How do you determine if the following series 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?






convergence







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$endgroup$












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










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












  • $begingroup$
    I cannot believe how horrible my intuition is with this stuff especially given how old I am.
    $endgroup$
    – Randall
    11 mins 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?






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
    47 mins ago










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












  • $begingroup$
    I cannot believe how horrible my intuition is with this stuff especially given how old I am.
    $endgroup$
    – Randall
    11 mins 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?






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?






convergence




convergence






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share|cite|improve this question











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share|cite|improve this question










asked 57 mins ago









JayJay

334




334












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










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












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


















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










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












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
















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




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












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






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














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




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










4 Answers
4






active

oldest

votes


















0












$begingroup$

HINT:



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



Can you finish?






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    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









    $endgroup$





















      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












        $begingroup$

        $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









        $endgroup$













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          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

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          0












          $begingroup$

          HINT:



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



          Can you finish?






          share|cite|improve this answer









          $endgroup$


















            0












            $begingroup$

            HINT:



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



            Can you finish?






            share|cite|improve this answer









            $endgroup$
















              0












              0








              0





              $begingroup$

              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 42 mins ago









              Mark ViolaMark Viola

              132k1277174




              132k1277174























                  3












                  $begingroup$

                  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









                  $endgroup$


















                    3












                    $begingroup$

                    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









                    $endgroup$
















                      3












                      3








                      3





                      $begingroup$

                      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 46 mins ago









                      Theo BenditTheo Bendit

                      18.8k12253




                      18.8k12253























                          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 33 mins ago









                              Claude LeiboviciClaude Leibovici

                              122k1157134




                              122k1157134























                                  0












                                  $begingroup$

                                  $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












                                    $begingroup$

                                    $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












                                      0








                                      0





                                      $begingroup$

                                      $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 22 mins ago









                                      marty cohenmarty cohen

                                      73.8k549128




                                      73.8k549128






























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