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?










share|cite|improve this question









$endgroup$












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










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












  • $begingroup$
    I cannot believe how horrible my intuition is with this stuff especially given how old I am.
    $endgroup$
    – Randall
    31 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?










share|cite|improve this question









$endgroup$












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










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












  • $begingroup$
    I cannot believe how horrible my intuition is with this stuff especially given how old I am.
    $endgroup$
    – Randall
    31 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?










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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 1 hour ago









JayJay

334




334












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










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












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


















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










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












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
















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




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












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






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














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




$begingroup$
I cannot believe how horrible my intuition is with this stuff especially given how old I am.
$endgroup$
– Randall
31 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









    $endgroup$





















      1












      $begingroup$

      $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









        $endgroup$













          Your Answer





          StackExchange.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          });
          });
          }, "mathjax-editing");

          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "69"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3123320%2fhow-do-you-determine-if-the-following-series-converges%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          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 1 hour 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 1 hour ago









                      Theo BenditTheo Bendit

                      18.8k12253




                      18.8k12253























                          1












                          $begingroup$

                          $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









                          $endgroup$


















                            1












                            $begingroup$

                            $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









                            $endgroup$
















                              1












                              1








                              1





                              $begingroup$

                              $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 43 mins 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 54 mins ago









                                      Claude LeiboviciClaude Leibovici

                                      122k1157134




                                      122k1157134






























                                          draft saved

                                          draft discarded




















































                                          Thanks for contributing an answer to Mathematics Stack Exchange!


                                          • Please be sure to answer the question. Provide details and share your research!

                                          But avoid



                                          • Asking for help, clarification, or responding to other answers.

                                          • Making statements based on opinion; back them up with references or personal experience.


                                          Use MathJax to format equations. MathJax reference.


                                          To learn more, see our tips on writing great answers.




                                          draft saved


                                          draft discarded














                                          StackExchange.ready(
                                          function () {
                                          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3123320%2fhow-do-you-determine-if-the-following-series-converges%23new-answer', 'question_page');
                                          }
                                          );

                                          Post as a guest















                                          Required, but never shown





















































                                          Required, but never shown














                                          Required, but never shown












                                          Required, but never shown







                                          Required, but never shown

































                                          Required, but never shown














                                          Required, but never shown












                                          Required, but never shown







                                          Required, but never shown







                                          Popular posts from this blog

                                          Knooppunt Holsloot

                                          Altaar (religie)

                                          Gregoriusmis