Determine whether or not the following series converge.












1












$begingroup$


$$sum_{k=1}^inftyleft(frac k{k+1}right)^{k^2}$$



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










  • $begingroup$
    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$$left(frac{k}{k+1}right)^{k^2}le e^{-k/2}$$
    $endgroup$
    – Mark Viola
    2 hours ago
















1












$begingroup$


$$sum_{k=1}^inftyleft(frac k{k+1}right)^{k^2}$$



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










  • $begingroup$
    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$$left(frac{k}{k+1}right)^{k^2}le e^{-k/2}$$
    $endgroup$
    – Mark Viola
    2 hours ago














1












1








1





$begingroup$


$$sum_{k=1}^inftyleft(frac k{k+1}right)^{k^2}$$



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











$endgroup$




$$sum_{k=1}^inftyleft(frac k{k+1}right)^{k^2}$$



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 hours ago









Simply Beautiful Art

50.9k580186




50.9k580186










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










  • $begingroup$
    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$$left(frac{k}{k+1}right)^{k^2}le e^{-k/2}$$
    $endgroup$
    – Mark Viola
    2 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










  • $begingroup$
    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$$left(frac{k}{k+1}right)^{k^2}le e^{-k/2}$$
    $endgroup$
    – Mark Viola
    2 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












$begingroup$
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$$left(frac{k}{k+1}right)^{k^2}le e^{-k/2}$$
$endgroup$
– Mark Viola
2 hours ago




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










3 Answers
3






active

oldest

votes


















2












$begingroup$

By Root test, $$limsup_{n to infty} sqrt[n]{left(frac{n}{n+1}right)^{n^2}}=limsup_{n to infty} {left(frac{n}{n+1}right)^{n}}=limsup_{n to infty} {left({1+frac{1}{n}}right)^{-n}}=frac{1}{e}<1$$



So your series converges!






share|cite|improve this answer











$endgroup$













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










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



















1












$begingroup$

Hint: $$left( frac{k}{k+1} right)^k sim e^{-1} $$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

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



    $$log(a_{k+1})-log(a_k)=(k+1)^2 log left(frac{k+1}{k+2}right)-k^2 log left(frac{k}{k+1}right)$$ Using Taylor expansions for large $k$
    $$log(a_{k+1})-log(a_k)=-1+frac{1}{3 k^2}+Oleft(frac{1}{k^3}right)$$
    $$frac {a_{k+1}}{a_k}=e^{log(a_{k+1})-log(a_k)}=frac 1 e left(1+frac{1}{3 k^2}+Oleft(frac{1}{k^3}right)right)to frac 1 e $$






    share|cite|improve this answer









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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, $$limsup_{n to infty} sqrt[n]{left(frac{n}{n+1}right)^{n^2}}=limsup_{n to infty} {left(frac{n}{n+1}right)^{n}}=limsup_{n to infty} {left({1+frac{1}{n}}right)^{-n}}=frac{1}{e}<1$$



      So your series converges!






      share|cite|improve this answer











      $endgroup$













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










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
















      2












      $begingroup$

      By Root test, $$limsup_{n to infty} sqrt[n]{left(frac{n}{n+1}right)^{n^2}}=limsup_{n to infty} {left(frac{n}{n+1}right)^{n}}=limsup_{n to infty} {left({1+frac{1}{n}}right)^{-n}}=frac{1}{e}<1$$



      So your series converges!






      share|cite|improve this answer











      $endgroup$













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










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














      2












      2








      2





      $begingroup$

      By Root test, $$limsup_{n to infty} sqrt[n]{left(frac{n}{n+1}right)^{n^2}}=limsup_{n to infty} {left(frac{n}{n+1}right)^{n}}=limsup_{n to infty} {left({1+frac{1}{n}}right)^{-n}}=frac{1}{e}<1$$



      So your series converges!






      share|cite|improve this answer











      $endgroup$



      By Root test, $$limsup_{n to infty} sqrt[n]{left(frac{n}{n+1}right)^{n^2}}=limsup_{n to infty} {left(frac{n}{n+1}right)^{n}}=limsup_{n to infty} {left({1+frac{1}{n}}right)^{-n}}=frac{1}{e}<1$$



      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










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


















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










      • $begingroup$
        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












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




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











      1












      $begingroup$

      Hint: $$left( frac{k}{k+1} right)^k sim e^{-1} $$






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Hint: $$left( frac{k}{k+1} right)^k sim e^{-1} $$






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Hint: $$left( frac{k}{k+1} right)^k sim e^{-1} $$






          share|cite|improve this answer









          $endgroup$



          Hint: $$left( frac{k}{k+1} right)^k sim e^{-1} $$







          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












              $begingroup$

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



              $$log(a_{k+1})-log(a_k)=(k+1)^2 log left(frac{k+1}{k+2}right)-k^2 log left(frac{k}{k+1}right)$$ Using Taylor expansions for large $k$
              $$log(a_{k+1})-log(a_k)=-1+frac{1}{3 k^2}+Oleft(frac{1}{k^3}right)$$
              $$frac {a_{k+1}}{a_k}=e^{log(a_{k+1})-log(a_k)}=frac 1 e left(1+frac{1}{3 k^2}+Oleft(frac{1}{k^3}right)right)to frac 1 e $$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

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



                $$log(a_{k+1})-log(a_k)=(k+1)^2 log left(frac{k+1}{k+2}right)-k^2 log left(frac{k}{k+1}right)$$ Using Taylor expansions for large $k$
                $$log(a_{k+1})-log(a_k)=-1+frac{1}{3 k^2}+Oleft(frac{1}{k^3}right)$$
                $$frac {a_{k+1}}{a_k}=e^{log(a_{k+1})-log(a_k)}=frac 1 e left(1+frac{1}{3 k^2}+Oleft(frac{1}{k^3}right)right)to frac 1 e $$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

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



                  $$log(a_{k+1})-log(a_k)=(k+1)^2 log left(frac{k+1}{k+2}right)-k^2 log left(frac{k}{k+1}right)$$ Using Taylor expansions for large $k$
                  $$log(a_{k+1})-log(a_k)=-1+frac{1}{3 k^2}+Oleft(frac{1}{k^3}right)$$
                  $$frac {a_{k+1}}{a_k}=e^{log(a_{k+1})-log(a_k)}=frac 1 e left(1+frac{1}{3 k^2}+Oleft(frac{1}{k^3}right)right)to frac 1 e $$






                  share|cite|improve this answer









                  $endgroup$



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



                  $$log(a_{k+1})-log(a_k)=(k+1)^2 log left(frac{k+1}{k+2}right)-k^2 log left(frac{k}{k+1}right)$$ Using Taylor expansions for large $k$
                  $$log(a_{k+1})-log(a_k)=-1+frac{1}{3 k^2}+Oleft(frac{1}{k^3}right)$$
                  $$frac {a_{k+1}}{a_k}=e^{log(a_{k+1})-log(a_k)}=frac 1 e left(1+frac{1}{3 k^2}+Oleft(frac{1}{k^3}right)right)to frac 1 e $$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 47 mins ago









                  Claude LeiboviciClaude Leibovici

                  126k1158135




                  126k1158135






























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