How do you determine if the series $sumlimits_k=1^infty left(1-frac1kright)^k^2$ converges? [duplicate]

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This question already has an answer here:



  • Convergence of $sum _k=1^infty (1-frac1k)^k^2$ [closed]

    3 answers



$$sum_k=1^infty mathrm(1-frac1k)^mathrmk^2$$
I tried using the limit comparison test with $$sum_k=1^infty mathrm(1-frac1k)^mathrmk^$$ but this leads to a limit of 0, which doesn't help. I think this may involve some use of
$mathrme^x$, but I don't know where else to start. Any suggestions?






sequences-and-series convergence







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marked as duplicate by rtybase, Xander Henderson, stressed out, K.Power, Leucippus Feb 24 at 2:28


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

















  • $begingroup$
    This is susceptible to the same approach as my answer to a different question.
    $endgroup$
    – T. Bongers
    Feb 23 at 3:16










  • $begingroup$
    Do you know what $left(1-frac1kright)^k$ converges to?
    $endgroup$
    – JavaMan
    Feb 23 at 3:22











  • $begingroup$
    I cannot believe how horrible my intuition is with this stuff especially given how old I am.
    $endgroup$
    – Randall
    Feb 23 at 3:52






  • 2




    $begingroup$
    See also: Convergence of $sum _k=1^infty (1-frac1k)^k^2$
    $endgroup$
    – Martin Sleziak
    Feb 23 at 6:18















4












$begingroup$



This question already has an answer here:



  • Convergence of $sum _k=1^infty (1-frac1k)^k^2$ [closed]

    3 answers



$$sum_k=1^infty mathrm(1-frac1k)^mathrmk^2$$
I tried using the limit comparison test with $$sum_k=1^infty mathrm(1-frac1k)^mathrmk^$$ but this leads to a limit of 0, which doesn't help. I think this may involve some use of
$mathrme^x$, but I don't know where else to start. Any suggestions?






sequences-and-series convergence







share|cite|improve this question











$endgroup$



marked as duplicate by rtybase, Xander Henderson, stressed out, K.Power, Leucippus Feb 24 at 2:28


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

















  • $begingroup$
    This is susceptible to the same approach as my answer to a different question.
    $endgroup$
    – T. Bongers
    Feb 23 at 3:16










  • $begingroup$
    Do you know what $left(1-frac1kright)^k$ converges to?
    $endgroup$
    – JavaMan
    Feb 23 at 3:22











  • $begingroup$
    I cannot believe how horrible my intuition is with this stuff especially given how old I am.
    $endgroup$
    – Randall
    Feb 23 at 3:52






  • 2




    $begingroup$
    See also: Convergence of $sum _k=1^infty (1-frac1k)^k^2$
    $endgroup$
    – Martin Sleziak
    Feb 23 at 6:18













4












4








4





$begingroup$



This question already has an answer here:



  • Convergence of $sum _k=1^infty (1-frac1k)^k^2$ [closed]

    3 answers



$$sum_k=1^infty mathrm(1-frac1k)^mathrmk^2$$
I tried using the limit comparison test with $$sum_k=1^infty mathrm(1-frac1k)^mathrmk^$$ but this leads to a limit of 0, which doesn't help. I think this may involve some use of
$mathrme^x$, but I don't know where else to start. Any suggestions?






sequences-and-series convergence







share|cite|improve this question











$endgroup$





This question already has an answer here:



  • Convergence of $sum _k=1^infty (1-frac1k)^k^2$ [closed]

    3 answers



$$sum_k=1^infty mathrm(1-frac1k)^mathrmk^2$$
I tried using the limit comparison test with $$sum_k=1^infty mathrm(1-frac1k)^mathrmk^$$ but this leads to a limit of 0, which doesn't help. I think this may involve some use of
$mathrme^x$, but I don't know where else to start. Any suggestions?





This question already has an answer here:



  • Convergence of $sum _k=1^infty (1-frac1k)^k^2$ [closed]

    3 answers






sequences-and-series convergence




sequences-and-series convergence






share|cite|improve this question















share|cite|improve this question













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share|cite|improve this question








edited Feb 23 at 6:15









Martin Sleziak

44.9k10121274




44.9k10121274










asked Feb 23 at 3:05









JayJay

435




435




marked as duplicate by rtybase, Xander Henderson, stressed out, K.Power, Leucippus Feb 24 at 2:28


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.









marked as duplicate by rtybase, Xander Henderson, stressed out, K.Power, Leucippus Feb 24 at 2:28


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.













  • $begingroup$
    This is susceptible to the same approach as my answer to a different question.
    $endgroup$
    – T. Bongers
    Feb 23 at 3:16










  • $begingroup$
    Do you know what $left(1-frac1kright)^k$ converges to?
    $endgroup$
    – JavaMan
    Feb 23 at 3:22











  • $begingroup$
    I cannot believe how horrible my intuition is with this stuff especially given how old I am.
    $endgroup$
    – Randall
    Feb 23 at 3:52






  • 2




    $begingroup$
    See also: Convergence of $sum _k=1^infty (1-frac1k)^k^2$
    $endgroup$
    – Martin Sleziak
    Feb 23 at 6:18
















  • $begingroup$
    This is susceptible to the same approach as my answer to a different question.
    $endgroup$
    – T. Bongers
    Feb 23 at 3:16










  • $begingroup$
    Do you know what $left(1-frac1kright)^k$ converges to?
    $endgroup$
    – JavaMan
    Feb 23 at 3:22











  • $begingroup$
    I cannot believe how horrible my intuition is with this stuff especially given how old I am.
    $endgroup$
    – Randall
    Feb 23 at 3:52






  • 2




    $begingroup$
    See also: Convergence of $sum _k=1^infty (1-frac1k)^k^2$
    $endgroup$
    – Martin Sleziak
    Feb 23 at 6:18















$begingroup$
This is susceptible to the same approach as my answer to a different question.
$endgroup$
– T. Bongers
Feb 23 at 3:16




$begingroup$
This is susceptible to the same approach as my answer to a different question.
$endgroup$
– T. Bongers
Feb 23 at 3:16












$begingroup$
Do you know what $left(1-frac1kright)^k$ converges to?
$endgroup$
– JavaMan
Feb 23 at 3:22





$begingroup$
Do you know what $left(1-frac1kright)^k$ converges to?
$endgroup$
– JavaMan
Feb 23 at 3:22













$begingroup$
I cannot believe how horrible my intuition is with this stuff especially given how old I am.
$endgroup$
– Randall
Feb 23 at 3:52




$begingroup$
I cannot believe how horrible my intuition is with this stuff especially given how old I am.
$endgroup$
– Randall
Feb 23 at 3:52




2




2




$begingroup$
See also: Convergence of $sum _k=1^infty (1-frac1k)^k^2$
$endgroup$
– Martin Sleziak
Feb 23 at 6:18




$begingroup$
See also: Convergence of $sum _k=1^infty (1-frac1k)^k^2$
$endgroup$
– Martin Sleziak
Feb 23 at 6:18










4 Answers
4






active

oldest

votes


















1












$begingroup$

HINT:



Note that $$left( 1-frac1k right)^kle e^-1$$



Can you finish?






share|cite|improve this answer









$endgroup$




















    4












    $begingroup$

    The root test works. Consider
    $$lim sup sqrt[k]left(1 - frac1kright)^k^2 = lim sup left(1 - frac1kright)^k = e^-1 < 1,$$
    hence the series converges.






    share|cite|improve this answer









    $endgroup$




















      2












      $begingroup$

      $beginarray\
      (1-frac1k)^k^2
      &=(frack-1k)^k^2\
      &=dfrac1(frackk-1)^k^2\
      &=dfrac1(1+frac1k-1)^k^2\
      &=dfrac1((1+frac1k-1)^k)^k\
      &<dfrac1(1+frackk-1)^k
      qquadtextby Bernoulli\
      &=dfrac1(frac2k-1k-1)^k\
      &<dfrac1(frac2k-2k-1)^k\
      &=dfrac12^k\
      endarray
      $



      and the sum of this converges.






      share|cite|improve this answer









      $endgroup$




















        0












        $begingroup$

        The ratio test is also interesting
        $$a_k=left(1-frac1kright)^k^2implies log(a_k)=k ^2 logleft(1-frac1kright)$$
        $$log(a_k+1)-log(a_k)=(k+1) ^2 logleft(1-frac1k+1right)-k ^2 logleft(1-frac1kright)$$



        Develop as a Taylor series for large values of $k$ to get
        $$log(a_k+1)-log(a_k)=-1+frac13 k^2+O(left(frac1k^3right)$$
        Continue with Taylor
        $$fraca_k+1a_k=e^log(a_k+1)-log(a_k)=frac 1 eleft(1+frac13 k^2+Oleft(frac1k^3right)right)to frac 1 e$$






        share|cite|improve this answer









        $endgroup$



















          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          HINT:



          Note that $$left( 1-frac1k right)^kle e^-1$$



          Can you finish?






          share|cite|improve this answer









          $endgroup$

















            1












            $begingroup$

            HINT:



            Note that $$left( 1-frac1k right)^kle e^-1$$



            Can you finish?






            share|cite|improve this answer









            $endgroup$















              1












              1








              1





              $begingroup$

              HINT:



              Note that $$left( 1-frac1k right)^kle e^-1$$



              Can you finish?






              share|cite|improve this answer









              $endgroup$



              HINT:



              Note that $$left( 1-frac1k right)^kle e^-1$$



              Can you finish?







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Feb 23 at 3:20









              Mark ViolaMark Viola

              133k1278176




              133k1278176





















                  4












                  $begingroup$

                  The root test works. Consider
                  $$lim sup sqrt[k]left(1 - frac1kright)^k^2 = lim sup left(1 - frac1kright)^k = e^-1 < 1,$$
                  hence the series converges.






                  share|cite|improve this answer









                  $endgroup$

















                    4












                    $begingroup$

                    The root test works. Consider
                    $$lim sup sqrt[k]left(1 - frac1kright)^k^2 = lim sup left(1 - frac1kright)^k = e^-1 < 1,$$
                    hence the series converges.






                    share|cite|improve this answer









                    $endgroup$















                      4












                      4








                      4





                      $begingroup$

                      The root test works. Consider
                      $$lim sup sqrt[k]left(1 - frac1kright)^k^2 = lim sup left(1 - frac1kright)^k = e^-1 < 1,$$
                      hence the series converges.






                      share|cite|improve this answer









                      $endgroup$



                      The root test works. Consider
                      $$lim sup sqrt[k]left(1 - frac1kright)^k^2 = lim sup left(1 - frac1kright)^k = e^-1 < 1,$$
                      hence the series converges.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Feb 23 at 3:16









                      Theo BenditTheo Bendit

                      19.6k12354




                      19.6k12354





















                          2












                          $begingroup$

                          $beginarray\
                          (1-frac1k)^k^2
                          &=(frack-1k)^k^2\
                          &=dfrac1(frackk-1)^k^2\
                          &=dfrac1(1+frac1k-1)^k^2\
                          &=dfrac1((1+frac1k-1)^k)^k\
                          &<dfrac1(1+frackk-1)^k
                          qquadtextby Bernoulli\
                          &=dfrac1(frac2k-1k-1)^k\
                          &<dfrac1(frac2k-2k-1)^k\
                          &=dfrac12^k\
                          endarray
                          $



                          and the sum of this converges.






                          share|cite|improve this answer









                          $endgroup$

















                            2












                            $begingroup$

                            $beginarray\
                            (1-frac1k)^k^2
                            &=(frack-1k)^k^2\
                            &=dfrac1(frackk-1)^k^2\
                            &=dfrac1(1+frac1k-1)^k^2\
                            &=dfrac1((1+frac1k-1)^k)^k\
                            &<dfrac1(1+frackk-1)^k
                            qquadtextby Bernoulli\
                            &=dfrac1(frac2k-1k-1)^k\
                            &<dfrac1(frac2k-2k-1)^k\
                            &=dfrac12^k\
                            endarray
                            $



                            and the sum of this converges.






                            share|cite|improve this answer









                            $endgroup$















                              2












                              2








                              2





                              $begingroup$

                              $beginarray\
                              (1-frac1k)^k^2
                              &=(frack-1k)^k^2\
                              &=dfrac1(frackk-1)^k^2\
                              &=dfrac1(1+frac1k-1)^k^2\
                              &=dfrac1((1+frac1k-1)^k)^k\
                              &<dfrac1(1+frackk-1)^k
                              qquadtextby Bernoulli\
                              &=dfrac1(frac2k-1k-1)^k\
                              &<dfrac1(frac2k-2k-1)^k\
                              &=dfrac12^k\
                              endarray
                              $



                              and the sum of this converges.






                              share|cite|improve this answer









                              $endgroup$



                              $beginarray\
                              (1-frac1k)^k^2
                              &=(frack-1k)^k^2\
                              &=dfrac1(frackk-1)^k^2\
                              &=dfrac1(1+frac1k-1)^k^2\
                              &=dfrac1((1+frac1k-1)^k)^k\
                              &<dfrac1(1+frackk-1)^k
                              qquadtextby Bernoulli\
                              &=dfrac1(frac2k-1k-1)^k\
                              &<dfrac1(frac2k-2k-1)^k\
                              &=dfrac12^k\
                              endarray
                              $



                              and the sum of this converges.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Feb 23 at 3:41









                              marty cohenmarty cohen

                              74.4k549129




                              74.4k549129





















                                  0












                                  $begingroup$

                                  The ratio test is also interesting
                                  $$a_k=left(1-frac1kright)^k^2implies log(a_k)=k ^2 logleft(1-frac1kright)$$
                                  $$log(a_k+1)-log(a_k)=(k+1) ^2 logleft(1-frac1k+1right)-k ^2 logleft(1-frac1kright)$$



                                  Develop as a Taylor series for large values of $k$ to get
                                  $$log(a_k+1)-log(a_k)=-1+frac13 k^2+O(left(frac1k^3right)$$
                                  Continue with Taylor
                                  $$fraca_k+1a_k=e^log(a_k+1)-log(a_k)=frac 1 eleft(1+frac13 k^2+Oleft(frac1k^3right)right)to frac 1 e$$






                                  share|cite|improve this answer









                                  $endgroup$

















                                    0












                                    $begingroup$

                                    The ratio test is also interesting
                                    $$a_k=left(1-frac1kright)^k^2implies log(a_k)=k ^2 logleft(1-frac1kright)$$
                                    $$log(a_k+1)-log(a_k)=(k+1) ^2 logleft(1-frac1k+1right)-k ^2 logleft(1-frac1kright)$$



                                    Develop as a Taylor series for large values of $k$ to get
                                    $$log(a_k+1)-log(a_k)=-1+frac13 k^2+O(left(frac1k^3right)$$
                                    Continue with Taylor
                                    $$fraca_k+1a_k=e^log(a_k+1)-log(a_k)=frac 1 eleft(1+frac13 k^2+Oleft(frac1k^3right)right)to frac 1 e$$






                                    share|cite|improve this answer









                                    $endgroup$















                                      0












                                      0








                                      0





                                      $begingroup$

                                      The ratio test is also interesting
                                      $$a_k=left(1-frac1kright)^k^2implies log(a_k)=k ^2 logleft(1-frac1kright)$$
                                      $$log(a_k+1)-log(a_k)=(k+1) ^2 logleft(1-frac1k+1right)-k ^2 logleft(1-frac1kright)$$



                                      Develop as a Taylor series for large values of $k$ to get
                                      $$log(a_k+1)-log(a_k)=-1+frac13 k^2+O(left(frac1k^3right)$$
                                      Continue with Taylor
                                      $$fraca_k+1a_k=e^log(a_k+1)-log(a_k)=frac 1 eleft(1+frac13 k^2+Oleft(frac1k^3right)right)to frac 1 e$$






                                      share|cite|improve this answer









                                      $endgroup$



                                      The ratio test is also interesting
                                      $$a_k=left(1-frac1kright)^k^2implies log(a_k)=k ^2 logleft(1-frac1kright)$$
                                      $$log(a_k+1)-log(a_k)=(k+1) ^2 logleft(1-frac1k+1right)-k ^2 logleft(1-frac1kright)$$



                                      Develop as a Taylor series for large values of $k$ to get
                                      $$log(a_k+1)-log(a_k)=-1+frac13 k^2+O(left(frac1k^3right)$$
                                      Continue with Taylor
                                      $$fraca_k+1a_k=e^log(a_k+1)-log(a_k)=frac 1 eleft(1+frac13 k^2+Oleft(frac1k^3right)right)to frac 1 e$$







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Feb 23 at 3:30









                                      Claude LeiboviciClaude Leibovici

                                      124k1158135




                                      124k1158135












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