Determine whether or not $sum_k=1^inftyleft(frac kk+1right)^k^2$ converges.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 $Decide whether the series $sum_n=1^infty frac1+5^n1+6^n$ converges or divergesDetermine whether the series converges or diverges.To 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.


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 $Decide whether the series $sum_n=1^infty frac1+5^n1+6^n$ converges or divergesDetermine whether the series converges or diverges.To 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 $sum_n=1^inftyfraclnn!n^3$






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








2












$begingroup$


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



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
    Apr 15 at 2:38










  • $begingroup$
    Since its squared, do I take the k^2 root
    $endgroup$
    – radius
    Apr 15 at 2:40






  • 1




    $begingroup$
    Does the root test say you take the $k^2$-th root?
    $endgroup$
    – Simply Beautiful Art
    Apr 15 at 2:44






  • 1




    $begingroup$
    For $kge 1$, we have$$left(frackk+1right)^k^2le e^-k/2$$
    $endgroup$
    – Mark Viola
    Apr 15 at 3:14

















2












$begingroup$


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



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
    Apr 15 at 2:38










  • $begingroup$
    Since its squared, do I take the k^2 root
    $endgroup$
    – radius
    Apr 15 at 2:40






  • 1




    $begingroup$
    Does the root test say you take the $k^2$-th root?
    $endgroup$
    – Simply Beautiful Art
    Apr 15 at 2:44






  • 1




    $begingroup$
    For $kge 1$, we have$$left(frackk+1right)^k^2le e^-k/2$$
    $endgroup$
    – Mark Viola
    Apr 15 at 3:14













2












2








2





$begingroup$


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



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 Apr 15 at 6:19









user21820

42k5 gold badges48 silver badges170 bronze badges




42k5 gold badges48 silver badges170 bronze badges










asked Apr 15 at 2:33









radiusradius

612 bronze badges




612 bronze badges














  • $begingroup$
    You have an exponent with $k$. Your instinct should be to remove it with the root test.
    $endgroup$
    – Simply Beautiful Art
    Apr 15 at 2:38










  • $begingroup$
    Since its squared, do I take the k^2 root
    $endgroup$
    – radius
    Apr 15 at 2:40






  • 1




    $begingroup$
    Does the root test say you take the $k^2$-th root?
    $endgroup$
    – Simply Beautiful Art
    Apr 15 at 2:44






  • 1




    $begingroup$
    For $kge 1$, we have$$left(frackk+1right)^k^2le e^-k/2$$
    $endgroup$
    – Mark Viola
    Apr 15 at 3:14
















  • $begingroup$
    You have an exponent with $k$. Your instinct should be to remove it with the root test.
    $endgroup$
    – Simply Beautiful Art
    Apr 15 at 2:38










  • $begingroup$
    Since its squared, do I take the k^2 root
    $endgroup$
    – radius
    Apr 15 at 2:40






  • 1




    $begingroup$
    Does the root test say you take the $k^2$-th root?
    $endgroup$
    – Simply Beautiful Art
    Apr 15 at 2:44






  • 1




    $begingroup$
    For $kge 1$, we have$$left(frackk+1right)^k^2le e^-k/2$$
    $endgroup$
    – Mark Viola
    Apr 15 at 3:14















$begingroup$
You have an exponent with $k$. Your instinct should be to remove it with the root test.
$endgroup$
– Simply Beautiful Art
Apr 15 at 2:38




$begingroup$
You have an exponent with $k$. Your instinct should be to remove it with the root test.
$endgroup$
– Simply Beautiful Art
Apr 15 at 2:38












$begingroup$
Since its squared, do I take the k^2 root
$endgroup$
– radius
Apr 15 at 2:40




$begingroup$
Since its squared, do I take the k^2 root
$endgroup$
– radius
Apr 15 at 2:40




1




1




$begingroup$
Does the root test say you take the $k^2$-th root?
$endgroup$
– Simply Beautiful Art
Apr 15 at 2:44




$begingroup$
Does the root test say you take the $k^2$-th root?
$endgroup$
– Simply Beautiful Art
Apr 15 at 2:44




1




1




$begingroup$
For $kge 1$, we have$$left(frackk+1right)^k^2le e^-k/2$$
$endgroup$
– Mark Viola
Apr 15 at 3:14




$begingroup$
For $kge 1$, we have$$left(frackk+1right)^k^2le e^-k/2$$
$endgroup$
– Mark Viola
Apr 15 at 3:14










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
    $endgroup$
    – radius
    Apr 15 at 2:51






  • 1




    $begingroup$
    See the wiki link!
    $endgroup$
    – Chinnapparaj R
    Apr 15 at 2:54


















1












$begingroup$

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






share|cite|improve this answer









$endgroup$






















    1












    $begingroup$

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



    $$log(a_k+1)-log(a_k)=(k+1)^2 log left(frack+1k+2right)-k^2 log left(frackk+1right)$$ 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
        $endgroup$
        – radius
        Apr 15 at 2:51






      • 1




        $begingroup$
        See the wiki link!
        $endgroup$
        – Chinnapparaj R
        Apr 15 at 2:54















      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
        $endgroup$
        – radius
        Apr 15 at 2:51






      • 1




        $begingroup$
        See the wiki link!
        $endgroup$
        – Chinnapparaj R
        Apr 15 at 2:54













      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 Apr 15 at 2:53

























      answered Apr 15 at 2:46









      Chinnapparaj RChinnapparaj R

      8,7612 gold badges10 silver badges32 bronze badges




      8,7612 gold badges10 silver badges32 bronze badges














      • $begingroup$
        How did you know to take the supremum
        $endgroup$
        – radius
        Apr 15 at 2:51






      • 1




        $begingroup$
        See the wiki link!
        $endgroup$
        – Chinnapparaj R
        Apr 15 at 2:54
















      • $begingroup$
        How did you know to take the supremum
        $endgroup$
        – radius
        Apr 15 at 2:51






      • 1




        $begingroup$
        See the wiki link!
        $endgroup$
        – Chinnapparaj R
        Apr 15 at 2:54















      $begingroup$
      How did you know to take the supremum
      $endgroup$
      – radius
      Apr 15 at 2:51




      $begingroup$
      How did you know to take the supremum
      $endgroup$
      – radius
      Apr 15 at 2:51




      1




      1




      $begingroup$
      See the wiki link!
      $endgroup$
      – Chinnapparaj R
      Apr 15 at 2:54




      $begingroup$
      See the wiki link!
      $endgroup$
      – Chinnapparaj R
      Apr 15 at 2:54













      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 Apr 15 at 2:38









          Robert IsraelRobert Israel

          345k23 gold badges241 silver badges503 bronze badges




          345k23 gold badges241 silver badges503 bronze badges
























              1












              $begingroup$

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



              $$log(a_k+1)-log(a_k)=(k+1)^2 log left(frack+1k+2right)-k^2 log left(frackk+1right)$$ 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









              $endgroup$



















                1












                $begingroup$

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



                $$log(a_k+1)-log(a_k)=(k+1)^2 log left(frack+1k+2right)-k^2 log left(frackk+1right)$$ 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









                $endgroup$

















                  1












                  1








                  1





                  $begingroup$

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



                  $$log(a_k+1)-log(a_k)=(k+1)^2 log left(frack+1k+2right)-k^2 log left(frackk+1right)$$ 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 Apr 15 at 5:25









                  Claude LeiboviciClaude Leibovici

                  133k11 gold badges61 silver badges142 bronze badges




                  133k11 gold badges61 silver badges142 bronze badges






























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