How do you determine if the series $sumlimits_{k=1}^infty left(1-frac1kright)^{k^2}$ converges?












3












$begingroup$


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






sequences-and-series convergence







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$endgroup$












  • $begingroup$
    This is susceptible to the same approach as my answer to a different question.
    $endgroup$
    – T. Bongers
    4 hours ago










  • $begingroup$
    Do you know what $left(1-frac{1}{k}right)^{k}$ converges to?
    $endgroup$
    – JavaMan
    4 hours ago












  • $begingroup$
    I cannot believe how horrible my intuition is with this stuff especially given how old I am.
    $endgroup$
    – Randall
    3 hours ago






  • 1




    $begingroup$
    See also: Convergence of $sum _{k=1}^infty (1-frac{1}{k})^{k^2}$
    $endgroup$
    – Martin Sleziak
    1 hour ago
















3












$begingroup$


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






sequences-and-series convergence







share|cite|improve this question











$endgroup$












  • $begingroup$
    This is susceptible to the same approach as my answer to a different question.
    $endgroup$
    – T. Bongers
    4 hours ago










  • $begingroup$
    Do you know what $left(1-frac{1}{k}right)^{k}$ converges to?
    $endgroup$
    – JavaMan
    4 hours ago












  • $begingroup$
    I cannot believe how horrible my intuition is with this stuff especially given how old I am.
    $endgroup$
    – Randall
    3 hours ago






  • 1




    $begingroup$
    See also: Convergence of $sum _{k=1}^infty (1-frac{1}{k})^{k^2}$
    $endgroup$
    – Martin Sleziak
    1 hour ago














3












3








3





$begingroup$


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






sequences-and-series convergence







share|cite|improve this question











$endgroup$




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






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 1 hour ago









Martin Sleziak

44.7k10119272




44.7k10119272










asked 4 hours ago









JayJay

334




334












  • $begingroup$
    This is susceptible to the same approach as my answer to a different question.
    $endgroup$
    – T. Bongers
    4 hours ago










  • $begingroup$
    Do you know what $left(1-frac{1}{k}right)^{k}$ converges to?
    $endgroup$
    – JavaMan
    4 hours ago












  • $begingroup$
    I cannot believe how horrible my intuition is with this stuff especially given how old I am.
    $endgroup$
    – Randall
    3 hours ago






  • 1




    $begingroup$
    See also: Convergence of $sum _{k=1}^infty (1-frac{1}{k})^{k^2}$
    $endgroup$
    – Martin Sleziak
    1 hour ago


















  • $begingroup$
    This is susceptible to the same approach as my answer to a different question.
    $endgroup$
    – T. Bongers
    4 hours ago










  • $begingroup$
    Do you know what $left(1-frac{1}{k}right)^{k}$ converges to?
    $endgroup$
    – JavaMan
    4 hours ago












  • $begingroup$
    I cannot believe how horrible my intuition is with this stuff especially given how old I am.
    $endgroup$
    – Randall
    3 hours ago






  • 1




    $begingroup$
    See also: Convergence of $sum _{k=1}^infty (1-frac{1}{k})^{k^2}$
    $endgroup$
    – Martin Sleziak
    1 hour ago
















$begingroup$
This is susceptible to the same approach as my answer to a different question.
$endgroup$
– T. Bongers
4 hours ago




$begingroup$
This is susceptible to the same approach as my answer to a different question.
$endgroup$
– T. Bongers
4 hours ago












$begingroup$
Do you know what $left(1-frac{1}{k}right)^{k}$ converges to?
$endgroup$
– JavaMan
4 hours ago






$begingroup$
Do you know what $left(1-frac{1}{k}right)^{k}$ converges to?
$endgroup$
– JavaMan
4 hours ago














$begingroup$
I cannot believe how horrible my intuition is with this stuff especially given how old I am.
$endgroup$
– Randall
3 hours ago




$begingroup$
I cannot believe how horrible my intuition is with this stuff especially given how old I am.
$endgroup$
– Randall
3 hours ago




1




1




$begingroup$
See also: Convergence of $sum _{k=1}^infty (1-frac{1}{k})^{k^2}$
$endgroup$
– Martin Sleziak
1 hour ago




$begingroup$
See also: Convergence of $sum _{k=1}^infty (1-frac{1}{k})^{k^2}$
$endgroup$
– Martin Sleziak
1 hour ago










4 Answers
4






active

oldest

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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
    $$lim sup sqrt[k]{left(1 - frac{1}{k}right)^{k^2}} = lim sup left(1 - frac{1}{k}right)^k = e^{-1} < 1,$$
    hence the series converges.






    share|cite|improve this answer









    $endgroup$





















      2












      $begingroup$

      $begin{array}\
      (1-frac1{k})^{k^2}
      &=(frac{k-1}{k})^{k^2}\
      &=dfrac1{(frac{k}{k-1})^{k^2}}\
      &=dfrac1{(1+frac{1}{k-1})^{k^2}}\
      &=dfrac1{((1+frac{1}{k-1})^{k})^k}\
      &<dfrac1{(1+frac{k}{k-1})^k}
      qquadtext{by Bernoulli}\
      &=dfrac1{(frac{2k-1}{k-1})^k}\
      &<dfrac1{(frac{2k-2}{k-1})^k}\
      &=dfrac1{2^k}\
      end{array}
      $



      and the sum of this converges.






      share|cite|improve this answer









      $endgroup$





















        0












        $begingroup$

        The ratio test is also interesting
        $$a_k=left(1-frac{1}{k}right)^{k^2}implies log(a_k)=k ^2 logleft(1-frac{1}{k}right)$$
        $$log(a_{k+1})-log(a_k)=(k+1) ^2 logleft(1-frac{1}{k+1}right)-k ^2 logleft(1-frac{1}{k}right)$$



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






        share|cite|improve this answer









        $endgroup$













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          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 4 hours ago









              Mark ViolaMark Viola

              132k1277174




              132k1277174























                  4












                  $begingroup$

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






                  share|cite|improve this answer









                  $endgroup$


















                    4












                    $begingroup$

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






                    share|cite|improve this answer









                    $endgroup$
















                      4












                      4








                      4





                      $begingroup$

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






                      share|cite|improve this answer









                      $endgroup$



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







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered 4 hours ago









                      Theo BenditTheo Bendit

                      18.8k12253




                      18.8k12253























                          2












                          $begingroup$

                          $begin{array}\
                          (1-frac1{k})^{k^2}
                          &=(frac{k-1}{k})^{k^2}\
                          &=dfrac1{(frac{k}{k-1})^{k^2}}\
                          &=dfrac1{(1+frac{1}{k-1})^{k^2}}\
                          &=dfrac1{((1+frac{1}{k-1})^{k})^k}\
                          &<dfrac1{(1+frac{k}{k-1})^k}
                          qquadtext{by Bernoulli}\
                          &=dfrac1{(frac{2k-1}{k-1})^k}\
                          &<dfrac1{(frac{2k-2}{k-1})^k}\
                          &=dfrac1{2^k}\
                          end{array}
                          $



                          and the sum of this converges.






                          share|cite|improve this answer









                          $endgroup$


















                            2












                            $begingroup$

                            $begin{array}\
                            (1-frac1{k})^{k^2}
                            &=(frac{k-1}{k})^{k^2}\
                            &=dfrac1{(frac{k}{k-1})^{k^2}}\
                            &=dfrac1{(1+frac{1}{k-1})^{k^2}}\
                            &=dfrac1{((1+frac{1}{k-1})^{k})^k}\
                            &<dfrac1{(1+frac{k}{k-1})^k}
                            qquadtext{by Bernoulli}\
                            &=dfrac1{(frac{2k-1}{k-1})^k}\
                            &<dfrac1{(frac{2k-2}{k-1})^k}\
                            &=dfrac1{2^k}\
                            end{array}
                            $



                            and the sum of this converges.






                            share|cite|improve this answer









                            $endgroup$
















                              2












                              2








                              2





                              $begingroup$

                              $begin{array}\
                              (1-frac1{k})^{k^2}
                              &=(frac{k-1}{k})^{k^2}\
                              &=dfrac1{(frac{k}{k-1})^{k^2}}\
                              &=dfrac1{(1+frac{1}{k-1})^{k^2}}\
                              &=dfrac1{((1+frac{1}{k-1})^{k})^k}\
                              &<dfrac1{(1+frac{k}{k-1})^k}
                              qquadtext{by Bernoulli}\
                              &=dfrac1{(frac{2k-1}{k-1})^k}\
                              &<dfrac1{(frac{2k-2}{k-1})^k}\
                              &=dfrac1{2^k}\
                              end{array}
                              $



                              and the sum of this converges.






                              share|cite|improve this answer









                              $endgroup$



                              $begin{array}\
                              (1-frac1{k})^{k^2}
                              &=(frac{k-1}{k})^{k^2}\
                              &=dfrac1{(frac{k}{k-1})^{k^2}}\
                              &=dfrac1{(1+frac{1}{k-1})^{k^2}}\
                              &=dfrac1{((1+frac{1}{k-1})^{k})^k}\
                              &<dfrac1{(1+frac{k}{k-1})^k}
                              qquadtext{by Bernoulli}\
                              &=dfrac1{(frac{2k-1}{k-1})^k}\
                              &<dfrac1{(frac{2k-2}{k-1})^k}\
                              &=dfrac1{2^k}\
                              end{array}
                              $



                              and the sum of this converges.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered 3 hours ago









                              marty cohenmarty cohen

                              73.8k549128




                              73.8k549128























                                  0












                                  $begingroup$

                                  The ratio test is also interesting
                                  $$a_k=left(1-frac{1}{k}right)^{k^2}implies log(a_k)=k ^2 logleft(1-frac{1}{k}right)$$
                                  $$log(a_{k+1})-log(a_k)=(k+1) ^2 logleft(1-frac{1}{k+1}right)-k ^2 logleft(1-frac{1}{k}right)$$



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






                                  share|cite|improve this answer









                                  $endgroup$


















                                    0












                                    $begingroup$

                                    The ratio test is also interesting
                                    $$a_k=left(1-frac{1}{k}right)^{k^2}implies log(a_k)=k ^2 logleft(1-frac{1}{k}right)$$
                                    $$log(a_{k+1})-log(a_k)=(k+1) ^2 logleft(1-frac{1}{k+1}right)-k ^2 logleft(1-frac{1}{k}right)$$



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






                                    share|cite|improve this answer









                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$

                                      The ratio test is also interesting
                                      $$a_k=left(1-frac{1}{k}right)^{k^2}implies log(a_k)=k ^2 logleft(1-frac{1}{k}right)$$
                                      $$log(a_{k+1})-log(a_k)=(k+1) ^2 logleft(1-frac{1}{k+1}right)-k ^2 logleft(1-frac{1}{k}right)$$



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






                                      share|cite|improve this answer









                                      $endgroup$



                                      The ratio test is also interesting
                                      $$a_k=left(1-frac{1}{k}right)^{k^2}implies log(a_k)=k ^2 logleft(1-frac{1}{k}right)$$
                                      $$log(a_{k+1})-log(a_k)=(k+1) ^2 logleft(1-frac{1}{k+1}right)-k ^2 logleft(1-frac{1}{k}right)$$



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







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered 4 hours ago









                                      Claude LeiboviciClaude Leibovici

                                      122k1157134




                                      122k1157134






























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