Does the series $sum_{n=1}^infty frac{n}{sqrt[3]{8n^5-1}}$ Converge?












1












$begingroup$


$$sum_{n=1}^infty frac{n}{sqrt[3]{8n^5-1}}$$



From the tests that I know of:



Divergence Test: The limit is ≠ to a constant, so inconclusive.



Geometric series: I don't think this could be written in that manner.



Comparison Test/Lim Comparison: Compare to $$frac{n}{8n^{frac{5}{3}}}$$



Integral Test: I can't think of a integration method that would work here.



Alternating Series/Root Test don't apply.



Ratio Test: The limit is 1 so inconclusive.



Perhaps I'm making a mistake throughout the methods I've tried, but I'm lost. Using these tests, is it possible to find whether or not it converges or diverges?






sequences-and-series







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$endgroup$












  • $begingroup$
    it's same as $frac{1}{n^{2/3}}$, so no, it doesn't converge
    $endgroup$
    – mathworker21
    Dec 12 '18 at 5:36
















1












$begingroup$


$$sum_{n=1}^infty frac{n}{sqrt[3]{8n^5-1}}$$



From the tests that I know of:



Divergence Test: The limit is ≠ to a constant, so inconclusive.



Geometric series: I don't think this could be written in that manner.



Comparison Test/Lim Comparison: Compare to $$frac{n}{8n^{frac{5}{3}}}$$



Integral Test: I can't think of a integration method that would work here.



Alternating Series/Root Test don't apply.



Ratio Test: The limit is 1 so inconclusive.



Perhaps I'm making a mistake throughout the methods I've tried, but I'm lost. Using these tests, is it possible to find whether or not it converges or diverges?






sequences-and-series







share|cite|improve this question









$endgroup$












  • $begingroup$
    it's same as $frac{1}{n^{2/3}}$, so no, it doesn't converge
    $endgroup$
    – mathworker21
    Dec 12 '18 at 5:36














1












1








1





$begingroup$


$$sum_{n=1}^infty frac{n}{sqrt[3]{8n^5-1}}$$



From the tests that I know of:



Divergence Test: The limit is ≠ to a constant, so inconclusive.



Geometric series: I don't think this could be written in that manner.



Comparison Test/Lim Comparison: Compare to $$frac{n}{8n^{frac{5}{3}}}$$



Integral Test: I can't think of a integration method that would work here.



Alternating Series/Root Test don't apply.



Ratio Test: The limit is 1 so inconclusive.



Perhaps I'm making a mistake throughout the methods I've tried, but I'm lost. Using these tests, is it possible to find whether or not it converges or diverges?






sequences-and-series







share|cite|improve this question









$endgroup$




$$sum_{n=1}^infty frac{n}{sqrt[3]{8n^5-1}}$$



From the tests that I know of:



Divergence Test: The limit is ≠ to a constant, so inconclusive.



Geometric series: I don't think this could be written in that manner.



Comparison Test/Lim Comparison: Compare to $$frac{n}{8n^{frac{5}{3}}}$$



Integral Test: I can't think of a integration method that would work here.



Alternating Series/Root Test don't apply.



Ratio Test: The limit is 1 so inconclusive.



Perhaps I'm making a mistake throughout the methods I've tried, but I'm lost. Using these tests, is it possible to find whether or not it converges or diverges?






sequences-and-series




sequences-and-series






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 12 '18 at 5:34









user623028user623028

235




235












  • $begingroup$
    it's same as $frac{1}{n^{2/3}}$, so no, it doesn't converge
    $endgroup$
    – mathworker21
    Dec 12 '18 at 5:36


















  • $begingroup$
    it's same as $frac{1}{n^{2/3}}$, so no, it doesn't converge
    $endgroup$
    – mathworker21
    Dec 12 '18 at 5:36
















$begingroup$
it's same as $frac{1}{n^{2/3}}$, so no, it doesn't converge
$endgroup$
– mathworker21
Dec 12 '18 at 5:36




$begingroup$
it's same as $frac{1}{n^{2/3}}$, so no, it doesn't converge
$endgroup$
– mathworker21
Dec 12 '18 at 5:36










2 Answers
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$begingroup$

Use comparison test,$$frac{n}{sqrt[3]{8n^5-1}} ge frac{n}{sqrt[3]{8n^5}} = frac{1}{2n^{frac53-1}}=frac{1}{2n^{frac23}}$$



Now, use $p$-series to make conclusion that it diverges.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    We have that



    $$frac{n}{sqrt[3]{8n^5-1}}sim frac1{2n^{2/3}}$$



    therefore we can conclude that the given series diverges by limit comparison test with $sum frac1{n^{2/3}}$.






    share|cite|improve this answer









    $endgroup$













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      2 Answers
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      2 Answers
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      4












      $begingroup$

      Use comparison test,$$frac{n}{sqrt[3]{8n^5-1}} ge frac{n}{sqrt[3]{8n^5}} = frac{1}{2n^{frac53-1}}=frac{1}{2n^{frac23}}$$



      Now, use $p$-series to make conclusion that it diverges.






      share|cite|improve this answer









      $endgroup$


















        4












        $begingroup$

        Use comparison test,$$frac{n}{sqrt[3]{8n^5-1}} ge frac{n}{sqrt[3]{8n^5}} = frac{1}{2n^{frac53-1}}=frac{1}{2n^{frac23}}$$



        Now, use $p$-series to make conclusion that it diverges.






        share|cite|improve this answer









        $endgroup$
















          4












          4








          4





          $begingroup$

          Use comparison test,$$frac{n}{sqrt[3]{8n^5-1}} ge frac{n}{sqrt[3]{8n^5}} = frac{1}{2n^{frac53-1}}=frac{1}{2n^{frac23}}$$



          Now, use $p$-series to make conclusion that it diverges.






          share|cite|improve this answer









          $endgroup$



          Use comparison test,$$frac{n}{sqrt[3]{8n^5-1}} ge frac{n}{sqrt[3]{8n^5}} = frac{1}{2n^{frac53-1}}=frac{1}{2n^{frac23}}$$



          Now, use $p$-series to make conclusion that it diverges.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 12 '18 at 5:40









          Siong Thye GohSiong Thye Goh

          102k1466118




          102k1466118























              0












              $begingroup$

              We have that



              $$frac{n}{sqrt[3]{8n^5-1}}sim frac1{2n^{2/3}}$$



              therefore we can conclude that the given series diverges by limit comparison test with $sum frac1{n^{2/3}}$.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                We have that



                $$frac{n}{sqrt[3]{8n^5-1}}sim frac1{2n^{2/3}}$$



                therefore we can conclude that the given series diverges by limit comparison test with $sum frac1{n^{2/3}}$.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  We have that



                  $$frac{n}{sqrt[3]{8n^5-1}}sim frac1{2n^{2/3}}$$



                  therefore we can conclude that the given series diverges by limit comparison test with $sum frac1{n^{2/3}}$.






                  share|cite|improve this answer









                  $endgroup$



                  We have that



                  $$frac{n}{sqrt[3]{8n^5-1}}sim frac1{2n^{2/3}}$$



                  therefore we can conclude that the given series diverges by limit comparison test with $sum frac1{n^{2/3}}$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 12 '18 at 7:16









                  gimusigimusi

                  92.9k84494




                  92.9k84494






























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