How does this series diverge by limit comparison test?











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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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    up vote
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    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






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      edited 49 mins ago









      Chinnapparaj R

      4,9951825




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      asked 53 mins ago









      Luke D

      615




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          2 Answers
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          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
            45 mins ago












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












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










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










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











          Your Answer





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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

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          active

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          active

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          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
            45 mins ago












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












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










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












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












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










          • Yeah im curious to see the limit comparison test. But regardless, thanks for the soln.
            – Luke D
            38 mins 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 49 mins 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
            45 mins ago












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












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










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












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












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










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














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






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














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




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












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




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










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










          • If your teacher said the series diverges, yes, that's wrong.
            – David
            39 mins 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 41 mins ago









          David

          67.6k663126




          67.6k663126












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










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


















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










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
















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




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












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




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


















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