Determine whether or not $sum_k=1^inftyleft(frac kk+1right)^k^2$ converges. Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Determine whether the series $sum_n=1^infty left ( fracpi2-arctan n right )$ converges or not.How to determine whether $sum_n=1^inftylnleft(fracn+2n+1right)$ converges or diverges.Determine whether the series $sum_n=1^+inftyleft(1+frac1nright)a_n$ is convergent or divergentConvergence for $sum _n=1^infty :fracsqrt[4]n^2-1sqrtn^4-1$Converge? $sum_k=1^inftyfrac sin left(frac1kright) k $Determine whether the series converges or diverges.Determine whether the series $sum_n=1^inftyleft(fracnn+1right)^n^2$ convergesTo test whether $sum_n=1^inftyfracn+22^n+3sinleft[(n+frac12)piright]$ convergesDetermining whether the series: $sum_n=1^infty tanleft(frac1nright) $ convergesDetermine the convergence/divergence of $sum_n=1^inftyfraclnn!n^3$

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Determine whether or not $sum_k=1^inftyleft(frac kk+1right)^k^2$ converges.



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Determine whether the series $sum_n=1^infty left ( fracpi2-arctan n right )$ converges or not.How to determine whether $sum_n=1^inftylnleft(fracn+2n+1right)$ converges or diverges.Determine whether the series $sum_n=1^+inftyleft(1+frac1nright)a_n$ is convergent or divergentConvergence for $sum _n=1^infty :fracsqrt[4]n^2-1sqrtn^4-1$Converge? $sum_k=1^inftyfrac sin left(frac1kright) k $Determine whether the series converges or diverges.Determine whether the series $sum_n=1^inftyleft(fracnn+1right)^n^2$ convergesTo test whether $sum_n=1^inftyfracn+22^n+3sinleft[(n+frac12)piright]$ convergesDetermining whether the series: $sum_n=1^infty tanleft(frac1nright) $ convergesDetermine the convergence/divergence of $sum_n=1^inftyfraclnn!n^3$










1












$begingroup$


$$sum_k=1^inftyleft(frac kk+1right)^k^2$$



Determine whether or not the following series converge.
I am not sure what test to use. I am pretty sure I am unable to use ratio test. Maybe comparison or Kummer, or Raabe. However I am not sure how to start it.






sequences-and-series convergence







share|cite|improve this question











$endgroup$











  • $begingroup$
    You have an exponent with $k$. Your instinct should be to remove it with the root test.
    $endgroup$
    – Simply Beautiful Art
    3 hours ago










  • $begingroup$
    Since its squared, do I take the k^2 root
    $endgroup$
    – MD3
    3 hours ago






  • 1




    $begingroup$
    Does the root test say you take the $k^2$-th root?
    $endgroup$
    – Simply Beautiful Art
    3 hours ago






  • 1




    $begingroup$
    For $kge 1$, we have$$left(frackk+1right)^k^2le e^-k/2$$
    $endgroup$
    – Mark Viola
    3 hours ago















1












$begingroup$


$$sum_k=1^inftyleft(frac kk+1right)^k^2$$



Determine whether or not the following series converge.
I am not sure what test to use. I am pretty sure I am unable to use ratio test. Maybe comparison or Kummer, or Raabe. However I am not sure how to start it.






sequences-and-series convergence







share|cite|improve this question











$endgroup$











  • $begingroup$
    You have an exponent with $k$. Your instinct should be to remove it with the root test.
    $endgroup$
    – Simply Beautiful Art
    3 hours ago










  • $begingroup$
    Since its squared, do I take the k^2 root
    $endgroup$
    – MD3
    3 hours ago






  • 1




    $begingroup$
    Does the root test say you take the $k^2$-th root?
    $endgroup$
    – Simply Beautiful Art
    3 hours ago






  • 1




    $begingroup$
    For $kge 1$, we have$$left(frackk+1right)^k^2le e^-k/2$$
    $endgroup$
    – Mark Viola
    3 hours ago













1












1








1





$begingroup$


$$sum_k=1^inftyleft(frac kk+1right)^k^2$$



Determine whether or not the following series converge.
I am not sure what test to use. I am pretty sure I am unable to use ratio test. Maybe comparison or Kummer, or Raabe. However I am not sure how to start it.






sequences-and-series convergence







share|cite|improve this question











$endgroup$




$$sum_k=1^inftyleft(frac kk+1right)^k^2$$



Determine whether or not the following series converge.
I am not sure what test to use. I am pretty sure I am unable to use ratio test. Maybe comparison or Kummer, or Raabe. However I am not sure how to start it.






sequences-and-series convergence




sequences-and-series convergence






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 3 mins ago









user21820

40.2k544163




40.2k544163










asked 3 hours ago









MD3MD3

462




462











  • $begingroup$
    You have an exponent with $k$. Your instinct should be to remove it with the root test.
    $endgroup$
    – Simply Beautiful Art
    3 hours ago










  • $begingroup$
    Since its squared, do I take the k^2 root
    $endgroup$
    – MD3
    3 hours ago






  • 1




    $begingroup$
    Does the root test say you take the $k^2$-th root?
    $endgroup$
    – Simply Beautiful Art
    3 hours ago






  • 1




    $begingroup$
    For $kge 1$, we have$$left(frackk+1right)^k^2le e^-k/2$$
    $endgroup$
    – Mark Viola
    3 hours ago
















  • $begingroup$
    You have an exponent with $k$. Your instinct should be to remove it with the root test.
    $endgroup$
    – Simply Beautiful Art
    3 hours ago










  • $begingroup$
    Since its squared, do I take the k^2 root
    $endgroup$
    – MD3
    3 hours ago






  • 1




    $begingroup$
    Does the root test say you take the $k^2$-th root?
    $endgroup$
    – Simply Beautiful Art
    3 hours ago






  • 1




    $begingroup$
    For $kge 1$, we have$$left(frackk+1right)^k^2le e^-k/2$$
    $endgroup$
    – Mark Viola
    3 hours ago















$begingroup$
You have an exponent with $k$. Your instinct should be to remove it with the root test.
$endgroup$
– Simply Beautiful Art
3 hours ago




$begingroup$
You have an exponent with $k$. Your instinct should be to remove it with the root test.
$endgroup$
– Simply Beautiful Art
3 hours ago












$begingroup$
Since its squared, do I take the k^2 root
$endgroup$
– MD3
3 hours ago




$begingroup$
Since its squared, do I take the k^2 root
$endgroup$
– MD3
3 hours ago




1




1




$begingroup$
Does the root test say you take the $k^2$-th root?
$endgroup$
– Simply Beautiful Art
3 hours ago




$begingroup$
Does the root test say you take the $k^2$-th root?
$endgroup$
– Simply Beautiful Art
3 hours ago




1




1




$begingroup$
For $kge 1$, we have$$left(frackk+1right)^k^2le e^-k/2$$
$endgroup$
– Mark Viola
3 hours ago




$begingroup$
For $kge 1$, we have$$left(frackk+1right)^k^2le e^-k/2$$
$endgroup$
– Mark Viola
3 hours ago










3 Answers
3






active

oldest

votes


















2












$begingroup$

By Root test, $$limsup_n to infty sqrt[n]left(fracnn+1right)^n^2=limsup_n to infty left(fracnn+1right)^n=limsup_n to infty left(1+frac1nright)^-n=frac1e<1$$



So your series converges!






share|cite|improve this answer











$endgroup$












  • $begingroup$
    How did you know to take the supremum
    $endgroup$
    – MD3
    3 hours ago






  • 1




    $begingroup$
    See the wiki link!
    $endgroup$
    – Chinnapparaj R
    3 hours ago


















1












$begingroup$

Hint: $$left( frackk+1 right)^k sim e^-1 $$






share|cite|improve this answer









$endgroup$




















    0












    $begingroup$

    $$a_k=left(frac kk+1right)^k^2implies log(a_k)=k^2 logleft(frac kk+1right)$$



    $$log(a_k+1)-log(a_k)=(k+1)^2 log left(frack+1k+2right)-k^2 log left(frackk+1right)$$ Using Taylor expansions for large $k$
    $$log(a_k+1)-log(a_k)=-1+frac13 k^2+Oleft(frac1k^3right)$$
    $$frac a_k+1a_k=e^log(a_k+1)-log(a_k)=frac 1 e left(1+frac13 k^2+Oleft(frac1k^3right)right)to frac 1 e $$






    share|cite|improve this answer









    $endgroup$













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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      By Root test, $$limsup_n to infty sqrt[n]left(fracnn+1right)^n^2=limsup_n to infty left(fracnn+1right)^n=limsup_n to infty left(1+frac1nright)^-n=frac1e<1$$



      So your series converges!






      share|cite|improve this answer











      $endgroup$












      • $begingroup$
        How did you know to take the supremum
        $endgroup$
        – MD3
        3 hours ago






      • 1




        $begingroup$
        See the wiki link!
        $endgroup$
        – Chinnapparaj R
        3 hours ago















      2












      $begingroup$

      By Root test, $$limsup_n to infty sqrt[n]left(fracnn+1right)^n^2=limsup_n to infty left(fracnn+1right)^n=limsup_n to infty left(1+frac1nright)^-n=frac1e<1$$



      So your series converges!






      share|cite|improve this answer











      $endgroup$












      • $begingroup$
        How did you know to take the supremum
        $endgroup$
        – MD3
        3 hours ago






      • 1




        $begingroup$
        See the wiki link!
        $endgroup$
        – Chinnapparaj R
        3 hours ago













      2












      2








      2





      $begingroup$

      By Root test, $$limsup_n to infty sqrt[n]left(fracnn+1right)^n^2=limsup_n to infty left(fracnn+1right)^n=limsup_n to infty left(1+frac1nright)^-n=frac1e<1$$



      So your series converges!






      share|cite|improve this answer











      $endgroup$



      By Root test, $$limsup_n to infty sqrt[n]left(fracnn+1right)^n^2=limsup_n to infty left(fracnn+1right)^n=limsup_n to infty left(1+frac1nright)^-n=frac1e<1$$



      So your series converges!







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited 3 hours ago

























      answered 3 hours ago









      Chinnapparaj RChinnapparaj R

      6,54021029




      6,54021029











      • $begingroup$
        How did you know to take the supremum
        $endgroup$
        – MD3
        3 hours ago






      • 1




        $begingroup$
        See the wiki link!
        $endgroup$
        – Chinnapparaj R
        3 hours ago
















      • $begingroup$
        How did you know to take the supremum
        $endgroup$
        – MD3
        3 hours ago






      • 1




        $begingroup$
        See the wiki link!
        $endgroup$
        – Chinnapparaj R
        3 hours ago















      $begingroup$
      How did you know to take the supremum
      $endgroup$
      – MD3
      3 hours ago




      $begingroup$
      How did you know to take the supremum
      $endgroup$
      – MD3
      3 hours ago




      1




      1




      $begingroup$
      See the wiki link!
      $endgroup$
      – Chinnapparaj R
      3 hours ago




      $begingroup$
      See the wiki link!
      $endgroup$
      – Chinnapparaj R
      3 hours ago











      1












      $begingroup$

      Hint: $$left( frackk+1 right)^k sim e^-1 $$






      share|cite|improve this answer









      $endgroup$

















        1












        $begingroup$

        Hint: $$left( frackk+1 right)^k sim e^-1 $$






        share|cite|improve this answer









        $endgroup$















          1












          1








          1





          $begingroup$

          Hint: $$left( frackk+1 right)^k sim e^-1 $$






          share|cite|improve this answer









          $endgroup$



          Hint: $$left( frackk+1 right)^k sim e^-1 $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 3 hours ago









          Robert IsraelRobert Israel

          331k23221478




          331k23221478





















              0












              $begingroup$

              $$a_k=left(frac kk+1right)^k^2implies log(a_k)=k^2 logleft(frac kk+1right)$$



              $$log(a_k+1)-log(a_k)=(k+1)^2 log left(frack+1k+2right)-k^2 log left(frackk+1right)$$ Using Taylor expansions for large $k$
              $$log(a_k+1)-log(a_k)=-1+frac13 k^2+Oleft(frac1k^3right)$$
              $$frac a_k+1a_k=e^log(a_k+1)-log(a_k)=frac 1 e left(1+frac13 k^2+Oleft(frac1k^3right)right)to frac 1 e $$






              share|cite|improve this answer









              $endgroup$

















                0












                $begingroup$

                $$a_k=left(frac kk+1right)^k^2implies log(a_k)=k^2 logleft(frac kk+1right)$$



                $$log(a_k+1)-log(a_k)=(k+1)^2 log left(frack+1k+2right)-k^2 log left(frackk+1right)$$ Using Taylor expansions for large $k$
                $$log(a_k+1)-log(a_k)=-1+frac13 k^2+Oleft(frac1k^3right)$$
                $$frac a_k+1a_k=e^log(a_k+1)-log(a_k)=frac 1 e left(1+frac13 k^2+Oleft(frac1k^3right)right)to frac 1 e $$






                share|cite|improve this answer









                $endgroup$















                  0












                  0








                  0





                  $begingroup$

                  $$a_k=left(frac kk+1right)^k^2implies log(a_k)=k^2 logleft(frac kk+1right)$$



                  $$log(a_k+1)-log(a_k)=(k+1)^2 log left(frack+1k+2right)-k^2 log left(frackk+1right)$$ Using Taylor expansions for large $k$
                  $$log(a_k+1)-log(a_k)=-1+frac13 k^2+Oleft(frac1k^3right)$$
                  $$frac a_k+1a_k=e^log(a_k+1)-log(a_k)=frac 1 e left(1+frac13 k^2+Oleft(frac1k^3right)right)to frac 1 e $$






                  share|cite|improve this answer









                  $endgroup$



                  $$a_k=left(frac kk+1right)^k^2implies log(a_k)=k^2 logleft(frac kk+1right)$$



                  $$log(a_k+1)-log(a_k)=(k+1)^2 log left(frack+1k+2right)-k^2 log left(frackk+1right)$$ Using Taylor expansions for large $k$
                  $$log(a_k+1)-log(a_k)=-1+frac13 k^2+Oleft(frac1k^3right)$$
                  $$frac a_k+1a_k=e^log(a_k+1)-log(a_k)=frac 1 e left(1+frac13 k^2+Oleft(frac1k^3right)right)to frac 1 e $$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 57 mins ago









                  Claude LeiboviciClaude Leibovici

                  126k1158135




                  126k1158135



























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