nright)^{1/2}left(sum_{n= 1}^kfrac{1}{n^2a_n}right)^{1/2}
                            end{align*}



                            Since $sum_{n=1}^kfrac{1}{n}toinfty$ as $ktoinfty$, one of $sum a_n$ and $sum frac{1}{n^2a_n}$ must diverge.






                            share|cite|improve this answer









                            $endgroup$



                            The above answer shows that if the $a_n$s are allowed to change sign, then the sum may converge. However, if the $a_n$s are all positive, then by the Cauchy-Schwarz inequality,



                            begin{align*}
                            sum_{n=1}^kfrac{1}{n}&=sum_{n= 1}^kfrac{sqrt{a_n}}{sqrt{n^2a_n}}\
                            &leq left(sum_{n =1}^ka_nright)^{1/2}left(sum_{n= 1}^kfrac{1}{n^2a_n}right)^{1/2}
                            end{align*}



                            Since $sum_{n=1}^kfrac{1}{n}toinfty$ as $ktoinfty$, one of $sum a_n$ and $sum frac{1}{n^2a_n}$ must diverge.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 7 at 21:15









                            Nathaniel BNathaniel B

                            851616




                            851616






























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