How does this series diverge by limit comparison test?











up vote
1
down vote

favorite












How does this series diverge by limit comparison test? $$sum_{n=1}^infty sqrt{frac{n+4}{n^4+4}}$$



I origionally tried using $frac{1}{n^2}$ for the comparison, but I'm pretty sure it has to be $frac{n}{n^2}$ to properly compare.






sequences-and-series divergent-series







share|cite|improve this question




























    up vote
    1
    down vote

    favorite












    How does this series diverge by limit comparison test? $$sum_{n=1}^infty sqrt{frac{n+4}{n^4+4}}$$



    I origionally tried using $frac{1}{n^2}$ for the comparison, but I'm pretty sure it has to be $frac{n}{n^2}$ to properly compare.






    sequences-and-series divergent-series







    share|cite|improve this question


























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      How does this series diverge by limit comparison test? $$sum_{n=1}^infty sqrt{frac{n+4}{n^4+4}}$$



      I origionally tried using $frac{1}{n^2}$ for the comparison, but I'm pretty sure it has to be $frac{n}{n^2}$ to properly compare.






      sequences-and-series divergent-series







      share|cite|improve this question















      How does this series diverge by limit comparison test? $$sum_{n=1}^infty sqrt{frac{n+4}{n^4+4}}$$



      I origionally tried using $frac{1}{n^2}$ for the comparison, but I'm pretty sure it has to be $frac{n}{n^2}$ to properly compare.






      sequences-and-series divergent-series




      sequences-and-series divergent-series






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 1 hour ago









      Chinnapparaj R

      4,9951825




      4,9951825










      asked 1 hour ago









      Luke D

      615




      615






















          2 Answers
          2






          active

          oldest

          votes

















          up vote
          4
          down vote



          accepted










          $frac{n+4}{n^4+4} leq frac{n+4n}{n^4} leq frac{5}{n^3}$,



          hence, $sqrt{frac{n+4}{n^4+4}} leq frac{sqrt{5}}{n^{3/2}}$



          so the series converges by comparison with convergent p-series $sum frac{1}{n^{3/2}}$






          share|cite|improve this answer





















          • Wont c be equal to 0 though once you take the limit? c must be greater than 0 to conclude the test with b sub n
            – Luke D
            1 hour ago












          • @LukeD I don't get $c = 0$. For example, if you use my argument above you get $c = sqrt{5}$.
            – Mustafa Said
            1 hour ago












          • This can also be done with the limit comparison test, which is what the OP asked for.
            – David
            1 hour ago










          • Yeah im curious to see the limit comparison test. But regardless, thanks for the soln.
            – Luke D
            1 hour ago


















          up vote
          3
          down vote













          For this sort of thing it is strongly advised to do a rough calculation first. We have
          $$sqrt{frac{n+4}{n^4+4}}approxsqrt{frac{n}{n^4}}=frac1{n^{3/2}} ,$$
          which suggests comparing with
          $$sumfrac1{n^{3/2}} .$$
          We have
          $$sqrt{frac{n+4}{n^4+4}}bigg/frac1{n^{3/2}}=sqrt{frac{n^4+4n^3}{n^4+4}}
          =sqrt{frac{1+4n^{-1}}{1+4n^{-4}}}to1quadhbox{as $ntoinfty$} .$$
          Since this limit exists and is finite and not zero, and we know that
          $$sumfrac1{n^{3/2}}$$
          converges, your series converges too. (Doesn't diverge!!!)






          share|cite|improve this answer





















          • So the answer our teacher gave us is wrong... thanks for the heads up.
            – Luke D
            1 hour ago










          • If your teacher said the series diverges, yes, that's wrong.
            – David
            1 hour ago











          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',
          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%2f3044811%2fhow-does-this-series-diverge-by-limit-comparison-test%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          4
          down vote



          accepted










          $frac{n+4}{n^4+4} leq frac{n+4n}{n^4} leq frac{5}{n^3}$,



          hence, $sqrt{frac{n+4}{n^4+4}} leq frac{sqrt{5}}{n^{3/2}}$



          so the series converges by comparison with convergent p-series $sum frac{1}{n^{3/2}}$






          share|cite|improve this answer





















          • Wont c be equal to 0 though once you take the limit? c must be greater than 0 to conclude the test with b sub n
            – Luke D
            1 hour ago












          • @LukeD I don't get $c = 0$. For example, if you use my argument above you get $c = sqrt{5}$.
            – Mustafa Said
            1 hour ago












          • This can also be done with the limit comparison test, which is what the OP asked for.
            – David
            1 hour ago










          • Yeah im curious to see the limit comparison test. But regardless, thanks for the soln.
            – Luke D
            1 hour ago















          up vote
          4
          down vote



          accepted










          $frac{n+4}{n^4+4} leq frac{n+4n}{n^4} leq frac{5}{n^3}$,



          hence, $sqrt{frac{n+4}{n^4+4}} leq frac{sqrt{5}}{n^{3/2}}$



          so the series converges by comparison with convergent p-series $sum frac{1}{n^{3/2}}$






          share|cite|improve this answer





















          • Wont c be equal to 0 though once you take the limit? c must be greater than 0 to conclude the test with b sub n
            – Luke D
            1 hour ago












          • @LukeD I don't get $c = 0$. For example, if you use my argument above you get $c = sqrt{5}$.
            – Mustafa Said
            1 hour ago












          • This can also be done with the limit comparison test, which is what the OP asked for.
            – David
            1 hour ago










          • Yeah im curious to see the limit comparison test. But regardless, thanks for the soln.
            – Luke D
            1 hour ago













          up vote
          4
          down vote



          accepted







          up vote
          4
          down vote



          accepted






          $frac{n+4}{n^4+4} leq frac{n+4n}{n^4} leq frac{5}{n^3}$,



          hence, $sqrt{frac{n+4}{n^4+4}} leq frac{sqrt{5}}{n^{3/2}}$



          so the series converges by comparison with convergent p-series $sum frac{1}{n^{3/2}}$






          share|cite|improve this answer












          $frac{n+4}{n^4+4} leq frac{n+4n}{n^4} leq frac{5}{n^3}$,



          hence, $sqrt{frac{n+4}{n^4+4}} leq frac{sqrt{5}}{n^{3/2}}$



          so the series converges by comparison with convergent p-series $sum frac{1}{n^{3/2}}$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 1 hour ago









          Mustafa Said

          2,8611913




          2,8611913












          • Wont c be equal to 0 though once you take the limit? c must be greater than 0 to conclude the test with b sub n
            – Luke D
            1 hour ago












          • @LukeD I don't get $c = 0$. For example, if you use my argument above you get $c = sqrt{5}$.
            – Mustafa Said
            1 hour ago












          • This can also be done with the limit comparison test, which is what the OP asked for.
            – David
            1 hour ago










          • Yeah im curious to see the limit comparison test. But regardless, thanks for the soln.
            – Luke D
            1 hour ago


















          • Wont c be equal to 0 though once you take the limit? c must be greater than 0 to conclude the test with b sub n
            – Luke D
            1 hour ago












          • @LukeD I don't get $c = 0$. For example, if you use my argument above you get $c = sqrt{5}$.
            – Mustafa Said
            1 hour ago












          • This can also be done with the limit comparison test, which is what the OP asked for.
            – David
            1 hour ago










          • Yeah im curious to see the limit comparison test. But regardless, thanks for the soln.
            – Luke D
            1 hour ago
















          Wont c be equal to 0 though once you take the limit? c must be greater than 0 to conclude the test with b sub n
          – Luke D
          1 hour ago






          Wont c be equal to 0 though once you take the limit? c must be greater than 0 to conclude the test with b sub n
          – Luke D
          1 hour ago














          @LukeD I don't get $c = 0$. For example, if you use my argument above you get $c = sqrt{5}$.
          – Mustafa Said
          1 hour ago






          @LukeD I don't get $c = 0$. For example, if you use my argument above you get $c = sqrt{5}$.
          – Mustafa Said
          1 hour ago














          This can also be done with the limit comparison test, which is what the OP asked for.
          – David
          1 hour ago




          This can also be done with the limit comparison test, which is what the OP asked for.
          – David
          1 hour ago












          Yeah im curious to see the limit comparison test. But regardless, thanks for the soln.
          – Luke D
          1 hour ago




          Yeah im curious to see the limit comparison test. But regardless, thanks for the soln.
          – Luke D
          1 hour ago










          up vote
          3
          down vote













          For this sort of thing it is strongly advised to do a rough calculation first. We have
          $$sqrt{frac{n+4}{n^4+4}}approxsqrt{frac{n}{n^4}}=frac1{n^{3/2}} ,$$
          which suggests comparing with
          $$sumfrac1{n^{3/2}} .$$
          We have
          $$sqrt{frac{n+4}{n^4+4}}bigg/frac1{n^{3/2}}=sqrt{frac{n^4+4n^3}{n^4+4}}
          =sqrt{frac{1+4n^{-1}}{1+4n^{-4}}}to1quadhbox{as $ntoinfty$} .$$
          Since this limit exists and is finite and not zero, and we know that
          $$sumfrac1{n^{3/2}}$$
          converges, your series converges too. (Doesn't diverge!!!)






          share|cite|improve this answer





















          • So the answer our teacher gave us is wrong... thanks for the heads up.
            – Luke D
            1 hour ago










          • If your teacher said the series diverges, yes, that's wrong.
            – David
            1 hour ago















          up vote
          3
          down vote













          For this sort of thing it is strongly advised to do a rough calculation first. We have
          $$sqrt{frac{n+4}{n^4+4}}approxsqrt{frac{n}{n^4}}=frac1{n^{3/2}} ,$$
          which suggests comparing with
          $$sumfrac1{n^{3/2}} .$$
          We have
          $$sqrt{frac{n+4}{n^4+4}}bigg/frac1{n^{3/2}}=sqrt{frac{n^4+4n^3}{n^4+4}}
          =sqrt{frac{1+4n^{-1}}{1+4n^{-4}}}to1quadhbox{as $ntoinfty$} .$$
          Since this limit exists and is finite and not zero, and we know that
          $$sumfrac1{n^{3/2}}$$
          converges, your series converges too. (Doesn't diverge!!!)






          share|cite|improve this answer





















          • So the answer our teacher gave us is wrong... thanks for the heads up.
            – Luke D
            1 hour ago










          • If your teacher said the series diverges, yes, that's wrong.
            – David
            1 hour ago













          up vote
          3
          down vote










          up vote
          3
          down vote









          For this sort of thing it is strongly advised to do a rough calculation first. We have
          $$sqrt{frac{n+4}{n^4+4}}approxsqrt{frac{n}{n^4}}=frac1{n^{3/2}} ,$$
          which suggests comparing with
          $$sumfrac1{n^{3/2}} .$$
          We have
          $$sqrt{frac{n+4}{n^4+4}}bigg/frac1{n^{3/2}}=sqrt{frac{n^4+4n^3}{n^4+4}}
          =sqrt{frac{1+4n^{-1}}{1+4n^{-4}}}to1quadhbox{as $ntoinfty$} .$$
          Since this limit exists and is finite and not zero, and we know that
          $$sumfrac1{n^{3/2}}$$
          converges, your series converges too. (Doesn't diverge!!!)






          share|cite|improve this answer












          For this sort of thing it is strongly advised to do a rough calculation first. We have
          $$sqrt{frac{n+4}{n^4+4}}approxsqrt{frac{n}{n^4}}=frac1{n^{3/2}} ,$$
          which suggests comparing with
          $$sumfrac1{n^{3/2}} .$$
          We have
          $$sqrt{frac{n+4}{n^4+4}}bigg/frac1{n^{3/2}}=sqrt{frac{n^4+4n^3}{n^4+4}}
          =sqrt{frac{1+4n^{-1}}{1+4n^{-4}}}to1quadhbox{as $ntoinfty$} .$$
          Since this limit exists and is finite and not zero, and we know that
          $$sumfrac1{n^{3/2}}$$
          converges, your series converges too. (Doesn't diverge!!!)







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 1 hour ago









          David

          67.6k663126




          67.6k663126












          • So the answer our teacher gave us is wrong... thanks for the heads up.
            – Luke D
            1 hour ago










          • If your teacher said the series diverges, yes, that's wrong.
            – David
            1 hour ago


















          • So the answer our teacher gave us is wrong... thanks for the heads up.
            – Luke D
            1 hour ago










          • If your teacher said the series diverges, yes, that's wrong.
            – David
            1 hour ago
















          So the answer our teacher gave us is wrong... thanks for the heads up.
          – Luke D
          1 hour ago




          So the answer our teacher gave us is wrong... thanks for the heads up.
          – Luke D
          1 hour ago












          If your teacher said the series diverges, yes, that's wrong.
          – David
          1 hour ago




          If your teacher said the series diverges, yes, that's wrong.
          – David
          1 hour ago


















          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.





          Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


          Please pay close attention to the following guidance:


          • 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.


          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%2f3044811%2fhow-does-this-series-diverge-by-limit-comparison-test%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

          404 Error Contact Form 7 ajax form submitting

          Image
          0 The contact form does not work, and affects loading after one try. The chrome console shows the error message: POST http://example.com/wp-json/contact-form-7/v1/contact-forms/50/feedback 404 (Not Found) send @ jquery.js?ver=1.12.4:4 ajax @ jquery.js?ver=1.12.4:4 wpcf7.submit @ scripts.js?ver=5.0.3:346 (anonymous) @ scripts.js?ver=5.0.3:53 dispatch @ jquery.js?ver=1.12.4:3 r.handle @ jquery.js?ver=1.12.4:3 How to solve this issue? thanks in advance javascript wordpress contact-form-7 share | improve this question edited Jul 17 '18 at 5:39 Kiran Shahi 4,745 3 16 43 ...
          Read more

          How to know if a Active Directory user can login interactively

          Image
          1 I would like to know if and how is it possible to know if a AD user can login interactively (on a server) in a domain. I need to know if I can find it out using a LDAP search. Thanks. windows active-directory login user-permissions user-profile share | improve this question edited 1 hour ago yagmoth555 ♦ 11.6k 3 17 42 asked 1 hour ago Luigi Luigi 6 1 New contributor ...
          Read more

          Refactoring coordinates for Minecraft Pi buildings written in Python

          Image
          0 $begingroup$ I'm fairly new to coding and I wrote this script in Python for a house with a roof in Minecraft Pi. I've found different ways to write the coordinates to make the code base shorter, but I'm wondering if it is possible for me to refactor it any further so I don't have to list each coordinate line by line. import mcpi.minecraft as minecraft import mcpi.block as block mc = minecraft.Minecraft.create() p = mc.player.getTilePos() mc.setBlocks(p.x+1, p.y, p.z+1, #Main structure starting coordinates (DO NOT ALTER) p.x+19, p.y+4, p.z+6, #Structure size/coordinates (Alters cube size) block.STONE_BRICK) mc.setBlocks(p.x+2, p.y+1, p.z+2, #Hollowing out main structure interior p.x+18, p.y+3, p.z+5, block.AIR) mc.setBlocks(p.x, p.y...
          Read more