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


Why does this pattern fail (sometimes) for the continued fraction convergents of $sqrt2$?Does this sequence always give a square number?Does the sequence $A_n = fracsin (n)sqrtn$ converge to $0$?Do this series converge or diverge?Does this sequence converge $a_n=frac 3^n+25^n $Given: $sum n a_n$ is convergent. To prove: The sequence $a_n$ convergesHow do you find the value of $sum_r=1^infty frac6^r(3^r-2^r)(3^r+1 - 2^r+1) $?Confirming that the sequence $a_n = fracsqrtncosnsqrtn^3-1$ convergesDoes this non-monotonic sequence converge?Finding the limit of a sequence involving n-th roots.













2












$begingroup$


Say I have this sequence:



$$a_n = fracn^2sqrtn^3 + 4n$$



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:



$$a_n = frac1sqrtfracn^3n^4 + frac4n$$






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
    7 hours ago
















2












$begingroup$


Say I have this sequence:



$$a_n = fracn^2sqrtn^3 + 4n$$



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:



$$a_n = frac1sqrtfracn^3n^4 + frac4n$$






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
    7 hours ago














2












2








2





$begingroup$


Say I have this sequence:



$$a_n = fracn^2sqrtn^3 + 4n$$



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:



$$a_n = frac1sqrtfracn^3n^4 + frac4n$$






sequences-and-series







share|cite|improve this question











$endgroup$




Say I have this sequence:



$$a_n = fracn^2sqrtn^3 + 4n$$



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:



$$a_n = frac1sqrtfracn^3n^4 + frac4n$$






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 7 hours ago







Jwan622

















asked 7 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
    7 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
    7 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
7 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
7 hours ago











3 Answers
3






active

oldest

votes


















3












$begingroup$

You can easily find a divergent minorant:



$$fracn^2sqrtn^3 + 4n ge fracn^2sqrtn^3 + 4n^colorblue3 = sqrtfracn5 to +infty$$






share|cite|improve this answer









$endgroup$




















    3












    $begingroup$

    Hint: It is $$sqrtfracn^4n^3+4n$$ and this is divergent.






    share|cite|improve this answer











    $endgroup$




















      1












      $begingroup$

      We have:



      $$a_n = fracsqrtn sqrt1 + frac4n^2$$



      You can see that the denominator tends to 1, so that $a_n$ clearly diverges, behaving asymptotically as $sqrtn$.






      share|cite|improve this answer









      $endgroup$












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










      • $begingroup$
        Multiply by $fracn^1.5n^1.5$.
        $endgroup$
        – Matthew Masarik
        7 hours ago










      • $begingroup$
        Can you flesh that out a bit? Don't you mean divide top and bottom by $n^1.5$
        $endgroup$
        – Jwan622
        6 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:



      $$fracn^2sqrtn^3 + 4n ge fracn^2sqrtn^3 + 4n^colorblue3 = sqrtfracn5 to +infty$$






      share|cite|improve this answer









      $endgroup$

















        3












        $begingroup$

        You can easily find a divergent minorant:



        $$fracn^2sqrtn^3 + 4n ge fracn^2sqrtn^3 + 4n^colorblue3 = sqrtfracn5 to +infty$$






        share|cite|improve this answer









        $endgroup$















          3












          3








          3





          $begingroup$

          You can easily find a divergent minorant:



          $$fracn^2sqrtn^3 + 4n ge fracn^2sqrtn^3 + 4n^colorblue3 = sqrtfracn5 to +infty$$






          share|cite|improve this answer









          $endgroup$



          You can easily find a divergent minorant:



          $$fracn^2sqrtn^3 + 4n ge fracn^2sqrtn^3 + 4n^colorblue3 = sqrtfracn5 to +infty$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 7 hours ago









          StackTDStackTD

          24.1k2254




          24.1k2254





















              3












              $begingroup$

              Hint: It is $$sqrtfracn^4n^3+4n$$ and this is divergent.






              share|cite|improve this answer











              $endgroup$

















                3












                $begingroup$

                Hint: It is $$sqrtfracn^4n^3+4n$$ and this is divergent.






                share|cite|improve this answer











                $endgroup$















                  3












                  3








                  3





                  $begingroup$

                  Hint: It is $$sqrtfracn^4n^3+4n$$ and this is divergent.






                  share|cite|improve this answer











                  $endgroup$



                  Hint: It is $$sqrtfracn^4n^3+4n$$ and this is divergent.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited 7 hours ago

























                  answered 7 hours ago









                  Dr. Sonnhard GraubnerDr. Sonnhard Graubner

                  78.1k42867




                  78.1k42867





















                      1












                      $begingroup$

                      We have:



                      $$a_n = fracsqrtn sqrt1 + frac4n^2$$



                      You can see that the denominator tends to 1, so that $a_n$ clearly diverges, behaving asymptotically as $sqrtn$.






                      share|cite|improve this answer









                      $endgroup$












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










                      • $begingroup$
                        Multiply by $fracn^1.5n^1.5$.
                        $endgroup$
                        – Matthew Masarik
                        7 hours ago










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
















                      1












                      $begingroup$

                      We have:



                      $$a_n = fracsqrtn sqrt1 + frac4n^2$$



                      You can see that the denominator tends to 1, so that $a_n$ clearly diverges, behaving asymptotically as $sqrtn$.






                      share|cite|improve this answer









                      $endgroup$












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










                      • $begingroup$
                        Multiply by $fracn^1.5n^1.5$.
                        $endgroup$
                        – Matthew Masarik
                        7 hours ago










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














                      1












                      1








                      1





                      $begingroup$

                      We have:



                      $$a_n = fracsqrtn sqrt1 + frac4n^2$$



                      You can see that the denominator tends to 1, so that $a_n$ clearly diverges, behaving asymptotically as $sqrtn$.






                      share|cite|improve this answer









                      $endgroup$



                      We have:



                      $$a_n = fracsqrtn sqrt1 + frac4n^2$$



                      You can see that the denominator tends to 1, so that $a_n$ clearly diverges, behaving asymptotically as $sqrtn$.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered 7 hours ago









                      Matthew MasarikMatthew Masarik

                      111




                      111











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










                      • $begingroup$
                        Multiply by $fracn^1.5n^1.5$.
                        $endgroup$
                        – Matthew Masarik
                        7 hours ago










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

















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










                      • $begingroup$
                        Multiply by $fracn^1.5n^1.5$.
                        $endgroup$
                        – Matthew Masarik
                        7 hours ago










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
















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




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












                      $begingroup$
                      Multiply by $fracn^1.5n^1.5$.
                      $endgroup$
                      – Matthew Masarik
                      7 hours ago




                      $begingroup$
                      Multiply by $fracn^1.5n^1.5$.
                      $endgroup$
                      – Matthew Masarik
                      7 hours ago












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





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


















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