Suppose $limlimits_{n rightarrow infty } n^2 a_n =1$, then $ sumlimits _{n=1} ^{infty} a_n$ is convergent? Or...












2












$begingroup$



Suppose $lim_{n rightarrow infty } n^2 a_n =1$ then $ sum _{n=1} ^{infty} a_n$ is convergent or divergent?




We use the definition of the limit, then for $varepsilon =1$ there must exist an $n_0geq n$ such that $$ |n^2 a_n -1|<1$$ so we can say that:
$$ 0 < n^2 a_n < 2 $$
We divide by:
$$ 0 < a_n < frac{2}{n^2} $$
We now know if we take the infinite sum we get:
$$ 0 < sum _{n=1} ^{infty} a_n < sum _{n=1} ^{infty} frac{2}{n^2} $$
We know that the right side converges, so the sum is bounded, but how would I prove that is converges? Or can we find a counterexample that diverges, bur stays within these bounds.






real-analysis sequences-and-series







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    you argument is correct
    $endgroup$
    – Guy Fsone
    Dec 1 '18 at 14:52






  • 1




    $begingroup$
    note that $a_nge 0$, hence the series converges, that is, the sequence defined by $s_n:=sum_{k=0}^na_k$ is increasing and bounded, now apply Bolzano-Weierstrass theorem
    $endgroup$
    – Masacroso
    Dec 1 '18 at 14:54








  • 1




    $begingroup$
    @Masacroso The sign of the terms $a_n$ is offtopic and not needed to conclude: if $|a_n|leqslant c/n^2$ then $sum a_n$ converges.
    $endgroup$
    – Did
    Dec 1 '18 at 15:09










  • $begingroup$
    Of course, it is bounded AND increasing, therefore it converges!
    $endgroup$
    – Wesley Strik
    Dec 1 '18 at 15:11






  • 1




    $begingroup$
    @Masacroso Why not give to the OP the proper tools they are lacking, from the onset, rather than letting them get only a fugitive glimpse of the picture? Anyway, as regards the case $b_n=(-1)^n$, your comment is misleading since then, the series $sum b_n$ diverges hence one should not write things like $sumlimits_{n=0}^infty b_n$. (Actually, rereading your comment, I wonder if you are aware of the difference of nature between the series $sum b_n$, which always exists, and the number $sumlimits_{n=0}^infty b_n$, which may or may not exist...)
    $endgroup$
    – Did
    Dec 1 '18 at 15:31


















2












$begingroup$



Suppose $lim_{n rightarrow infty } n^2 a_n =1$ then $ sum _{n=1} ^{infty} a_n$ is convergent or divergent?




We use the definition of the limit, then for $varepsilon =1$ there must exist an $n_0geq n$ such that $$ |n^2 a_n -1|<1$$ so we can say that:
$$ 0 < n^2 a_n < 2 $$
We divide by:
$$ 0 < a_n < frac{2}{n^2} $$
We now know if we take the infinite sum we get:
$$ 0 < sum _{n=1} ^{infty} a_n < sum _{n=1} ^{infty} frac{2}{n^2} $$
We know that the right side converges, so the sum is bounded, but how would I prove that is converges? Or can we find a counterexample that diverges, bur stays within these bounds.






real-analysis sequences-and-series







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    you argument is correct
    $endgroup$
    – Guy Fsone
    Dec 1 '18 at 14:52






  • 1




    $begingroup$
    note that $a_nge 0$, hence the series converges, that is, the sequence defined by $s_n:=sum_{k=0}^na_k$ is increasing and bounded, now apply Bolzano-Weierstrass theorem
    $endgroup$
    – Masacroso
    Dec 1 '18 at 14:54








  • 1




    $begingroup$
    @Masacroso The sign of the terms $a_n$ is offtopic and not needed to conclude: if $|a_n|leqslant c/n^2$ then $sum a_n$ converges.
    $endgroup$
    – Did
    Dec 1 '18 at 15:09










  • $begingroup$
    Of course, it is bounded AND increasing, therefore it converges!
    $endgroup$
    – Wesley Strik
    Dec 1 '18 at 15:11






  • 1




    $begingroup$
    @Masacroso Why not give to the OP the proper tools they are lacking, from the onset, rather than letting them get only a fugitive glimpse of the picture? Anyway, as regards the case $b_n=(-1)^n$, your comment is misleading since then, the series $sum b_n$ diverges hence one should not write things like $sumlimits_{n=0}^infty b_n$. (Actually, rereading your comment, I wonder if you are aware of the difference of nature between the series $sum b_n$, which always exists, and the number $sumlimits_{n=0}^infty b_n$, which may or may not exist...)
    $endgroup$
    – Did
    Dec 1 '18 at 15:31
















2












2








2





$begingroup$



Suppose $lim_{n rightarrow infty } n^2 a_n =1$ then $ sum _{n=1} ^{infty} a_n$ is convergent or divergent?




We use the definition of the limit, then for $varepsilon =1$ there must exist an $n_0geq n$ such that $$ |n^2 a_n -1|<1$$ so we can say that:
$$ 0 < n^2 a_n < 2 $$
We divide by:
$$ 0 < a_n < frac{2}{n^2} $$
We now know if we take the infinite sum we get:
$$ 0 < sum _{n=1} ^{infty} a_n < sum _{n=1} ^{infty} frac{2}{n^2} $$
We know that the right side converges, so the sum is bounded, but how would I prove that is converges? Or can we find a counterexample that diverges, bur stays within these bounds.






real-analysis sequences-and-series







share|cite|improve this question











$endgroup$





Suppose $lim_{n rightarrow infty } n^2 a_n =1$ then $ sum _{n=1} ^{infty} a_n$ is convergent or divergent?




We use the definition of the limit, then for $varepsilon =1$ there must exist an $n_0geq n$ such that $$ |n^2 a_n -1|<1$$ so we can say that:
$$ 0 < n^2 a_n < 2 $$
We divide by:
$$ 0 < a_n < frac{2}{n^2} $$
We now know if we take the infinite sum we get:
$$ 0 < sum _{n=1} ^{infty} a_n < sum _{n=1} ^{infty} frac{2}{n^2} $$
We know that the right side converges, so the sum is bounded, but how would I prove that is converges? Or can we find a counterexample that diverges, bur stays within these bounds.






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 Dec 1 '18 at 15:12







Wesley Strik

















asked Dec 1 '18 at 14:47









Wesley StrikWesley Strik

1,653423




1,653423








  • 2




    $begingroup$
    you argument is correct
    $endgroup$
    – Guy Fsone
    Dec 1 '18 at 14:52






  • 1




    $begingroup$
    note that $a_nge 0$, hence the series converges, that is, the sequence defined by $s_n:=sum_{k=0}^na_k$ is increasing and bounded, now apply Bolzano-Weierstrass theorem
    $endgroup$
    – Masacroso
    Dec 1 '18 at 14:54








  • 1




    $begingroup$
    @Masacroso The sign of the terms $a_n$ is offtopic and not needed to conclude: if $|a_n|leqslant c/n^2$ then $sum a_n$ converges.
    $endgroup$
    – Did
    Dec 1 '18 at 15:09










  • $begingroup$
    Of course, it is bounded AND increasing, therefore it converges!
    $endgroup$
    – Wesley Strik
    Dec 1 '18 at 15:11






  • 1




    $begingroup$
    @Masacroso Why not give to the OP the proper tools they are lacking, from the onset, rather than letting them get only a fugitive glimpse of the picture? Anyway, as regards the case $b_n=(-1)^n$, your comment is misleading since then, the series $sum b_n$ diverges hence one should not write things like $sumlimits_{n=0}^infty b_n$. (Actually, rereading your comment, I wonder if you are aware of the difference of nature between the series $sum b_n$, which always exists, and the number $sumlimits_{n=0}^infty b_n$, which may or may not exist...)
    $endgroup$
    – Did
    Dec 1 '18 at 15:31
















  • 2




    $begingroup$
    you argument is correct
    $endgroup$
    – Guy Fsone
    Dec 1 '18 at 14:52






  • 1




    $begingroup$
    note that $a_nge 0$, hence the series converges, that is, the sequence defined by $s_n:=sum_{k=0}^na_k$ is increasing and bounded, now apply Bolzano-Weierstrass theorem
    $endgroup$
    – Masacroso
    Dec 1 '18 at 14:54








  • 1




    $begingroup$
    @Masacroso The sign of the terms $a_n$ is offtopic and not needed to conclude: if $|a_n|leqslant c/n^2$ then $sum a_n$ converges.
    $endgroup$
    – Did
    Dec 1 '18 at 15:09










  • $begingroup$
    Of course, it is bounded AND increasing, therefore it converges!
    $endgroup$
    – Wesley Strik
    Dec 1 '18 at 15:11






  • 1




    $begingroup$
    @Masacroso Why not give to the OP the proper tools they are lacking, from the onset, rather than letting them get only a fugitive glimpse of the picture? Anyway, as regards the case $b_n=(-1)^n$, your comment is misleading since then, the series $sum b_n$ diverges hence one should not write things like $sumlimits_{n=0}^infty b_n$. (Actually, rereading your comment, I wonder if you are aware of the difference of nature between the series $sum b_n$, which always exists, and the number $sumlimits_{n=0}^infty b_n$, which may or may not exist...)
    $endgroup$
    – Did
    Dec 1 '18 at 15:31










2




2




$begingroup$
you argument is correct
$endgroup$
– Guy Fsone
Dec 1 '18 at 14:52




$begingroup$
you argument is correct
$endgroup$
– Guy Fsone
Dec 1 '18 at 14:52




1




1




$begingroup$
note that $a_nge 0$, hence the series converges, that is, the sequence defined by $s_n:=sum_{k=0}^na_k$ is increasing and bounded, now apply Bolzano-Weierstrass theorem
$endgroup$
– Masacroso
Dec 1 '18 at 14:54






$begingroup$
note that $a_nge 0$, hence the series converges, that is, the sequence defined by $s_n:=sum_{k=0}^na_k$ is increasing and bounded, now apply Bolzano-Weierstrass theorem
$endgroup$
– Masacroso
Dec 1 '18 at 14:54






1




1




$begingroup$
@Masacroso The sign of the terms $a_n$ is offtopic and not needed to conclude: if $|a_n|leqslant c/n^2$ then $sum a_n$ converges.
$endgroup$
– Did
Dec 1 '18 at 15:09




$begingroup$
@Masacroso The sign of the terms $a_n$ is offtopic and not needed to conclude: if $|a_n|leqslant c/n^2$ then $sum a_n$ converges.
$endgroup$
– Did
Dec 1 '18 at 15:09












$begingroup$
Of course, it is bounded AND increasing, therefore it converges!
$endgroup$
– Wesley Strik
Dec 1 '18 at 15:11




$begingroup$
Of course, it is bounded AND increasing, therefore it converges!
$endgroup$
– Wesley Strik
Dec 1 '18 at 15:11




1




1




$begingroup$
@Masacroso Why not give to the OP the proper tools they are lacking, from the onset, rather than letting them get only a fugitive glimpse of the picture? Anyway, as regards the case $b_n=(-1)^n$, your comment is misleading since then, the series $sum b_n$ diverges hence one should not write things like $sumlimits_{n=0}^infty b_n$. (Actually, rereading your comment, I wonder if you are aware of the difference of nature between the series $sum b_n$, which always exists, and the number $sumlimits_{n=0}^infty b_n$, which may or may not exist...)
$endgroup$
– Did
Dec 1 '18 at 15:31






$begingroup$
@Masacroso Why not give to the OP the proper tools they are lacking, from the onset, rather than letting them get only a fugitive glimpse of the picture? Anyway, as regards the case $b_n=(-1)^n$, your comment is misleading since then, the series $sum b_n$ diverges hence one should not write things like $sumlimits_{n=0}^infty b_n$. (Actually, rereading your comment, I wonder if you are aware of the difference of nature between the series $sum b_n$, which always exists, and the number $sumlimits_{n=0}^infty b_n$, which may or may not exist...)
$endgroup$
– Did
Dec 1 '18 at 15:31












4 Answers
4






active

oldest

votes


















2












$begingroup$

Option.



$lim_{n rightarrow infty}n^2a_n=1.$



This Implies for $n in mathbb{Z^+}$ :



$n^2|a_n| lt M$, positive real number.



$|a_n| lt M/n^2.$



By comparison test $sum |a_n|$ is convergent,



hence $sum a_n$ is convergent.






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    A series converges iff its remainder term $sumlimits_N^infty a_n rightarrow 0$ as $Nrightarrow infty$. Your argument shows that you can, for large enough $N$, bound $|sumlimits_N^infty a_n|$ by $sumlimits_N^infty frac{1}{n^2}$. The latter term approaches zero as $Nrightarrow infty$, since $sum frac{1}{n^2}$ converges.






    share|cite|improve this answer









    $endgroup$





















      2












      $begingroup$

      The hypothesis means first that, if $n$ is large enough, $a_n>0$, and also that $a_n$ is asymptotically equivalent to $dfrac1{n^2}$.



      Now two series with (eventually positive) equivalent terms both converge or both diverge.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        It is increasing and bounded, therefore it converges!
        $endgroup$
        – Wesley Strik
        Dec 1 '18 at 15:10






      • 1




        $begingroup$
        Of course, but equivalence gives a faster proof. Bondedness is implicit in the definition of equivalence.
        $endgroup$
        – Bernard
        Dec 1 '18 at 15:16



















      1












      $begingroup$

      Your argument is correct, more simply we can refer directly to limit comparison test with $$sum frac1{n^2}$$



      indeed



      $$frac{a_n}{ frac1{n^2}}=n^2a_nto 1$$



      therefore $sum a_n$ converges.



      It is important to recognize that since, more in general, when we solve a problem we don't need everytime to prove all the results that we have already proved in a general way.



      Therefore if you are not requested to use esplicitely the $epsilon-delta$ definition a solution by limit comparison test is fine.






      share|cite|improve this answer











      $endgroup$









      • 1




        $begingroup$
        The test only applies to positive series, so a small step is needed before applying it.
        $endgroup$
        – Andrés E. Caicedo
        Dec 1 '18 at 15:10






      • 1




        $begingroup$
        @AndresE.Caicedo yes of course we can assume $a_n$ strictly positive for n sufficiently large.
        $endgroup$
        – gimusi
        Dec 1 '18 at 15:20











      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%2f3021429%2fsuppose-lim-limits-n-rightarrow-infty-n2-a-n-1-then-sum-limits-n%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      Option.



      $lim_{n rightarrow infty}n^2a_n=1.$



      This Implies for $n in mathbb{Z^+}$ :



      $n^2|a_n| lt M$, positive real number.



      $|a_n| lt M/n^2.$



      By comparison test $sum |a_n|$ is convergent,



      hence $sum a_n$ is convergent.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        Option.



        $lim_{n rightarrow infty}n^2a_n=1.$



        This Implies for $n in mathbb{Z^+}$ :



        $n^2|a_n| lt M$, positive real number.



        $|a_n| lt M/n^2.$



        By comparison test $sum |a_n|$ is convergent,



        hence $sum a_n$ is convergent.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          Option.



          $lim_{n rightarrow infty}n^2a_n=1.$



          This Implies for $n in mathbb{Z^+}$ :



          $n^2|a_n| lt M$, positive real number.



          $|a_n| lt M/n^2.$



          By comparison test $sum |a_n|$ is convergent,



          hence $sum a_n$ is convergent.






          share|cite|improve this answer









          $endgroup$



          Option.



          $lim_{n rightarrow infty}n^2a_n=1.$



          This Implies for $n in mathbb{Z^+}$ :



          $n^2|a_n| lt M$, positive real number.



          $|a_n| lt M/n^2.$



          By comparison test $sum |a_n|$ is convergent,



          hence $sum a_n$ is convergent.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 1 '18 at 15:49









          Peter SzilasPeter Szilas

          10.9k2721




          10.9k2721























              3












              $begingroup$

              A series converges iff its remainder term $sumlimits_N^infty a_n rightarrow 0$ as $Nrightarrow infty$. Your argument shows that you can, for large enough $N$, bound $|sumlimits_N^infty a_n|$ by $sumlimits_N^infty frac{1}{n^2}$. The latter term approaches zero as $Nrightarrow infty$, since $sum frac{1}{n^2}$ converges.






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                A series converges iff its remainder term $sumlimits_N^infty a_n rightarrow 0$ as $Nrightarrow infty$. Your argument shows that you can, for large enough $N$, bound $|sumlimits_N^infty a_n|$ by $sumlimits_N^infty frac{1}{n^2}$. The latter term approaches zero as $Nrightarrow infty$, since $sum frac{1}{n^2}$ converges.






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  A series converges iff its remainder term $sumlimits_N^infty a_n rightarrow 0$ as $Nrightarrow infty$. Your argument shows that you can, for large enough $N$, bound $|sumlimits_N^infty a_n|$ by $sumlimits_N^infty frac{1}{n^2}$. The latter term approaches zero as $Nrightarrow infty$, since $sum frac{1}{n^2}$ converges.






                  share|cite|improve this answer









                  $endgroup$



                  A series converges iff its remainder term $sumlimits_N^infty a_n rightarrow 0$ as $Nrightarrow infty$. Your argument shows that you can, for large enough $N$, bound $|sumlimits_N^infty a_n|$ by $sumlimits_N^infty frac{1}{n^2}$. The latter term approaches zero as $Nrightarrow infty$, since $sum frac{1}{n^2}$ converges.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 1 '18 at 14:56









                  user25959user25959

                  1,573816




                  1,573816























                      2












                      $begingroup$

                      The hypothesis means first that, if $n$ is large enough, $a_n>0$, and also that $a_n$ is asymptotically equivalent to $dfrac1{n^2}$.



                      Now two series with (eventually positive) equivalent terms both converge or both diverge.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        It is increasing and bounded, therefore it converges!
                        $endgroup$
                        – Wesley Strik
                        Dec 1 '18 at 15:10






                      • 1




                        $begingroup$
                        Of course, but equivalence gives a faster proof. Bondedness is implicit in the definition of equivalence.
                        $endgroup$
                        – Bernard
                        Dec 1 '18 at 15:16
















                      2












                      $begingroup$

                      The hypothesis means first that, if $n$ is large enough, $a_n>0$, and also that $a_n$ is asymptotically equivalent to $dfrac1{n^2}$.



                      Now two series with (eventually positive) equivalent terms both converge or both diverge.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        It is increasing and bounded, therefore it converges!
                        $endgroup$
                        – Wesley Strik
                        Dec 1 '18 at 15:10






                      • 1




                        $begingroup$
                        Of course, but equivalence gives a faster proof. Bondedness is implicit in the definition of equivalence.
                        $endgroup$
                        – Bernard
                        Dec 1 '18 at 15:16














                      2












                      2








                      2





                      $begingroup$

                      The hypothesis means first that, if $n$ is large enough, $a_n>0$, and also that $a_n$ is asymptotically equivalent to $dfrac1{n^2}$.



                      Now two series with (eventually positive) equivalent terms both converge or both diverge.






                      share|cite|improve this answer









                      $endgroup$



                      The hypothesis means first that, if $n$ is large enough, $a_n>0$, and also that $a_n$ is asymptotically equivalent to $dfrac1{n^2}$.



                      Now two series with (eventually positive) equivalent terms both converge or both diverge.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Dec 1 '18 at 15:07









                      BernardBernard

                      119k639112




                      119k639112












                      • $begingroup$
                        It is increasing and bounded, therefore it converges!
                        $endgroup$
                        – Wesley Strik
                        Dec 1 '18 at 15:10






                      • 1




                        $begingroup$
                        Of course, but equivalence gives a faster proof. Bondedness is implicit in the definition of equivalence.
                        $endgroup$
                        – Bernard
                        Dec 1 '18 at 15:16


















                      • $begingroup$
                        It is increasing and bounded, therefore it converges!
                        $endgroup$
                        – Wesley Strik
                        Dec 1 '18 at 15:10






                      • 1




                        $begingroup$
                        Of course, but equivalence gives a faster proof. Bondedness is implicit in the definition of equivalence.
                        $endgroup$
                        – Bernard
                        Dec 1 '18 at 15:16
















                      $begingroup$
                      It is increasing and bounded, therefore it converges!
                      $endgroup$
                      – Wesley Strik
                      Dec 1 '18 at 15:10




                      $begingroup$
                      It is increasing and bounded, therefore it converges!
                      $endgroup$
                      – Wesley Strik
                      Dec 1 '18 at 15:10




                      1




                      1




                      $begingroup$
                      Of course, but equivalence gives a faster proof. Bondedness is implicit in the definition of equivalence.
                      $endgroup$
                      – Bernard
                      Dec 1 '18 at 15:16




                      $begingroup$
                      Of course, but equivalence gives a faster proof. Bondedness is implicit in the definition of equivalence.
                      $endgroup$
                      – Bernard
                      Dec 1 '18 at 15:16











                      1












                      $begingroup$

                      Your argument is correct, more simply we can refer directly to limit comparison test with $$sum frac1{n^2}$$



                      indeed



                      $$frac{a_n}{ frac1{n^2}}=n^2a_nto 1$$



                      therefore $sum a_n$ converges.



                      It is important to recognize that since, more in general, when we solve a problem we don't need everytime to prove all the results that we have already proved in a general way.



                      Therefore if you are not requested to use esplicitely the $epsilon-delta$ definition a solution by limit comparison test is fine.






                      share|cite|improve this answer











                      $endgroup$









                      • 1




                        $begingroup$
                        The test only applies to positive series, so a small step is needed before applying it.
                        $endgroup$
                        – Andrés E. Caicedo
                        Dec 1 '18 at 15:10






                      • 1




                        $begingroup$
                        @AndresE.Caicedo yes of course we can assume $a_n$ strictly positive for n sufficiently large.
                        $endgroup$
                        – gimusi
                        Dec 1 '18 at 15:20
















                      1












                      $begingroup$

                      Your argument is correct, more simply we can refer directly to limit comparison test with $$sum frac1{n^2}$$



                      indeed



                      $$frac{a_n}{ frac1{n^2}}=n^2a_nto 1$$



                      therefore $sum a_n$ converges.



                      It is important to recognize that since, more in general, when we solve a problem we don't need everytime to prove all the results that we have already proved in a general way.



                      Therefore if you are not requested to use esplicitely the $epsilon-delta$ definition a solution by limit comparison test is fine.






                      share|cite|improve this answer











                      $endgroup$









                      • 1




                        $begingroup$
                        The test only applies to positive series, so a small step is needed before applying it.
                        $endgroup$
                        – Andrés E. Caicedo
                        Dec 1 '18 at 15:10






                      • 1




                        $begingroup$
                        @AndresE.Caicedo yes of course we can assume $a_n$ strictly positive for n sufficiently large.
                        $endgroup$
                        – gimusi
                        Dec 1 '18 at 15:20














                      1












                      1








                      1





                      $begingroup$

                      Your argument is correct, more simply we can refer directly to limit comparison test with $$sum frac1{n^2}$$



                      indeed



                      $$frac{a_n}{ frac1{n^2}}=n^2a_nto 1$$



                      therefore $sum a_n$ converges.



                      It is important to recognize that since, more in general, when we solve a problem we don't need everytime to prove all the results that we have already proved in a general way.



                      Therefore if you are not requested to use esplicitely the $epsilon-delta$ definition a solution by limit comparison test is fine.






                      share|cite|improve this answer











                      $endgroup$



                      Your argument is correct, more simply we can refer directly to limit comparison test with $$sum frac1{n^2}$$



                      indeed



                      $$frac{a_n}{ frac1{n^2}}=n^2a_nto 1$$



                      therefore $sum a_n$ converges.



                      It is important to recognize that since, more in general, when we solve a problem we don't need everytime to prove all the results that we have already proved in a general way.



                      Therefore if you are not requested to use esplicitely the $epsilon-delta$ definition a solution by limit comparison test is fine.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Dec 1 '18 at 14:57

























                      answered Dec 1 '18 at 14:49









                      gimusigimusi

                      1




                      1








                      • 1




                        $begingroup$
                        The test only applies to positive series, so a small step is needed before applying it.
                        $endgroup$
                        – Andrés E. Caicedo
                        Dec 1 '18 at 15:10






                      • 1




                        $begingroup$
                        @AndresE.Caicedo yes of course we can assume $a_n$ strictly positive for n sufficiently large.
                        $endgroup$
                        – gimusi
                        Dec 1 '18 at 15:20














                      • 1




                        $begingroup$
                        The test only applies to positive series, so a small step is needed before applying it.
                        $endgroup$
                        – Andrés E. Caicedo
                        Dec 1 '18 at 15:10






                      • 1




                        $begingroup$
                        @AndresE.Caicedo yes of course we can assume $a_n$ strictly positive for n sufficiently large.
                        $endgroup$
                        – gimusi
                        Dec 1 '18 at 15:20








                      1




                      1




                      $begingroup$
                      The test only applies to positive series, so a small step is needed before applying it.
                      $endgroup$
                      – Andrés E. Caicedo
                      Dec 1 '18 at 15:10




                      $begingroup$
                      The test only applies to positive series, so a small step is needed before applying it.
                      $endgroup$
                      – Andrés E. Caicedo
                      Dec 1 '18 at 15:10




                      1




                      1




                      $begingroup$
                      @AndresE.Caicedo yes of course we can assume $a_n$ strictly positive for n sufficiently large.
                      $endgroup$
                      – gimusi
                      Dec 1 '18 at 15:20




                      $begingroup$
                      @AndresE.Caicedo yes of course we can assume $a_n$ strictly positive for n sufficiently large.
                      $endgroup$
                      – gimusi
                      Dec 1 '18 at 15:20


















                      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%2f3021429%2fsuppose-lim-limits-n-rightarrow-infty-n2-a-n-1-then-sum-limits-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

                      Bundesstraße 106

                      Image
                      Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück
                      Read more

                      Verónica Boquete

                      Image
                      Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl
                      Read more

                      Ida-Boy-Ed-Garten

                      Image
                      Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
                      Read more