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
    4 hours ago










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






  • 1




    $begingroup$
    Does the root test say you take the $k^2$-th root?
    $endgroup$
    – Simply Beautiful Art
    4 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
    4 hours ago










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






  • 1




    $begingroup$
    Does the root test say you take the $k^2$-th root?
    $endgroup$
    – Simply Beautiful Art
    4 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 25 mins ago









user21820

40.2k544163




40.2k544163










asked 4 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
    4 hours ago










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






  • 1




    $begingroup$
    Does the root test say you take the $k^2$-th root?
    $endgroup$
    – Simply Beautiful Art
    4 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
    4 hours ago










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






  • 1




    $begingroup$
    Does the root test say you take the $k^2$-th root?
    $endgroup$
    – Simply Beautiful Art
    4 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
4 hours ago




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












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




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




1




1




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




$begingroup$
Does the root test say you take the $k^2$-th root?
$endgroup$
– Simply Beautiful Art
4 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









$endgroup$




















    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
        $endgroup$
        – 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 4 hours ago









          Robert IsraelRobert Israel

          331k23221478




          331k23221478





















              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 1 hour ago









                  Claude LeiboviciClaude Leibovici

                  126k1158135




                  126k1158135



























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                      Sinh 1996Nhân vật còn sốngNam diễn viên Anh 1 tháng 61996người AnhNhà hát cung điện VictoriaLuân ĐônSpider-ManMarvel Cinematic UniverseKingston upon ThamesLuân ĐônIsle of ManIrelandDonheadCông giáo La MãWimbledonWimbledonCông giáo RômaNifty Feet Dance SchoolLynne PageBilly ElliotBilly Elliot the MusicalWest EndBilly Elliot The MusicalTanner PfluegerBilly ElliotsTanner PfluegerLayton WilliamsAngry DanceBilly Elliot the MusicalMyleene KlassBilly Elliot the Musical10 Downing StreetGordon BrownWolf HallBBC TwoGregory CromwellThomas CromwellMark RylanceBeneath a Scarlet SkyLip Sync BattleZendayaGiải Oscar cho hiệu ứng hình ảnh xuất sắc nhấtGiải thưởng Viện Hàn lâmTội phạm nhân bản 2049Jimmy Kimmel Live!Người nhệnGhibli StudioJuan Antonio BayonaNaomi WaHttsEwan McGregorLiên hoan phim Quốc tế TorontoLondon Film CriticsSaoirse RonanRon HowardPeter ParkerSpider-ManMarvelCharlie HunnamJames GraySamuel InsullAlfonso Gomez-RejonPatrick Ness Chaos WalkingDaisy RidleyDoug LimanNathan DrakeUnchartedN
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