Real Analysis: Provide example of two Series that…












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$begingroup$


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










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$endgroup$

















    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






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      asked 2 hours ago









      K KAK KA

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          $begingroup$

          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









            $endgroup$













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              $begingroup$

              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






























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