Rearrangement of Alternating Harmonic Series to be Infinity












0












$begingroup$


Our professor gave a problem asking to rearrange the alternating harmonic series:
$sum_{n=0}^{infty} frac{(-1)^n}{n} $

such that the rearrangement equals infinity.



So I was doing some searching and found this property that the rearranged sums of the alternating harmonic series sum to:



$ln(2) + frac{1}{2}ln(frac{p}{n}) $



Where $p$ is the number of positive terms listed followed by $n$ negative terms in the rearrangement.



So my idea is that in order for the rearrangement to go to infinity, either $p$ is going to have to be infinite, or $n$ is going to have to be $0$. Would this make sense for the problem? It almost seems like this would not be a valid rearrangement of the alternating harmonic series, since I would basically be rearranging it to be the normal harmonic series.






real-analysis sequences-and-series







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    0












    $begingroup$


    Our professor gave a problem asking to rearrange the alternating harmonic series:
    $sum_{n=0}^{infty} frac{(-1)^n}{n} $

    such that the rearrangement equals infinity.



    So I was doing some searching and found this property that the rearranged sums of the alternating harmonic series sum to:



    $ln(2) + frac{1}{2}ln(frac{p}{n}) $



    Where $p$ is the number of positive terms listed followed by $n$ negative terms in the rearrangement.



    So my idea is that in order for the rearrangement to go to infinity, either $p$ is going to have to be infinite, or $n$ is going to have to be $0$. Would this make sense for the problem? It almost seems like this would not be a valid rearrangement of the alternating harmonic series, since I would basically be rearranging it to be the normal harmonic series.






    real-analysis sequences-and-series







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Our professor gave a problem asking to rearrange the alternating harmonic series:
      $sum_{n=0}^{infty} frac{(-1)^n}{n} $

      such that the rearrangement equals infinity.



      So I was doing some searching and found this property that the rearranged sums of the alternating harmonic series sum to:



      $ln(2) + frac{1}{2}ln(frac{p}{n}) $



      Where $p$ is the number of positive terms listed followed by $n$ negative terms in the rearrangement.



      So my idea is that in order for the rearrangement to go to infinity, either $p$ is going to have to be infinite, or $n$ is going to have to be $0$. Would this make sense for the problem? It almost seems like this would not be a valid rearrangement of the alternating harmonic series, since I would basically be rearranging it to be the normal harmonic series.






      real-analysis sequences-and-series







      share|cite|improve this question









      $endgroup$




      Our professor gave a problem asking to rearrange the alternating harmonic series:
      $sum_{n=0}^{infty} frac{(-1)^n}{n} $

      such that the rearrangement equals infinity.



      So I was doing some searching and found this property that the rearranged sums of the alternating harmonic series sum to:



      $ln(2) + frac{1}{2}ln(frac{p}{n}) $



      Where $p$ is the number of positive terms listed followed by $n$ negative terms in the rearrangement.



      So my idea is that in order for the rearrangement to go to infinity, either $p$ is going to have to be infinite, or $n$ is going to have to be $0$. Would this make sense for the problem? It almost seems like this would not be a valid rearrangement of the alternating harmonic series, since I would basically be rearranging it to be the normal harmonic series.






      real-analysis sequences-and-series




      real-analysis sequences-and-series






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Oct 25 '18 at 0:42









      TynaTyna

      876




      876






















          2 Answers
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          $begingroup$

          Well. First, note that
          begin{align}
          sum^infty_{n=1} frac{1}{2n} =infty.
          end{align}
          Then we see that there exists $N_1$ such that
          begin{align}
          2<frac{1}{2}+frac{1}{4}+ldots +frac{1}{2N_1}<3
          end{align}
          then
          begin{align}
          1<frac{1}{2}+frac{1}{4}+ldots +frac{1}{2N_1}-1<2.
          end{align}
          Next, we can find an $N_2$ such that
          begin{align}
          3<frac{1}{2}+frac{1}{4}+ldots +frac{1}{2N_1}-1+frac{1}{2(N_1+1)}+ldots+frac{1}{2N_2}<4
          end{align}
          then
          begin{align}
          2<frac{1}{2}+frac{1}{4}+ldots +frac{1}{2N_1}-1+frac{1}{2(N_1+1)}+ldots+frac{1}{2N_2}-frac{1}{3}.
          end{align}
          Again, choose $N_3$ such that
          begin{align}
          4<frac{1}{2}+frac{1}{4}+ldots +frac{1}{2N_1}-1+frac{1}{2(N_1+1)}+ldots+frac{1}{2N_2}-frac{1}{3}+frac{1}{2(N_2+1)}+ldots+frac{1}{2N_3}<5
          end{align}
          which means
          begin{align}
          3<frac{1}{2}+frac{1}{4}+ldots +frac{1}{2N_1}-1+frac{1}{2(N_1+1)}+ldots+frac{1}{2N_2}-frac{1}{3}+frac{1}{2(N_2+1)}+ldots+frac{1}{2N_3}-frac{1}{5}.
          end{align}
          Applying this process, you will obtain a rearrangment of $sum (-1)^n/n$ such that the resulting series diverges.






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            Rearrange the as following$$left({1over 1}+{1over 3}-{1over 2}right)+left({1over 5}+{1over 7}-{1over 4}right)+left({1over 9}+{1over 11}-{1over 6}right)+left({1over 13}+{1over 15}-{1over 8}right)+cdots\> {1over 3}+{1over 7}+{1over 11}+{1over 15}+cdots =infty$$therefore the series with this kind of rearrangement, goes to $infty$






            share|cite|improve this answer









            $endgroup$













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              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

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              active

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              active

              oldest

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              0












              $begingroup$

              Well. First, note that
              begin{align}
              sum^infty_{n=1} frac{1}{2n} =infty.
              end{align}
              Then we see that there exists $N_1$ such that
              begin{align}
              2<frac{1}{2}+frac{1}{4}+ldots +frac{1}{2N_1}<3
              end{align}
              then
              begin{align}
              1<frac{1}{2}+frac{1}{4}+ldots +frac{1}{2N_1}-1<2.
              end{align}
              Next, we can find an $N_2$ such that
              begin{align}
              3<frac{1}{2}+frac{1}{4}+ldots +frac{1}{2N_1}-1+frac{1}{2(N_1+1)}+ldots+frac{1}{2N_2}<4
              end{align}
              then
              begin{align}
              2<frac{1}{2}+frac{1}{4}+ldots +frac{1}{2N_1}-1+frac{1}{2(N_1+1)}+ldots+frac{1}{2N_2}-frac{1}{3}.
              end{align}
              Again, choose $N_3$ such that
              begin{align}
              4<frac{1}{2}+frac{1}{4}+ldots +frac{1}{2N_1}-1+frac{1}{2(N_1+1)}+ldots+frac{1}{2N_2}-frac{1}{3}+frac{1}{2(N_2+1)}+ldots+frac{1}{2N_3}<5
              end{align}
              which means
              begin{align}
              3<frac{1}{2}+frac{1}{4}+ldots +frac{1}{2N_1}-1+frac{1}{2(N_1+1)}+ldots+frac{1}{2N_2}-frac{1}{3}+frac{1}{2(N_2+1)}+ldots+frac{1}{2N_3}-frac{1}{5}.
              end{align}
              Applying this process, you will obtain a rearrangment of $sum (-1)^n/n$ such that the resulting series diverges.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Well. First, note that
                begin{align}
                sum^infty_{n=1} frac{1}{2n} =infty.
                end{align}
                Then we see that there exists $N_1$ such that
                begin{align}
                2<frac{1}{2}+frac{1}{4}+ldots +frac{1}{2N_1}<3
                end{align}
                then
                begin{align}
                1<frac{1}{2}+frac{1}{4}+ldots +frac{1}{2N_1}-1<2.
                end{align}
                Next, we can find an $N_2$ such that
                begin{align}
                3<frac{1}{2}+frac{1}{4}+ldots +frac{1}{2N_1}-1+frac{1}{2(N_1+1)}+ldots+frac{1}{2N_2}<4
                end{align}
                then
                begin{align}
                2<frac{1}{2}+frac{1}{4}+ldots +frac{1}{2N_1}-1+frac{1}{2(N_1+1)}+ldots+frac{1}{2N_2}-frac{1}{3}.
                end{align}
                Again, choose $N_3$ such that
                begin{align}
                4<frac{1}{2}+frac{1}{4}+ldots +frac{1}{2N_1}-1+frac{1}{2(N_1+1)}+ldots+frac{1}{2N_2}-frac{1}{3}+frac{1}{2(N_2+1)}+ldots+frac{1}{2N_3}<5
                end{align}
                which means
                begin{align}
                3<frac{1}{2}+frac{1}{4}+ldots +frac{1}{2N_1}-1+frac{1}{2(N_1+1)}+ldots+frac{1}{2N_2}-frac{1}{3}+frac{1}{2(N_2+1)}+ldots+frac{1}{2N_3}-frac{1}{5}.
                end{align}
                Applying this process, you will obtain a rearrangment of $sum (-1)^n/n$ such that the resulting series diverges.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Well. First, note that
                  begin{align}
                  sum^infty_{n=1} frac{1}{2n} =infty.
                  end{align}
                  Then we see that there exists $N_1$ such that
                  begin{align}
                  2<frac{1}{2}+frac{1}{4}+ldots +frac{1}{2N_1}<3
                  end{align}
                  then
                  begin{align}
                  1<frac{1}{2}+frac{1}{4}+ldots +frac{1}{2N_1}-1<2.
                  end{align}
                  Next, we can find an $N_2$ such that
                  begin{align}
                  3<frac{1}{2}+frac{1}{4}+ldots +frac{1}{2N_1}-1+frac{1}{2(N_1+1)}+ldots+frac{1}{2N_2}<4
                  end{align}
                  then
                  begin{align}
                  2<frac{1}{2}+frac{1}{4}+ldots +frac{1}{2N_1}-1+frac{1}{2(N_1+1)}+ldots+frac{1}{2N_2}-frac{1}{3}.
                  end{align}
                  Again, choose $N_3$ such that
                  begin{align}
                  4<frac{1}{2}+frac{1}{4}+ldots +frac{1}{2N_1}-1+frac{1}{2(N_1+1)}+ldots+frac{1}{2N_2}-frac{1}{3}+frac{1}{2(N_2+1)}+ldots+frac{1}{2N_3}<5
                  end{align}
                  which means
                  begin{align}
                  3<frac{1}{2}+frac{1}{4}+ldots +frac{1}{2N_1}-1+frac{1}{2(N_1+1)}+ldots+frac{1}{2N_2}-frac{1}{3}+frac{1}{2(N_2+1)}+ldots+frac{1}{2N_3}-frac{1}{5}.
                  end{align}
                  Applying this process, you will obtain a rearrangment of $sum (-1)^n/n$ such that the resulting series diverges.






                  share|cite|improve this answer









                  $endgroup$



                  Well. First, note that
                  begin{align}
                  sum^infty_{n=1} frac{1}{2n} =infty.
                  end{align}
                  Then we see that there exists $N_1$ such that
                  begin{align}
                  2<frac{1}{2}+frac{1}{4}+ldots +frac{1}{2N_1}<3
                  end{align}
                  then
                  begin{align}
                  1<frac{1}{2}+frac{1}{4}+ldots +frac{1}{2N_1}-1<2.
                  end{align}
                  Next, we can find an $N_2$ such that
                  begin{align}
                  3<frac{1}{2}+frac{1}{4}+ldots +frac{1}{2N_1}-1+frac{1}{2(N_1+1)}+ldots+frac{1}{2N_2}<4
                  end{align}
                  then
                  begin{align}
                  2<frac{1}{2}+frac{1}{4}+ldots +frac{1}{2N_1}-1+frac{1}{2(N_1+1)}+ldots+frac{1}{2N_2}-frac{1}{3}.
                  end{align}
                  Again, choose $N_3$ such that
                  begin{align}
                  4<frac{1}{2}+frac{1}{4}+ldots +frac{1}{2N_1}-1+frac{1}{2(N_1+1)}+ldots+frac{1}{2N_2}-frac{1}{3}+frac{1}{2(N_2+1)}+ldots+frac{1}{2N_3}<5
                  end{align}
                  which means
                  begin{align}
                  3<frac{1}{2}+frac{1}{4}+ldots +frac{1}{2N_1}-1+frac{1}{2(N_1+1)}+ldots+frac{1}{2N_2}-frac{1}{3}+frac{1}{2(N_2+1)}+ldots+frac{1}{2N_3}-frac{1}{5}.
                  end{align}
                  Applying this process, you will obtain a rearrangment of $sum (-1)^n/n$ such that the resulting series diverges.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Oct 25 '18 at 1:40









                  Jacky ChongJacky Chong

                  17.8k21128




                  17.8k21128























                      0












                      $begingroup$

                      Rearrange the as following$$left({1over 1}+{1over 3}-{1over 2}right)+left({1over 5}+{1over 7}-{1over 4}right)+left({1over 9}+{1over 11}-{1over 6}right)+left({1over 13}+{1over 15}-{1over 8}right)+cdots\> {1over 3}+{1over 7}+{1over 11}+{1over 15}+cdots =infty$$therefore the series with this kind of rearrangement, goes to $infty$






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        Rearrange the as following$$left({1over 1}+{1over 3}-{1over 2}right)+left({1over 5}+{1over 7}-{1over 4}right)+left({1over 9}+{1over 11}-{1over 6}right)+left({1over 13}+{1over 15}-{1over 8}right)+cdots\> {1over 3}+{1over 7}+{1over 11}+{1over 15}+cdots =infty$$therefore the series with this kind of rearrangement, goes to $infty$






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          Rearrange the as following$$left({1over 1}+{1over 3}-{1over 2}right)+left({1over 5}+{1over 7}-{1over 4}right)+left({1over 9}+{1over 11}-{1over 6}right)+left({1over 13}+{1over 15}-{1over 8}right)+cdots\> {1over 3}+{1over 7}+{1over 11}+{1over 15}+cdots =infty$$therefore the series with this kind of rearrangement, goes to $infty$






                          share|cite|improve this answer









                          $endgroup$



                          Rearrange the as following$$left({1over 1}+{1over 3}-{1over 2}right)+left({1over 5}+{1over 7}-{1over 4}right)+left({1over 9}+{1over 11}-{1over 6}right)+left({1over 13}+{1over 15}-{1over 8}right)+cdots\> {1over 3}+{1over 7}+{1over 11}+{1over 15}+cdots =infty$$therefore the series with this kind of rearrangement, goes to $infty$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 30 '18 at 15:59









                          Mostafa AyazMostafa Ayaz

                          14.8k3938




                          14.8k3938






























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