How does this series diverge by limit comparison test?

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP












3














How does this series diverge by limit comparison test? $$sum_n=1^infty sqrtfracn+4n^4+4$$



I origionally tried using $frac1n^2$ for the comparison, but I'm pretty sure it has to be $fracnn^2$ to properly compare.










share|cite|improve this question




























    3














    How does this series diverge by limit comparison test? $$sum_n=1^infty sqrtfracn+4n^4+4$$



    I origionally tried using $frac1n^2$ for the comparison, but I'm pretty sure it has to be $fracnn^2$ to properly compare.










    share|cite|improve this question


























      3












      3








      3


      3





      How does this series diverge by limit comparison test? $$sum_n=1^infty sqrtfracn+4n^4+4$$



      I origionally tried using $frac1n^2$ for the comparison, but I'm pretty sure it has to be $fracnn^2$ to properly compare.










      share|cite|improve this question















      How does this series diverge by limit comparison test? $$sum_n=1^infty sqrtfracn+4n^4+4$$



      I origionally tried using $frac1n^2$ for the comparison, but I'm pretty sure it has to be $fracnn^2$ to properly compare.







      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 Dec 18 at 5:36









      Chinnapparaj R

      5,2751826




      5,2751826










      asked Dec 18 at 5:33









      Luke D

      796




      796




















          2 Answers
          2






          active

          oldest

          votes


















          9














          $fracn+4n^4+4 leq fracn+4nn^4 leq frac5n^3$,



          hence, $sqrtfracn+4n^4+4 leq fracsqrt5n^3/2$



          so the series converges by comparison with convergent p-series $sum frac1n^3/2$






          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
            Dec 18 at 5:41











          • @LukeD I don't get $c = 0$. For example, if you use my argument above you get $c = sqrt5$.
            – Mustafa Said
            Dec 18 at 5:43











          • This can also be done with the limit comparison test, which is what the OP asked for.
            – David
            Dec 18 at 5:45










          • Yeah im curious to see the limit comparison test. But regardless, thanks for the soln.
            – Luke D
            Dec 18 at 5:48


















          8














          For this sort of thing it is strongly advised to do a rough calculation first. We have
          $$sqrtfracn+4n^4+4approxsqrtfracnn^4=frac1n^3/2 ,$$
          which suggests comparing with
          $$sumfrac1n^3/2 .$$
          We have
          $$sqrtfracn+4n^4+4bigg/frac1n^3/2=sqrtfracn^4+4n^3n^4+4
          =sqrtfrac1+4n^-11+4n^-4to1quadhboxas $ntoinfty$ .$$

          Since this limit exists and is finite and not zero, and we know that
          $$sumfrac1n^3/2$$
          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
            Dec 18 at 5:46










          • If your teacher said the series diverges, yes, that's wrong.
            – David
            Dec 18 at 5:47










          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%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









          9














          $fracn+4n^4+4 leq fracn+4nn^4 leq frac5n^3$,



          hence, $sqrtfracn+4n^4+4 leq fracsqrt5n^3/2$



          so the series converges by comparison with convergent p-series $sum frac1n^3/2$






          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
            Dec 18 at 5:41











          • @LukeD I don't get $c = 0$. For example, if you use my argument above you get $c = sqrt5$.
            – Mustafa Said
            Dec 18 at 5:43











          • This can also be done with the limit comparison test, which is what the OP asked for.
            – David
            Dec 18 at 5:45










          • Yeah im curious to see the limit comparison test. But regardless, thanks for the soln.
            – Luke D
            Dec 18 at 5:48















          9














          $fracn+4n^4+4 leq fracn+4nn^4 leq frac5n^3$,



          hence, $sqrtfracn+4n^4+4 leq fracsqrt5n^3/2$



          so the series converges by comparison with convergent p-series $sum frac1n^3/2$






          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
            Dec 18 at 5:41











          • @LukeD I don't get $c = 0$. For example, if you use my argument above you get $c = sqrt5$.
            – Mustafa Said
            Dec 18 at 5:43











          • This can also be done with the limit comparison test, which is what the OP asked for.
            – David
            Dec 18 at 5:45










          • Yeah im curious to see the limit comparison test. But regardless, thanks for the soln.
            – Luke D
            Dec 18 at 5:48













          9












          9








          9






          $fracn+4n^4+4 leq fracn+4nn^4 leq frac5n^3$,



          hence, $sqrtfracn+4n^4+4 leq fracsqrt5n^3/2$



          so the series converges by comparison with convergent p-series $sum frac1n^3/2$






          share|cite|improve this answer












          $fracn+4n^4+4 leq fracn+4nn^4 leq frac5n^3$,



          hence, $sqrtfracn+4n^4+4 leq fracsqrt5n^3/2$



          so the series converges by comparison with convergent p-series $sum frac1n^3/2$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 18 at 5:37









          Mustafa Said

          2,9411913




          2,9411913











          • 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
            Dec 18 at 5:41











          • @LukeD I don't get $c = 0$. For example, if you use my argument above you get $c = sqrt5$.
            – Mustafa Said
            Dec 18 at 5:43











          • This can also be done with the limit comparison test, which is what the OP asked for.
            – David
            Dec 18 at 5:45










          • Yeah im curious to see the limit comparison test. But regardless, thanks for the soln.
            – Luke D
            Dec 18 at 5:48
















          • 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
            Dec 18 at 5:41











          • @LukeD I don't get $c = 0$. For example, if you use my argument above you get $c = sqrt5$.
            – Mustafa Said
            Dec 18 at 5:43











          • This can also be done with the limit comparison test, which is what the OP asked for.
            – David
            Dec 18 at 5:45










          • Yeah im curious to see the limit comparison test. But regardless, thanks for the soln.
            – Luke D
            Dec 18 at 5:48















          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
          Dec 18 at 5:41





          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
          Dec 18 at 5:41













          @LukeD I don't get $c = 0$. For example, if you use my argument above you get $c = sqrt5$.
          – Mustafa Said
          Dec 18 at 5:43





          @LukeD I don't get $c = 0$. For example, if you use my argument above you get $c = sqrt5$.
          – Mustafa Said
          Dec 18 at 5:43













          This can also be done with the limit comparison test, which is what the OP asked for.
          – David
          Dec 18 at 5:45




          This can also be done with the limit comparison test, which is what the OP asked for.
          – David
          Dec 18 at 5:45












          Yeah im curious to see the limit comparison test. But regardless, thanks for the soln.
          – Luke D
          Dec 18 at 5:48




          Yeah im curious to see the limit comparison test. But regardless, thanks for the soln.
          – Luke D
          Dec 18 at 5:48











          8














          For this sort of thing it is strongly advised to do a rough calculation first. We have
          $$sqrtfracn+4n^4+4approxsqrtfracnn^4=frac1n^3/2 ,$$
          which suggests comparing with
          $$sumfrac1n^3/2 .$$
          We have
          $$sqrtfracn+4n^4+4bigg/frac1n^3/2=sqrtfracn^4+4n^3n^4+4
          =sqrtfrac1+4n^-11+4n^-4to1quadhboxas $ntoinfty$ .$$

          Since this limit exists and is finite and not zero, and we know that
          $$sumfrac1n^3/2$$
          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
            Dec 18 at 5:46










          • If your teacher said the series diverges, yes, that's wrong.
            – David
            Dec 18 at 5:47















          8














          For this sort of thing it is strongly advised to do a rough calculation first. We have
          $$sqrtfracn+4n^4+4approxsqrtfracnn^4=frac1n^3/2 ,$$
          which suggests comparing with
          $$sumfrac1n^3/2 .$$
          We have
          $$sqrtfracn+4n^4+4bigg/frac1n^3/2=sqrtfracn^4+4n^3n^4+4
          =sqrtfrac1+4n^-11+4n^-4to1quadhboxas $ntoinfty$ .$$

          Since this limit exists and is finite and not zero, and we know that
          $$sumfrac1n^3/2$$
          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
            Dec 18 at 5:46










          • If your teacher said the series diverges, yes, that's wrong.
            – David
            Dec 18 at 5:47













          8












          8








          8






          For this sort of thing it is strongly advised to do a rough calculation first. We have
          $$sqrtfracn+4n^4+4approxsqrtfracnn^4=frac1n^3/2 ,$$
          which suggests comparing with
          $$sumfrac1n^3/2 .$$
          We have
          $$sqrtfracn+4n^4+4bigg/frac1n^3/2=sqrtfracn^4+4n^3n^4+4
          =sqrtfrac1+4n^-11+4n^-4to1quadhboxas $ntoinfty$ .$$

          Since this limit exists and is finite and not zero, and we know that
          $$sumfrac1n^3/2$$
          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
          $$sqrtfracn+4n^4+4approxsqrtfracnn^4=frac1n^3/2 ,$$
          which suggests comparing with
          $$sumfrac1n^3/2 .$$
          We have
          $$sqrtfracn+4n^4+4bigg/frac1n^3/2=sqrtfracn^4+4n^3n^4+4
          =sqrtfrac1+4n^-11+4n^-4to1quadhboxas $ntoinfty$ .$$

          Since this limit exists and is finite and not zero, and we know that
          $$sumfrac1n^3/2$$
          converges, your series converges too. (Doesn't diverge!!!)







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 18 at 5:44









          David

          67.7k663126




          67.7k663126











          • So the answer our teacher gave us is wrong... thanks for the heads up.
            – Luke D
            Dec 18 at 5:46










          • If your teacher said the series diverges, yes, that's wrong.
            – David
            Dec 18 at 5:47
















          • So the answer our teacher gave us is wrong... thanks for the heads up.
            – Luke D
            Dec 18 at 5:46










          • If your teacher said the series diverges, yes, that's wrong.
            – David
            Dec 18 at 5:47















          So the answer our teacher gave us is wrong... thanks for the heads up.
          – Luke D
          Dec 18 at 5:46




          So the answer our teacher gave us is wrong... thanks for the heads up.
          – Luke D
          Dec 18 at 5:46












          If your teacher said the series diverges, yes, that's wrong.
          – David
          Dec 18 at 5:47




          If your teacher said the series diverges, yes, that's wrong.
          – David
          Dec 18 at 5:47

















          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

          How to check contact read email or not when send email to Individual?

          Displaying single band from multi-band raster using QGIS

          How many registers does an x86_64 CPU actually have?