Does this power sequence converge or diverge? If it converges, what is the limit?












2












$begingroup$


Say I have this sequence:



$$a_n = frac{n^2}{sqrt{n^3 + 4n}}$$



Again, I don't think I can divide the numerator and denominator by $n^{1.5}$... that seems like it complicates things. What else can I do?



I can't square the top and bottom because that changes the value of the general sequence. Can I divide by $n^2$?



Is this valid:



$$a_n = frac{1}{sqrt{frac{n^3}{n^4} + frac{4}{n}}}$$






sequences-and-series







share|cite|improve this question











$endgroup$












  • $begingroup$
    What are you trying to do with the sequence? Are you trying to determine if it converges / find its limit? In your last identity, you should have $4/n^3$ in the denominator.
    $endgroup$
    – MisterRiemann
    9 hours ago


















2












$begingroup$


Say I have this sequence:



$$a_n = frac{n^2}{sqrt{n^3 + 4n}}$$



Again, I don't think I can divide the numerator and denominator by $n^{1.5}$... that seems like it complicates things. What else can I do?



I can't square the top and bottom because that changes the value of the general sequence. Can I divide by $n^2$?



Is this valid:



$$a_n = frac{1}{sqrt{frac{n^3}{n^4} + frac{4}{n}}}$$






sequences-and-series







share|cite|improve this question











$endgroup$












  • $begingroup$
    What are you trying to do with the sequence? Are you trying to determine if it converges / find its limit? In your last identity, you should have $4/n^3$ in the denominator.
    $endgroup$
    – MisterRiemann
    9 hours ago
















2












2








2





$begingroup$


Say I have this sequence:



$$a_n = frac{n^2}{sqrt{n^3 + 4n}}$$



Again, I don't think I can divide the numerator and denominator by $n^{1.5}$... that seems like it complicates things. What else can I do?



I can't square the top and bottom because that changes the value of the general sequence. Can I divide by $n^2$?



Is this valid:



$$a_n = frac{1}{sqrt{frac{n^3}{n^4} + frac{4}{n}}}$$






sequences-and-series







share|cite|improve this question











$endgroup$




Say I have this sequence:



$$a_n = frac{n^2}{sqrt{n^3 + 4n}}$$



Again, I don't think I can divide the numerator and denominator by $n^{1.5}$... that seems like it complicates things. What else can I do?



I can't square the top and bottom because that changes the value of the general sequence. Can I divide by $n^2$?



Is this valid:



$$a_n = frac{1}{sqrt{frac{n^3}{n^4} + frac{4}{n}}}$$






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








edited 9 hours ago







Jwan622

















asked 9 hours ago









Jwan622Jwan622

2,30211632




2,30211632












  • $begingroup$
    What are you trying to do with the sequence? Are you trying to determine if it converges / find its limit? In your last identity, you should have $4/n^3$ in the denominator.
    $endgroup$
    – MisterRiemann
    9 hours ago




















  • $begingroup$
    What are you trying to do with the sequence? Are you trying to determine if it converges / find its limit? In your last identity, you should have $4/n^3$ in the denominator.
    $endgroup$
    – MisterRiemann
    9 hours ago


















$begingroup$
What are you trying to do with the sequence? Are you trying to determine if it converges / find its limit? In your last identity, you should have $4/n^3$ in the denominator.
$endgroup$
– MisterRiemann
9 hours ago






$begingroup$
What are you trying to do with the sequence? Are you trying to determine if it converges / find its limit? In your last identity, you should have $4/n^3$ in the denominator.
$endgroup$
– MisterRiemann
9 hours ago












3 Answers
3






active

oldest

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3












$begingroup$

You can easily find a divergent minorant:



$$frac{n^2}{sqrt{n^3 + 4n}} ge frac{n^2}{sqrt{n^3 + 4n^color{blue}{3}}} = sqrt{frac{n}{5}} to +infty$$






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    Hint: It is $$sqrt{frac{n^4}{n^3+4n}}$$ and this is divergent.






    share|cite|improve this answer











    $endgroup$





















      1












      $begingroup$

      We have:



      $$a_n = frac{sqrt{n} }{sqrt{1 + frac{4}{n^2}}}$$



      You can see that the denominator tends to 1, so that $a_n$ clearly diverges, behaving asymptotically as $sqrt{n}$.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        How did you get to here?
        $endgroup$
        – Jwan622
        9 hours ago










      • $begingroup$
        Multiply by $frac{n^{1.5}}{n^{1.5}}$.
        $endgroup$
        – Matthew Masarik
        9 hours ago










      • $begingroup$
        Can you flesh that out a bit? Don't you mean divide top and bottom by $n^{1.5}$
        $endgroup$
        – Jwan622
        8 hours ago













      Your Answer





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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      You can easily find a divergent minorant:



      $$frac{n^2}{sqrt{n^3 + 4n}} ge frac{n^2}{sqrt{n^3 + 4n^color{blue}{3}}} = sqrt{frac{n}{5}} to +infty$$






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        You can easily find a divergent minorant:



        $$frac{n^2}{sqrt{n^3 + 4n}} ge frac{n^2}{sqrt{n^3 + 4n^color{blue}{3}}} = sqrt{frac{n}{5}} to +infty$$






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          You can easily find a divergent minorant:



          $$frac{n^2}{sqrt{n^3 + 4n}} ge frac{n^2}{sqrt{n^3 + 4n^color{blue}{3}}} = sqrt{frac{n}{5}} to +infty$$






          share|cite|improve this answer









          $endgroup$



          You can easily find a divergent minorant:



          $$frac{n^2}{sqrt{n^3 + 4n}} ge frac{n^2}{sqrt{n^3 + 4n^color{blue}{3}}} = sqrt{frac{n}{5}} to +infty$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 9 hours ago









          StackTDStackTD

          24.1k2254




          24.1k2254























              3












              $begingroup$

              Hint: It is $$sqrt{frac{n^4}{n^3+4n}}$$ and this is divergent.






              share|cite|improve this answer











              $endgroup$


















                3












                $begingroup$

                Hint: It is $$sqrt{frac{n^4}{n^3+4n}}$$ and this is divergent.






                share|cite|improve this answer











                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  Hint: It is $$sqrt{frac{n^4}{n^3+4n}}$$ and this is divergent.






                  share|cite|improve this answer











                  $endgroup$



                  Hint: It is $$sqrt{frac{n^4}{n^3+4n}}$$ and this is divergent.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited 8 hours ago

























                  answered 9 hours ago









                  Dr. Sonnhard GraubnerDr. Sonnhard Graubner

                  78.1k42867




                  78.1k42867























                      1












                      $begingroup$

                      We have:



                      $$a_n = frac{sqrt{n} }{sqrt{1 + frac{4}{n^2}}}$$



                      You can see that the denominator tends to 1, so that $a_n$ clearly diverges, behaving asymptotically as $sqrt{n}$.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        How did you get to here?
                        $endgroup$
                        – Jwan622
                        9 hours ago










                      • $begingroup$
                        Multiply by $frac{n^{1.5}}{n^{1.5}}$.
                        $endgroup$
                        – Matthew Masarik
                        9 hours ago










                      • $begingroup$
                        Can you flesh that out a bit? Don't you mean divide top and bottom by $n^{1.5}$
                        $endgroup$
                        – Jwan622
                        8 hours ago


















                      1












                      $begingroup$

                      We have:



                      $$a_n = frac{sqrt{n} }{sqrt{1 + frac{4}{n^2}}}$$



                      You can see that the denominator tends to 1, so that $a_n$ clearly diverges, behaving asymptotically as $sqrt{n}$.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        How did you get to here?
                        $endgroup$
                        – Jwan622
                        9 hours ago










                      • $begingroup$
                        Multiply by $frac{n^{1.5}}{n^{1.5}}$.
                        $endgroup$
                        – Matthew Masarik
                        9 hours ago










                      • $begingroup$
                        Can you flesh that out a bit? Don't you mean divide top and bottom by $n^{1.5}$
                        $endgroup$
                        – Jwan622
                        8 hours ago
















                      1












                      1








                      1





                      $begingroup$

                      We have:



                      $$a_n = frac{sqrt{n} }{sqrt{1 + frac{4}{n^2}}}$$



                      You can see that the denominator tends to 1, so that $a_n$ clearly diverges, behaving asymptotically as $sqrt{n}$.






                      share|cite|improve this answer









                      $endgroup$



                      We have:



                      $$a_n = frac{sqrt{n} }{sqrt{1 + frac{4}{n^2}}}$$



                      You can see that the denominator tends to 1, so that $a_n$ clearly diverges, behaving asymptotically as $sqrt{n}$.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered 9 hours ago









                      Matthew MasarikMatthew Masarik

                      111




                      111












                      • $begingroup$
                        How did you get to here?
                        $endgroup$
                        – Jwan622
                        9 hours ago










                      • $begingroup$
                        Multiply by $frac{n^{1.5}}{n^{1.5}}$.
                        $endgroup$
                        – Matthew Masarik
                        9 hours ago










                      • $begingroup$
                        Can you flesh that out a bit? Don't you mean divide top and bottom by $n^{1.5}$
                        $endgroup$
                        – Jwan622
                        8 hours ago




















                      • $begingroup$
                        How did you get to here?
                        $endgroup$
                        – Jwan622
                        9 hours ago










                      • $begingroup$
                        Multiply by $frac{n^{1.5}}{n^{1.5}}$.
                        $endgroup$
                        – Matthew Masarik
                        9 hours ago










                      • $begingroup$
                        Can you flesh that out a bit? Don't you mean divide top and bottom by $n^{1.5}$
                        $endgroup$
                        – Jwan622
                        8 hours ago


















                      $begingroup$
                      How did you get to here?
                      $endgroup$
                      – Jwan622
                      9 hours ago




                      $begingroup$
                      How did you get to here?
                      $endgroup$
                      – Jwan622
                      9 hours ago












                      $begingroup$
                      Multiply by $frac{n^{1.5}}{n^{1.5}}$.
                      $endgroup$
                      – Matthew Masarik
                      9 hours ago




                      $begingroup$
                      Multiply by $frac{n^{1.5}}{n^{1.5}}$.
                      $endgroup$
                      – Matthew Masarik
                      9 hours ago












                      $begingroup$
                      Can you flesh that out a bit? Don't you mean divide top and bottom by $n^{1.5}$
                      $endgroup$
                      – Jwan622
                      8 hours ago






                      $begingroup$
                      Can you flesh that out a bit? Don't you mean divide top and bottom by $n^{1.5}$
                      $endgroup$
                      – Jwan622
                      8 hours ago




















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