How does this series diverge by limit comparison test?












3












$begingroup$


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







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    3












    $begingroup$


    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











    $endgroup$















      3












      3








      3


      3



      $begingroup$


      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











      $endgroup$




      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






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      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 18 '18 at 5:36









      Chinnapparaj R

      5,3541828




      5,3541828










      asked Dec 18 '18 at 5:33









      Luke DLuke D

      796




      796






















          2 Answers
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          9












          $begingroup$

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



          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









          $endgroup$













          • $begingroup$
            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
            $endgroup$
            – Luke D
            Dec 18 '18 at 5:41












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












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










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



















          8












          $begingroup$

          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









          $endgroup$













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










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











          Your Answer





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          2 Answers
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          2 Answers
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          9












          $begingroup$

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



          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









          $endgroup$













          • $begingroup$
            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
            $endgroup$
            – Luke D
            Dec 18 '18 at 5:41












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












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










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
















          9












          $begingroup$

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



          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









          $endgroup$













          • $begingroup$
            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
            $endgroup$
            – Luke D
            Dec 18 '18 at 5:41












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












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










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














          9












          9








          9





          $begingroup$

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



          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









          $endgroup$



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



          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 Dec 18 '18 at 5:37









          Mustafa SaidMustafa Said

          2,9511913




          2,9511913












          • $begingroup$
            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
            $endgroup$
            – Luke D
            Dec 18 '18 at 5:41












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












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










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


















          • $begingroup$
            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
            $endgroup$
            – Luke D
            Dec 18 '18 at 5:41












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












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










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
















          $begingroup$
          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
          $endgroup$
          – Luke D
          Dec 18 '18 at 5:41






          $begingroup$
          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
          $endgroup$
          – Luke D
          Dec 18 '18 at 5:41














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






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














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




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












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




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











          8












          $begingroup$

          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









          $endgroup$













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










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
















          8












          $begingroup$

          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









          $endgroup$













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










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














          8












          8








          8





          $begingroup$

          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









          $endgroup$



          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 Dec 18 '18 at 5:44









          DavidDavid

          68.1k664126




          68.1k664126












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










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


















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










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
















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




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












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




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


















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