Is the series $a_n = (1 + (-1)^{n}3)^{-n} n^{2}$ absolutely convergent$?$












0












$begingroup$


The given series is $a_n = (1 + (-1)^{n}3)^{-n} n^2$



I need to check if it is absolutely convergent.



I partition the standard form of series to get -
$$a_n = frac{n^{2}}{(-2)^{n}} = frac{(-1)^{n}n^{2}}{2^n} $$, when $n$ is odd



And



$$a_n = frac{n^{2}}{4^{n}},$$ when n is even.



Since, both parts are absolutely convergent, can I conclude from this that the whole series is absolutely convergent$?$



Moreover, Is this series an alternating series ?



The series is changing its sign alternatively, but it's not in the form of $sum (-1)^{n}a_n$






sequences-and-series







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    The given series is $a_n = (1 + (-1)^{n}3)^{-n} n^2$



    I need to check if it is absolutely convergent.



    I partition the standard form of series to get -
    $$a_n = frac{n^{2}}{(-2)^{n}} = frac{(-1)^{n}n^{2}}{2^n} $$, when $n$ is odd



    And



    $$a_n = frac{n^{2}}{4^{n}},$$ when n is even.



    Since, both parts are absolutely convergent, can I conclude from this that the whole series is absolutely convergent$?$



    Moreover, Is this series an alternating series ?



    The series is changing its sign alternatively, but it's not in the form of $sum (-1)^{n}a_n$






    sequences-and-series







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      The given series is $a_n = (1 + (-1)^{n}3)^{-n} n^2$



      I need to check if it is absolutely convergent.



      I partition the standard form of series to get -
      $$a_n = frac{n^{2}}{(-2)^{n}} = frac{(-1)^{n}n^{2}}{2^n} $$, when $n$ is odd



      And



      $$a_n = frac{n^{2}}{4^{n}},$$ when n is even.



      Since, both parts are absolutely convergent, can I conclude from this that the whole series is absolutely convergent$?$



      Moreover, Is this series an alternating series ?



      The series is changing its sign alternatively, but it's not in the form of $sum (-1)^{n}a_n$






      sequences-and-series







      share|cite|improve this question









      $endgroup$




      The given series is $a_n = (1 + (-1)^{n}3)^{-n} n^2$



      I need to check if it is absolutely convergent.



      I partition the standard form of series to get -
      $$a_n = frac{n^{2}}{(-2)^{n}} = frac{(-1)^{n}n^{2}}{2^n} $$, when $n$ is odd



      And



      $$a_n = frac{n^{2}}{4^{n}},$$ when n is even.



      Since, both parts are absolutely convergent, can I conclude from this that the whole series is absolutely convergent$?$



      Moreover, Is this series an alternating series ?



      The series is changing its sign alternatively, but it's not in the form of $sum (-1)^{n}a_n$






      sequences-and-series




      sequences-and-series






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 26 '18 at 12:01









      MathsaddictMathsaddict

      3619




      3619






















          2 Answers
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          $begingroup$

          From what you have found you can say that $$|a_n|leq frac{n^2}{2^n}$$
          for all $n$. Since by ratio test $sum frac{n^2}{2^n}$ converges, hence $sum |a_n|$ converges, i.e. $sum a_n$ is absolutely convergent.






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            You can first use absolute convergence, remember the (reverse) triangle inequality and notice that $n^2/3^n leq 1/2^n$ from some $n$ (now by comparison with a geometric series you can conclude)






            share|cite|improve this answer











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              2 Answers
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              2 Answers
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              $begingroup$

              From what you have found you can say that $$|a_n|leq frac{n^2}{2^n}$$
              for all $n$. Since by ratio test $sum frac{n^2}{2^n}$ converges, hence $sum |a_n|$ converges, i.e. $sum a_n$ is absolutely convergent.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                From what you have found you can say that $$|a_n|leq frac{n^2}{2^n}$$
                for all $n$. Since by ratio test $sum frac{n^2}{2^n}$ converges, hence $sum |a_n|$ converges, i.e. $sum a_n$ is absolutely convergent.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  From what you have found you can say that $$|a_n|leq frac{n^2}{2^n}$$
                  for all $n$. Since by ratio test $sum frac{n^2}{2^n}$ converges, hence $sum |a_n|$ converges, i.e. $sum a_n$ is absolutely convergent.






                  share|cite|improve this answer









                  $endgroup$



                  From what you have found you can say that $$|a_n|leq frac{n^2}{2^n}$$
                  for all $n$. Since by ratio test $sum frac{n^2}{2^n}$ converges, hence $sum |a_n|$ converges, i.e. $sum a_n$ is absolutely convergent.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 26 '18 at 12:11









                  user9077user9077

                  1,239612




                  1,239612























                      0












                      $begingroup$

                      You can first use absolute convergence, remember the (reverse) triangle inequality and notice that $n^2/3^n leq 1/2^n$ from some $n$ (now by comparison with a geometric series you can conclude)






                      share|cite|improve this answer











                      $endgroup$


















                        0












                        $begingroup$

                        You can first use absolute convergence, remember the (reverse) triangle inequality and notice that $n^2/3^n leq 1/2^n$ from some $n$ (now by comparison with a geometric series you can conclude)






                        share|cite|improve this answer











                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          You can first use absolute convergence, remember the (reverse) triangle inequality and notice that $n^2/3^n leq 1/2^n$ from some $n$ (now by comparison with a geometric series you can conclude)






                          share|cite|improve this answer











                          $endgroup$



                          You can first use absolute convergence, remember the (reverse) triangle inequality and notice that $n^2/3^n leq 1/2^n$ from some $n$ (now by comparison with a geometric series you can conclude)







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Dec 26 '18 at 12:36

























                          answered Dec 26 '18 at 12:18









                          LeonardoLeonardo

                          3469




                          3469






























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