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

1486




1486












  • $begingroup$
    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









      $endgroup$














        Your Answer





        StackExchange.ifUsing("editor", function () {
        return StackExchange.using("mathjaxEditing", function () {
        StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
        StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
        });
        });
        }, "mathjax-editing");

        StackExchange.ready(function() {
        var channelOptions = {
        tags: "".split(" "),
        id: "69"
        };
        initTagRenderer("".split(" "), "".split(" "), channelOptions);

        StackExchange.using("externalEditor", function() {
        // Have to fire editor after snippets, if snippets enabled
        if (StackExchange.settings.snippets.snippetsEnabled) {
        StackExchange.using("snippets", function() {
        createEditor();
        });
        }
        else {
        createEditor();
        }
        });

        function createEditor() {
        StackExchange.prepareEditor({
        heartbeatType: 'answer',
        autoActivateHeartbeat: false,
        convertImagesToLinks: true,
        noModals: true,
        showLowRepImageUploadWarning: true,
        reputationToPostImages: 10,
        bindNavPrevention: true,
        postfix: "",
        imageUploader: {
        brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
        contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
        allowUrls: true
        },
        noCode: true, onDemand: true,
        discardSelector: ".discard-answer"
        ,immediatelyShowMarkdownHelp:true
        });


        }
        });














        draft saved

        draft discarded


















        StackExchange.ready(
        function () {
        StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3065483%2ffind-a-sequence-a-n-such-that-both-sum-a-n-and-sum-frac1n2-a-n%23new-answer', 'question_page');
        }
        );

        Post as a guest















        Required, but never shown

























        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

            28.2k22352




            28.2k22352























                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

                    11.7k2822




                    11.7k2822























                        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






























                                draft saved

                                draft discarded




















































                                Thanks for contributing an answer to Mathematics Stack Exchange!


                                • Please be sure to answer the question. Provide details and share your research!

                                But avoid …



                                • Asking for help, clarification, or responding to other answers.

                                • Making statements based on opinion; back them up with references or personal experience.


                                Use MathJax to format equations. MathJax reference.


                                To learn more, see our tips on writing great answers.




                                draft saved


                                draft discarded














                                StackExchange.ready(
                                function () {
                                StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3065483%2ffind-a-sequence-a-n-such-that-both-sum-a-n-and-sum-frac1n2-a-n%23new-answer', 'question_page');
                                }
                                );

                                Post as a guest















                                Required, but never shown





















































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown

































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown







                                -

                                Popular posts from this blog

                                Bressuire

                                Image
                                .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid
                                Read more

                                Cabo Verde

                                Image
                                  Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º)
                                Read more

                                Gyllenstierna

                                Image
                                .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid
                                Read more