Real Analysis: Provide example of two Series that…












3












$begingroup$


I've posted the solution for this problem and I'm trying to understand this.



In the end of the solution provided it says to continue this process. So, do we hold $a_n$ to be $frac{1}{n^2}$ and $b_n$ to be $frac{1}{900^2}$ for the next $900^2$ terms? And then hold $b_n$ to be $frac{1}{n^2}$ and $a_n$ to be $frac{1}{810000 ^2}$ for the next $810000 ^2$ terms? (because $900^2$ is $810000$)



And why do we have to add one to the sum of partial sums?



QUESTION AND SOLUTION:
Question And Solution










share|cite|improve this question









$endgroup$

















    3












    $begingroup$


    I've posted the solution for this problem and I'm trying to understand this.



    In the end of the solution provided it says to continue this process. So, do we hold $a_n$ to be $frac{1}{n^2}$ and $b_n$ to be $frac{1}{900^2}$ for the next $900^2$ terms? And then hold $b_n$ to be $frac{1}{n^2}$ and $a_n$ to be $frac{1}{810000 ^2}$ for the next $810000 ^2$ terms? (because $900^2$ is $810000$)



    And why do we have to add one to the sum of partial sums?



    QUESTION AND SOLUTION:
    Question And Solution










    share|cite|improve this question









    $endgroup$















      3












      3








      3





      $begingroup$


      I've posted the solution for this problem and I'm trying to understand this.



      In the end of the solution provided it says to continue this process. So, do we hold $a_n$ to be $frac{1}{n^2}$ and $b_n$ to be $frac{1}{900^2}$ for the next $900^2$ terms? And then hold $b_n$ to be $frac{1}{n^2}$ and $a_n$ to be $frac{1}{810000 ^2}$ for the next $810000 ^2$ terms? (because $900^2$ is $810000$)



      And why do we have to add one to the sum of partial sums?



      QUESTION AND SOLUTION:
      Question And Solution










      share|cite|improve this question









      $endgroup$




      I've posted the solution for this problem and I'm trying to understand this.



      In the end of the solution provided it says to continue this process. So, do we hold $a_n$ to be $frac{1}{n^2}$ and $b_n$ to be $frac{1}{900^2}$ for the next $900^2$ terms? And then hold $b_n$ to be $frac{1}{n^2}$ and $a_n$ to be $frac{1}{810000 ^2}$ for the next $810000 ^2$ terms? (because $900^2$ is $810000$)



      And why do we have to add one to the sum of partial sums?



      QUESTION AND SOLUTION:
      Question And Solution







      real-analysis






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 2 hours ago









      K KAK KA

      454




      454






















          2 Answers
          2






          active

          oldest

          votes


















          3












          $begingroup$

          Essentially, they're making $a_n$ and $b_n$ a positive monotone summable sequence (in this case, $frac{1}{n^2}$), and just "pausing" each sequence long enough that a $1$ is added to the partial sum, thereby forcing the sum to diverge.



          So, start with the same series
          begin{matrix}
          a_n = bigl(1 & frac{1}{2^2} & frac{1}{3^2} & frac{1}{4^2} & frac{1}{5^2} & frac{1}{6^2} & frac{1}{7^2} & frac{1}{8^2} & frac{1}{9^2} & frac{1}{10^2} & frac{1}{11^2} & frac{1}{12^2} & frac{1}{13^2} & frac{1}{14^2} & frac{1}{15^2} & frac{1}{16^2} & cdots &bigr) \
          b_n = bigl(1 & frac{1}{2^2} & frac{1}{3^2} & frac{1}{4^2} & frac{1}{5^2} & frac{1}{6^2} & frac{1}{7^2} & frac{1}{8^2} & frac{1}{9^2} & frac{1}{10^2} & frac{1}{11^2} & frac{1}{12^2} & frac{1}{13^2} & frac{1}{14^2} & frac{1}{15^2} & frac{1}{16^2} & cdots &bigr)
          end{matrix}

          and modify them like so:
          begin{matrix}
          a_n = bigl(1 & color{red}{frac{1}{4}} & color{red}{frac{1}{4}} & color{red}{frac{1}{4}} & color{red}{frac{1}{4}} & frac{1}{6^2} & frac{1}{7^2} & cdots & frac{1}{29^2} & color{red}{frac{1}{900}} & color{red}{frac{1}{900}} & color{red}{frac{1}{900}} & cdots & color{red}{frac{1}{900}} & color{red}{frac{1}{900}} & frac{1}{930^2} & cdots &bigr) \
          b_n = bigl(1 & frac{1}{2^2} & frac{1}{3^2} & frac{1}{4^2} & color{red}{frac{1}{25}} & color{red}{frac{1}{25}} & color{red}{frac{1}{25}} & cdots & color{red}{frac{1}{25}} & color{red}{frac{1}{25}} & frac{1}{31^2} & frac{1}{32^2} & cdots & frac{1}{928^2} & color{red}{frac{1}{929^2}} & color{red}{frac{1}{929^2}} & cdots &bigr).
          end{matrix}

          The streaks of red numbers add to $1$, and occur infinitely many often in both sequences, so each partial sum becomes unbounded and hence the series fails to converge. But, the minimum of the two sequences is always $frac{1}{n^2}$.






          share|cite|improve this answer









          $endgroup$





















            3












            $begingroup$

            The idea behind the more challenging version is to construct $(a_n)$ and $(b_n)$ such that $sum a_n,sum b_n$ both diverge, but $sum min{a_n,b_n}$ converges. The way the author of this solution has chosen to proceed is by making $(a_n)$ and $(b_n)$ such that for each $n$, we have $min{a_n,b_n} = 1/n^2$, and yet we add enough small constant terms to each sequence so that the partial sums eventually grow by $1$ if we wait long enough. This growth by $1$ repeated over and over again ensures that the series $sum a_n,sum b_n$ both diverge since their partial sums each grow without bound by merely waiting long enough.



            To find how many terms we add again, think about the pattern
            begin{align*}
            (1+1) - 1 &= 1^2 \
            (5 + 1) - 2 &= 2^2 \
            (30 + 1) - 6 &= 5^2 \
            (930 + 1) - 31 &= 30^2 \
            (865830 + 1) - 931 &= 930^2 \
            (x + 1) - 865831 &= 865830^2 \
            dotsb
            end{align*}






            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%2f3123303%2freal-analysis-provide-example-of-two-series-that%23new-answer', 'question_page');
              }
              );

              Post as a guest















              Required, but never shown

























              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              3












              $begingroup$

              Essentially, they're making $a_n$ and $b_n$ a positive monotone summable sequence (in this case, $frac{1}{n^2}$), and just "pausing" each sequence long enough that a $1$ is added to the partial sum, thereby forcing the sum to diverge.



              So, start with the same series
              begin{matrix}
              a_n = bigl(1 & frac{1}{2^2} & frac{1}{3^2} & frac{1}{4^2} & frac{1}{5^2} & frac{1}{6^2} & frac{1}{7^2} & frac{1}{8^2} & frac{1}{9^2} & frac{1}{10^2} & frac{1}{11^2} & frac{1}{12^2} & frac{1}{13^2} & frac{1}{14^2} & frac{1}{15^2} & frac{1}{16^2} & cdots &bigr) \
              b_n = bigl(1 & frac{1}{2^2} & frac{1}{3^2} & frac{1}{4^2} & frac{1}{5^2} & frac{1}{6^2} & frac{1}{7^2} & frac{1}{8^2} & frac{1}{9^2} & frac{1}{10^2} & frac{1}{11^2} & frac{1}{12^2} & frac{1}{13^2} & frac{1}{14^2} & frac{1}{15^2} & frac{1}{16^2} & cdots &bigr)
              end{matrix}

              and modify them like so:
              begin{matrix}
              a_n = bigl(1 & color{red}{frac{1}{4}} & color{red}{frac{1}{4}} & color{red}{frac{1}{4}} & color{red}{frac{1}{4}} & frac{1}{6^2} & frac{1}{7^2} & cdots & frac{1}{29^2} & color{red}{frac{1}{900}} & color{red}{frac{1}{900}} & color{red}{frac{1}{900}} & cdots & color{red}{frac{1}{900}} & color{red}{frac{1}{900}} & frac{1}{930^2} & cdots &bigr) \
              b_n = bigl(1 & frac{1}{2^2} & frac{1}{3^2} & frac{1}{4^2} & color{red}{frac{1}{25}} & color{red}{frac{1}{25}} & color{red}{frac{1}{25}} & cdots & color{red}{frac{1}{25}} & color{red}{frac{1}{25}} & frac{1}{31^2} & frac{1}{32^2} & cdots & frac{1}{928^2} & color{red}{frac{1}{929^2}} & color{red}{frac{1}{929^2}} & cdots &bigr).
              end{matrix}

              The streaks of red numbers add to $1$, and occur infinitely many often in both sequences, so each partial sum becomes unbounded and hence the series fails to converge. But, the minimum of the two sequences is always $frac{1}{n^2}$.






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                Essentially, they're making $a_n$ and $b_n$ a positive monotone summable sequence (in this case, $frac{1}{n^2}$), and just "pausing" each sequence long enough that a $1$ is added to the partial sum, thereby forcing the sum to diverge.



                So, start with the same series
                begin{matrix}
                a_n = bigl(1 & frac{1}{2^2} & frac{1}{3^2} & frac{1}{4^2} & frac{1}{5^2} & frac{1}{6^2} & frac{1}{7^2} & frac{1}{8^2} & frac{1}{9^2} & frac{1}{10^2} & frac{1}{11^2} & frac{1}{12^2} & frac{1}{13^2} & frac{1}{14^2} & frac{1}{15^2} & frac{1}{16^2} & cdots &bigr) \
                b_n = bigl(1 & frac{1}{2^2} & frac{1}{3^2} & frac{1}{4^2} & frac{1}{5^2} & frac{1}{6^2} & frac{1}{7^2} & frac{1}{8^2} & frac{1}{9^2} & frac{1}{10^2} & frac{1}{11^2} & frac{1}{12^2} & frac{1}{13^2} & frac{1}{14^2} & frac{1}{15^2} & frac{1}{16^2} & cdots &bigr)
                end{matrix}

                and modify them like so:
                begin{matrix}
                a_n = bigl(1 & color{red}{frac{1}{4}} & color{red}{frac{1}{4}} & color{red}{frac{1}{4}} & color{red}{frac{1}{4}} & frac{1}{6^2} & frac{1}{7^2} & cdots & frac{1}{29^2} & color{red}{frac{1}{900}} & color{red}{frac{1}{900}} & color{red}{frac{1}{900}} & cdots & color{red}{frac{1}{900}} & color{red}{frac{1}{900}} & frac{1}{930^2} & cdots &bigr) \
                b_n = bigl(1 & frac{1}{2^2} & frac{1}{3^2} & frac{1}{4^2} & color{red}{frac{1}{25}} & color{red}{frac{1}{25}} & color{red}{frac{1}{25}} & cdots & color{red}{frac{1}{25}} & color{red}{frac{1}{25}} & frac{1}{31^2} & frac{1}{32^2} & cdots & frac{1}{928^2} & color{red}{frac{1}{929^2}} & color{red}{frac{1}{929^2}} & cdots &bigr).
                end{matrix}

                The streaks of red numbers add to $1$, and occur infinitely many often in both sequences, so each partial sum becomes unbounded and hence the series fails to converge. But, the minimum of the two sequences is always $frac{1}{n^2}$.






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  Essentially, they're making $a_n$ and $b_n$ a positive monotone summable sequence (in this case, $frac{1}{n^2}$), and just "pausing" each sequence long enough that a $1$ is added to the partial sum, thereby forcing the sum to diverge.



                  So, start with the same series
                  begin{matrix}
                  a_n = bigl(1 & frac{1}{2^2} & frac{1}{3^2} & frac{1}{4^2} & frac{1}{5^2} & frac{1}{6^2} & frac{1}{7^2} & frac{1}{8^2} & frac{1}{9^2} & frac{1}{10^2} & frac{1}{11^2} & frac{1}{12^2} & frac{1}{13^2} & frac{1}{14^2} & frac{1}{15^2} & frac{1}{16^2} & cdots &bigr) \
                  b_n = bigl(1 & frac{1}{2^2} & frac{1}{3^2} & frac{1}{4^2} & frac{1}{5^2} & frac{1}{6^2} & frac{1}{7^2} & frac{1}{8^2} & frac{1}{9^2} & frac{1}{10^2} & frac{1}{11^2} & frac{1}{12^2} & frac{1}{13^2} & frac{1}{14^2} & frac{1}{15^2} & frac{1}{16^2} & cdots &bigr)
                  end{matrix}

                  and modify them like so:
                  begin{matrix}
                  a_n = bigl(1 & color{red}{frac{1}{4}} & color{red}{frac{1}{4}} & color{red}{frac{1}{4}} & color{red}{frac{1}{4}} & frac{1}{6^2} & frac{1}{7^2} & cdots & frac{1}{29^2} & color{red}{frac{1}{900}} & color{red}{frac{1}{900}} & color{red}{frac{1}{900}} & cdots & color{red}{frac{1}{900}} & color{red}{frac{1}{900}} & frac{1}{930^2} & cdots &bigr) \
                  b_n = bigl(1 & frac{1}{2^2} & frac{1}{3^2} & frac{1}{4^2} & color{red}{frac{1}{25}} & color{red}{frac{1}{25}} & color{red}{frac{1}{25}} & cdots & color{red}{frac{1}{25}} & color{red}{frac{1}{25}} & frac{1}{31^2} & frac{1}{32^2} & cdots & frac{1}{928^2} & color{red}{frac{1}{929^2}} & color{red}{frac{1}{929^2}} & cdots &bigr).
                  end{matrix}

                  The streaks of red numbers add to $1$, and occur infinitely many often in both sequences, so each partial sum becomes unbounded and hence the series fails to converge. But, the minimum of the two sequences is always $frac{1}{n^2}$.






                  share|cite|improve this answer









                  $endgroup$



                  Essentially, they're making $a_n$ and $b_n$ a positive monotone summable sequence (in this case, $frac{1}{n^2}$), and just "pausing" each sequence long enough that a $1$ is added to the partial sum, thereby forcing the sum to diverge.



                  So, start with the same series
                  begin{matrix}
                  a_n = bigl(1 & frac{1}{2^2} & frac{1}{3^2} & frac{1}{4^2} & frac{1}{5^2} & frac{1}{6^2} & frac{1}{7^2} & frac{1}{8^2} & frac{1}{9^2} & frac{1}{10^2} & frac{1}{11^2} & frac{1}{12^2} & frac{1}{13^2} & frac{1}{14^2} & frac{1}{15^2} & frac{1}{16^2} & cdots &bigr) \
                  b_n = bigl(1 & frac{1}{2^2} & frac{1}{3^2} & frac{1}{4^2} & frac{1}{5^2} & frac{1}{6^2} & frac{1}{7^2} & frac{1}{8^2} & frac{1}{9^2} & frac{1}{10^2} & frac{1}{11^2} & frac{1}{12^2} & frac{1}{13^2} & frac{1}{14^2} & frac{1}{15^2} & frac{1}{16^2} & cdots &bigr)
                  end{matrix}

                  and modify them like so:
                  begin{matrix}
                  a_n = bigl(1 & color{red}{frac{1}{4}} & color{red}{frac{1}{4}} & color{red}{frac{1}{4}} & color{red}{frac{1}{4}} & frac{1}{6^2} & frac{1}{7^2} & cdots & frac{1}{29^2} & color{red}{frac{1}{900}} & color{red}{frac{1}{900}} & color{red}{frac{1}{900}} & cdots & color{red}{frac{1}{900}} & color{red}{frac{1}{900}} & frac{1}{930^2} & cdots &bigr) \
                  b_n = bigl(1 & frac{1}{2^2} & frac{1}{3^2} & frac{1}{4^2} & color{red}{frac{1}{25}} & color{red}{frac{1}{25}} & color{red}{frac{1}{25}} & cdots & color{red}{frac{1}{25}} & color{red}{frac{1}{25}} & frac{1}{31^2} & frac{1}{32^2} & cdots & frac{1}{928^2} & color{red}{frac{1}{929^2}} & color{red}{frac{1}{929^2}} & cdots &bigr).
                  end{matrix}

                  The streaks of red numbers add to $1$, and occur infinitely many often in both sequences, so each partial sum becomes unbounded and hence the series fails to converge. But, the minimum of the two sequences is always $frac{1}{n^2}$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 2 hours ago









                  Theo BenditTheo Bendit

                  18.8k12253




                  18.8k12253























                      3












                      $begingroup$

                      The idea behind the more challenging version is to construct $(a_n)$ and $(b_n)$ such that $sum a_n,sum b_n$ both diverge, but $sum min{a_n,b_n}$ converges. The way the author of this solution has chosen to proceed is by making $(a_n)$ and $(b_n)$ such that for each $n$, we have $min{a_n,b_n} = 1/n^2$, and yet we add enough small constant terms to each sequence so that the partial sums eventually grow by $1$ if we wait long enough. This growth by $1$ repeated over and over again ensures that the series $sum a_n,sum b_n$ both diverge since their partial sums each grow without bound by merely waiting long enough.



                      To find how many terms we add again, think about the pattern
                      begin{align*}
                      (1+1) - 1 &= 1^2 \
                      (5 + 1) - 2 &= 2^2 \
                      (30 + 1) - 6 &= 5^2 \
                      (930 + 1) - 31 &= 30^2 \
                      (865830 + 1) - 931 &= 930^2 \
                      (x + 1) - 865831 &= 865830^2 \
                      dotsb
                      end{align*}






                      share|cite|improve this answer









                      $endgroup$


















                        3












                        $begingroup$

                        The idea behind the more challenging version is to construct $(a_n)$ and $(b_n)$ such that $sum a_n,sum b_n$ both diverge, but $sum min{a_n,b_n}$ converges. The way the author of this solution has chosen to proceed is by making $(a_n)$ and $(b_n)$ such that for each $n$, we have $min{a_n,b_n} = 1/n^2$, and yet we add enough small constant terms to each sequence so that the partial sums eventually grow by $1$ if we wait long enough. This growth by $1$ repeated over and over again ensures that the series $sum a_n,sum b_n$ both diverge since their partial sums each grow without bound by merely waiting long enough.



                        To find how many terms we add again, think about the pattern
                        begin{align*}
                        (1+1) - 1 &= 1^2 \
                        (5 + 1) - 2 &= 2^2 \
                        (30 + 1) - 6 &= 5^2 \
                        (930 + 1) - 31 &= 30^2 \
                        (865830 + 1) - 931 &= 930^2 \
                        (x + 1) - 865831 &= 865830^2 \
                        dotsb
                        end{align*}






                        share|cite|improve this answer









                        $endgroup$
















                          3












                          3








                          3





                          $begingroup$

                          The idea behind the more challenging version is to construct $(a_n)$ and $(b_n)$ such that $sum a_n,sum b_n$ both diverge, but $sum min{a_n,b_n}$ converges. The way the author of this solution has chosen to proceed is by making $(a_n)$ and $(b_n)$ such that for each $n$, we have $min{a_n,b_n} = 1/n^2$, and yet we add enough small constant terms to each sequence so that the partial sums eventually grow by $1$ if we wait long enough. This growth by $1$ repeated over and over again ensures that the series $sum a_n,sum b_n$ both diverge since their partial sums each grow without bound by merely waiting long enough.



                          To find how many terms we add again, think about the pattern
                          begin{align*}
                          (1+1) - 1 &= 1^2 \
                          (5 + 1) - 2 &= 2^2 \
                          (30 + 1) - 6 &= 5^2 \
                          (930 + 1) - 31 &= 30^2 \
                          (865830 + 1) - 931 &= 930^2 \
                          (x + 1) - 865831 &= 865830^2 \
                          dotsb
                          end{align*}






                          share|cite|improve this answer









                          $endgroup$



                          The idea behind the more challenging version is to construct $(a_n)$ and $(b_n)$ such that $sum a_n,sum b_n$ both diverge, but $sum min{a_n,b_n}$ converges. The way the author of this solution has chosen to proceed is by making $(a_n)$ and $(b_n)$ such that for each $n$, we have $min{a_n,b_n} = 1/n^2$, and yet we add enough small constant terms to each sequence so that the partial sums eventually grow by $1$ if we wait long enough. This growth by $1$ repeated over and over again ensures that the series $sum a_n,sum b_n$ both diverge since their partial sums each grow without bound by merely waiting long enough.



                          To find how many terms we add again, think about the pattern
                          begin{align*}
                          (1+1) - 1 &= 1^2 \
                          (5 + 1) - 2 &= 2^2 \
                          (30 + 1) - 6 &= 5^2 \
                          (930 + 1) - 31 &= 30^2 \
                          (865830 + 1) - 931 &= 930^2 \
                          (x + 1) - 865831 &= 865830^2 \
                          dotsb
                          end{align*}







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 2 hours ago









                          Alex OrtizAlex Ortiz

                          10.8k21441




                          10.8k21441






























                              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%2f3123303%2freal-analysis-provide-example-of-two-series-that%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

                              Liste der Baudenkmale in Friedland (Mecklenburg)

                              Single-Malt-Whisky

                              Czorneboh