Determine if the series converges absolutely,conditionally or diverges.












0












$begingroup$


The series is as follows
begin{equation}
sum_{n=1}^{infty} frac{(-1)^n}{(n+(-1)^n)^2}
end{equation}
And I need help for taking the absolute value of the denominator. It cannot be just $|(n+(-1)^n)^2|=(n+1)^2$ right?



Thanks in advance.






real-analysis sequences-and-series







share|cite|improve this question











$endgroup$












  • $begingroup$
    you want "series" in the title, not "sequences"
    $endgroup$
    – zhw.
    Jan 10 at 18:56










  • $begingroup$
    ofccccc. thanks
    $endgroup$
    – Allorja
    Jan 10 at 18:58










  • $begingroup$
    Use comparison, compare with $1/n^2.$
    $endgroup$
    – Will M.
    Jan 10 at 20:02










  • $begingroup$
    It would be better if you start the sum from n=2.
    $endgroup$
    – JV.Stalker
    Jan 11 at 4:02
















0












$begingroup$


The series is as follows
begin{equation}
sum_{n=1}^{infty} frac{(-1)^n}{(n+(-1)^n)^2}
end{equation}
And I need help for taking the absolute value of the denominator. It cannot be just $|(n+(-1)^n)^2|=(n+1)^2$ right?



Thanks in advance.






real-analysis sequences-and-series







share|cite|improve this question











$endgroup$












  • $begingroup$
    you want "series" in the title, not "sequences"
    $endgroup$
    – zhw.
    Jan 10 at 18:56










  • $begingroup$
    ofccccc. thanks
    $endgroup$
    – Allorja
    Jan 10 at 18:58










  • $begingroup$
    Use comparison, compare with $1/n^2.$
    $endgroup$
    – Will M.
    Jan 10 at 20:02










  • $begingroup$
    It would be better if you start the sum from n=2.
    $endgroup$
    – JV.Stalker
    Jan 11 at 4:02














0












0








0





$begingroup$


The series is as follows
begin{equation}
sum_{n=1}^{infty} frac{(-1)^n}{(n+(-1)^n)^2}
end{equation}
And I need help for taking the absolute value of the denominator. It cannot be just $|(n+(-1)^n)^2|=(n+1)^2$ right?



Thanks in advance.






real-analysis sequences-and-series







share|cite|improve this question











$endgroup$




The series is as follows
begin{equation}
sum_{n=1}^{infty} frac{(-1)^n}{(n+(-1)^n)^2}
end{equation}
And I need help for taking the absolute value of the denominator. It cannot be just $|(n+(-1)^n)^2|=(n+1)^2$ right?



Thanks in advance.






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








edited Jan 10 at 18:57







Allorja

















asked Jan 10 at 18:53









AllorjaAllorja

789




789












  • $begingroup$
    you want "series" in the title, not "sequences"
    $endgroup$
    – zhw.
    Jan 10 at 18:56










  • $begingroup$
    ofccccc. thanks
    $endgroup$
    – Allorja
    Jan 10 at 18:58










  • $begingroup$
    Use comparison, compare with $1/n^2.$
    $endgroup$
    – Will M.
    Jan 10 at 20:02










  • $begingroup$
    It would be better if you start the sum from n=2.
    $endgroup$
    – JV.Stalker
    Jan 11 at 4:02


















  • $begingroup$
    you want "series" in the title, not "sequences"
    $endgroup$
    – zhw.
    Jan 10 at 18:56










  • $begingroup$
    ofccccc. thanks
    $endgroup$
    – Allorja
    Jan 10 at 18:58










  • $begingroup$
    Use comparison, compare with $1/n^2.$
    $endgroup$
    – Will M.
    Jan 10 at 20:02










  • $begingroup$
    It would be better if you start the sum from n=2.
    $endgroup$
    – JV.Stalker
    Jan 11 at 4:02
















$begingroup$
you want "series" in the title, not "sequences"
$endgroup$
– zhw.
Jan 10 at 18:56




$begingroup$
you want "series" in the title, not "sequences"
$endgroup$
– zhw.
Jan 10 at 18:56












$begingroup$
ofccccc. thanks
$endgroup$
– Allorja
Jan 10 at 18:58




$begingroup$
ofccccc. thanks
$endgroup$
– Allorja
Jan 10 at 18:58












$begingroup$
Use comparison, compare with $1/n^2.$
$endgroup$
– Will M.
Jan 10 at 20:02




$begingroup$
Use comparison, compare with $1/n^2.$
$endgroup$
– Will M.
Jan 10 at 20:02












$begingroup$
It would be better if you start the sum from n=2.
$endgroup$
– JV.Stalker
Jan 11 at 4:02




$begingroup$
It would be better if you start the sum from n=2.
$endgroup$
– JV.Stalker
Jan 11 at 4:02










3 Answers
3






active

oldest

votes


















0












$begingroup$

Hint: $|n+(-1)^n|^2 ge |n-1|^2.$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Hint



    $$left|frac{(-1)^n}{(n+(-1)^n)^2} right|le {1over (n-1)^2}$$since $$(-1)^nge -1$$






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      begin{equation}
      S=sum_{n=1}^{infty} frac{(-1)^n}{(n+(-1)^n)^2}
      end{equation}       S is divergent because at $n=1$, $Srightarrow -infty$.



      If the summation is started from $n=2$ we get the followings:



      begin{equation}
      S^*=sum_{n=1}^{infty} frac{(-1)^{n+1}}{(n+(-1)^{n+1})^2}
      end{equation}



      After separation the even and odd terms we have that:



      $S^*=sumlimits_{k=1}^inftyfrac{1}{4k^2}-sumlimits_{k=1}^inftyfrac{1}{(2k+1)^2}=frac{1}{4}zeta(2)-sumlimits_{k=1}^inftyfrac{1}{(2k+1)^2}$



      Known that $sumlimits_{k=1}^inftyfrac{1}{(2k+1)^2}=frac{pi^2}{8}-1 $



      So begin{equation} S^*=1-frac{pi^2}{12} end{equation}



      The sum is absolutely convergent because:



      $|S^*|=sumlimits_{k=1}^inftyfrac{1}{4k^2}+sumlimits_{k=1}^inftyfrac{1}{(2k+1)^2}=frac{1}{4}zeta(2)+frac{pi^2}{8}-1=frac{pi^2}{6}-1$






      share|cite|improve this answer









      $endgroup$













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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        0












        $begingroup$

        Hint: $|n+(-1)^n|^2 ge |n-1|^2.$






        share|cite|improve this answer









        $endgroup$


















          0












          $begingroup$

          Hint: $|n+(-1)^n|^2 ge |n-1|^2.$






          share|cite|improve this answer









          $endgroup$
















            0












            0








            0





            $begingroup$

            Hint: $|n+(-1)^n|^2 ge |n-1|^2.$






            share|cite|improve this answer









            $endgroup$



            Hint: $|n+(-1)^n|^2 ge |n-1|^2.$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 10 at 18:58









            zhw.zhw.

            73.8k43175




            73.8k43175























                0












                $begingroup$

                Hint



                $$left|frac{(-1)^n}{(n+(-1)^n)^2} right|le {1over (n-1)^2}$$since $$(-1)^nge -1$$






                share|cite|improve this answer









                $endgroup$


















                  0












                  $begingroup$

                  Hint



                  $$left|frac{(-1)^n}{(n+(-1)^n)^2} right|le {1over (n-1)^2}$$since $$(-1)^nge -1$$






                  share|cite|improve this answer









                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    Hint



                    $$left|frac{(-1)^n}{(n+(-1)^n)^2} right|le {1over (n-1)^2}$$since $$(-1)^nge -1$$






                    share|cite|improve this answer









                    $endgroup$



                    Hint



                    $$left|frac{(-1)^n}{(n+(-1)^n)^2} right|le {1over (n-1)^2}$$since $$(-1)^nge -1$$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 10 at 19:59









                    Mostafa AyazMostafa Ayaz

                    15.7k3939




                    15.7k3939























                        0












                        $begingroup$

                        begin{equation}
                        S=sum_{n=1}^{infty} frac{(-1)^n}{(n+(-1)^n)^2}
                        end{equation}                         S is divergent because at $n=1$, $Srightarrow -infty$.



                        If the summation is started from $n=2$ we get the followings:



                        begin{equation}
                        S^*=sum_{n=1}^{infty} frac{(-1)^{n+1}}{(n+(-1)^{n+1})^2}
                        end{equation}



                        After separation the even and odd terms we have that:



                        $S^*=sumlimits_{k=1}^inftyfrac{1}{4k^2}-sumlimits_{k=1}^inftyfrac{1}{(2k+1)^2}=frac{1}{4}zeta(2)-sumlimits_{k=1}^inftyfrac{1}{(2k+1)^2}$



                        Known that $sumlimits_{k=1}^inftyfrac{1}{(2k+1)^2}=frac{pi^2}{8}-1 $



                        So begin{equation} S^*=1-frac{pi^2}{12} end{equation}



                        The sum is absolutely convergent because:



                        $|S^*|=sumlimits_{k=1}^inftyfrac{1}{4k^2}+sumlimits_{k=1}^inftyfrac{1}{(2k+1)^2}=frac{1}{4}zeta(2)+frac{pi^2}{8}-1=frac{pi^2}{6}-1$






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          begin{equation}
                          S=sum_{n=1}^{infty} frac{(-1)^n}{(n+(-1)^n)^2}
                          end{equation}                           S is divergent because at $n=1$, $Srightarrow -infty$.



                          If the summation is started from $n=2$ we get the followings:



                          begin{equation}
                          S^*=sum_{n=1}^{infty} frac{(-1)^{n+1}}{(n+(-1)^{n+1})^2}
                          end{equation}



                          After separation the even and odd terms we have that:



                          $S^*=sumlimits_{k=1}^inftyfrac{1}{4k^2}-sumlimits_{k=1}^inftyfrac{1}{(2k+1)^2}=frac{1}{4}zeta(2)-sumlimits_{k=1}^inftyfrac{1}{(2k+1)^2}$



                          Known that $sumlimits_{k=1}^inftyfrac{1}{(2k+1)^2}=frac{pi^2}{8}-1 $



                          So begin{equation} S^*=1-frac{pi^2}{12} end{equation}



                          The sum is absolutely convergent because:



                          $|S^*|=sumlimits_{k=1}^inftyfrac{1}{4k^2}+sumlimits_{k=1}^inftyfrac{1}{(2k+1)^2}=frac{1}{4}zeta(2)+frac{pi^2}{8}-1=frac{pi^2}{6}-1$






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            begin{equation}
                            S=sum_{n=1}^{infty} frac{(-1)^n}{(n+(-1)^n)^2}
                            end{equation}                             S is divergent because at $n=1$, $Srightarrow -infty$.



                            If the summation is started from $n=2$ we get the followings:



                            begin{equation}
                            S^*=sum_{n=1}^{infty} frac{(-1)^{n+1}}{(n+(-1)^{n+1})^2}
                            end{equation}



                            After separation the even and odd terms we have that:



                            $S^*=sumlimits_{k=1}^inftyfrac{1}{4k^2}-sumlimits_{k=1}^inftyfrac{1}{(2k+1)^2}=frac{1}{4}zeta(2)-sumlimits_{k=1}^inftyfrac{1}{(2k+1)^2}$



                            Known that $sumlimits_{k=1}^inftyfrac{1}{(2k+1)^2}=frac{pi^2}{8}-1 $



                            So begin{equation} S^*=1-frac{pi^2}{12} end{equation}



                            The sum is absolutely convergent because:



                            $|S^*|=sumlimits_{k=1}^inftyfrac{1}{4k^2}+sumlimits_{k=1}^inftyfrac{1}{(2k+1)^2}=frac{1}{4}zeta(2)+frac{pi^2}{8}-1=frac{pi^2}{6}-1$






                            share|cite|improve this answer









                            $endgroup$



                            begin{equation}
                            S=sum_{n=1}^{infty} frac{(-1)^n}{(n+(-1)^n)^2}
                            end{equation}                             S is divergent because at $n=1$, $Srightarrow -infty$.



                            If the summation is started from $n=2$ we get the followings:



                            begin{equation}
                            S^*=sum_{n=1}^{infty} frac{(-1)^{n+1}}{(n+(-1)^{n+1})^2}
                            end{equation}



                            After separation the even and odd terms we have that:



                            $S^*=sumlimits_{k=1}^inftyfrac{1}{4k^2}-sumlimits_{k=1}^inftyfrac{1}{(2k+1)^2}=frac{1}{4}zeta(2)-sumlimits_{k=1}^inftyfrac{1}{(2k+1)^2}$



                            Known that $sumlimits_{k=1}^inftyfrac{1}{(2k+1)^2}=frac{pi^2}{8}-1 $



                            So begin{equation} S^*=1-frac{pi^2}{12} end{equation}



                            The sum is absolutely convergent because:



                            $|S^*|=sumlimits_{k=1}^inftyfrac{1}{4k^2}+sumlimits_{k=1}^inftyfrac{1}{(2k+1)^2}=frac{1}{4}zeta(2)+frac{pi^2}{8}-1=frac{pi^2}{6}-1$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 11 at 7:24









                            JV.StalkerJV.Stalker

                            91649




                            91649






























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