Convergence of Sum of Sequences












4












$begingroup$


This week, I learned a bit more about limits, convergence and divergence.
I was given a sum of two sequences and asked to tell whether or not it is convergent, and what its limit is:



$a_n := (-1)^n + frac{1}{n^2 +1}$



which I re-wrote into



$lim_{nto infty}(-1)^n +lim_{nto infty}frac{1}{n^2 +1}$



I noticed that $lim_{nto infty}(-1)^n$ isn't convergent, whereas the latter is convergent and has the limit of $0$. That is why I'm not entirely sure whether $a_n$ is convergent or not, and got confused.



I hope someone can clear my doubts and explain their answer to me!
Thank you.






sequences-and-series limits convergence







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    To start with, what do you know about sums of sequences and their convergence?
    $endgroup$
    – Chris Culter
    Dec 5 '18 at 19:48
















4












$begingroup$


This week, I learned a bit more about limits, convergence and divergence.
I was given a sum of two sequences and asked to tell whether or not it is convergent, and what its limit is:



$a_n := (-1)^n + frac{1}{n^2 +1}$



which I re-wrote into



$lim_{nto infty}(-1)^n +lim_{nto infty}frac{1}{n^2 +1}$



I noticed that $lim_{nto infty}(-1)^n$ isn't convergent, whereas the latter is convergent and has the limit of $0$. That is why I'm not entirely sure whether $a_n$ is convergent or not, and got confused.



I hope someone can clear my doubts and explain their answer to me!
Thank you.






sequences-and-series limits convergence







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    To start with, what do you know about sums of sequences and their convergence?
    $endgroup$
    – Chris Culter
    Dec 5 '18 at 19:48














4












4








4





$begingroup$


This week, I learned a bit more about limits, convergence and divergence.
I was given a sum of two sequences and asked to tell whether or not it is convergent, and what its limit is:



$a_n := (-1)^n + frac{1}{n^2 +1}$



which I re-wrote into



$lim_{nto infty}(-1)^n +lim_{nto infty}frac{1}{n^2 +1}$



I noticed that $lim_{nto infty}(-1)^n$ isn't convergent, whereas the latter is convergent and has the limit of $0$. That is why I'm not entirely sure whether $a_n$ is convergent or not, and got confused.



I hope someone can clear my doubts and explain their answer to me!
Thank you.






sequences-and-series limits convergence







share|cite|improve this question









$endgroup$




This week, I learned a bit more about limits, convergence and divergence.
I was given a sum of two sequences and asked to tell whether or not it is convergent, and what its limit is:



$a_n := (-1)^n + frac{1}{n^2 +1}$



which I re-wrote into



$lim_{nto infty}(-1)^n +lim_{nto infty}frac{1}{n^2 +1}$



I noticed that $lim_{nto infty}(-1)^n$ isn't convergent, whereas the latter is convergent and has the limit of $0$. That is why I'm not entirely sure whether $a_n$ is convergent or not, and got confused.



I hope someone can clear my doubts and explain their answer to me!
Thank you.






sequences-and-series limits convergence




sequences-and-series limits convergence






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 5 '18 at 19:43









RikkRikk

494




494








  • 1




    $begingroup$
    To start with, what do you know about sums of sequences and their convergence?
    $endgroup$
    – Chris Culter
    Dec 5 '18 at 19:48














  • 1




    $begingroup$
    To start with, what do you know about sums of sequences and their convergence?
    $endgroup$
    – Chris Culter
    Dec 5 '18 at 19:48








1




1




$begingroup$
To start with, what do you know about sums of sequences and their convergence?
$endgroup$
– Chris Culter
Dec 5 '18 at 19:48




$begingroup$
To start with, what do you know about sums of sequences and their convergence?
$endgroup$
– Chris Culter
Dec 5 '18 at 19:48










3 Answers
3






active

oldest

votes


















2












$begingroup$

Let $u_n=(-1)^n$ and
$v_n=frac{1}{n^2+1}$.



$(u_n)$ is divergent since
$$lim u_{2n}ne lim u_{2n+1}$$



$(v_n)$ is convergent since



$$lim v_n=0.$$



the sum of a convergent sequence and a divergent one is DIVERGENT.






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    You should look at the global situation rather than focusing on the specific example.



    The sum of a convergent sequence and a divergent one diverges. If this wasn't the case the difference of the sum and the convergent sequence would converge. That can't be as this is equal to the divergent sequence.






    share|cite|improve this answer









    $endgroup$





















      2












      $begingroup$

      Let consider




      • n odd $implies a_n := (-1)^n + frac{1}{n^2 +1} to -1$


      • n even $implies a_n := (-1)^n + frac{1}{n^2 +1} to 1$



      what can we conclude, recalling that for a convergent sequence all the subsequence need to converge to the same limit?






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        We conclude is that $a_n$ cannot be convergent! Thank you.
        $endgroup$
        – Rikk
        Dec 5 '18 at 20:12










      • $begingroup$
        @Rikk Yes of course that the general way to prove that. You are welcome! Bye
        $endgroup$
        – gimusi
        Dec 5 '18 at 20:14











      Your Answer





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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      Let $u_n=(-1)^n$ and
      $v_n=frac{1}{n^2+1}$.



      $(u_n)$ is divergent since
      $$lim u_{2n}ne lim u_{2n+1}$$



      $(v_n)$ is convergent since



      $$lim v_n=0.$$



      the sum of a convergent sequence and a divergent one is DIVERGENT.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        Let $u_n=(-1)^n$ and
        $v_n=frac{1}{n^2+1}$.



        $(u_n)$ is divergent since
        $$lim u_{2n}ne lim u_{2n+1}$$



        $(v_n)$ is convergent since



        $$lim v_n=0.$$



        the sum of a convergent sequence and a divergent one is DIVERGENT.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          Let $u_n=(-1)^n$ and
          $v_n=frac{1}{n^2+1}$.



          $(u_n)$ is divergent since
          $$lim u_{2n}ne lim u_{2n+1}$$



          $(v_n)$ is convergent since



          $$lim v_n=0.$$



          the sum of a convergent sequence and a divergent one is DIVERGENT.






          share|cite|improve this answer









          $endgroup$



          Let $u_n=(-1)^n$ and
          $v_n=frac{1}{n^2+1}$.



          $(u_n)$ is divergent since
          $$lim u_{2n}ne lim u_{2n+1}$$



          $(v_n)$ is convergent since



          $$lim v_n=0.$$



          the sum of a convergent sequence and a divergent one is DIVERGENT.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 5 '18 at 19:50









          hamam_Abdallahhamam_Abdallah

          38k21634




          38k21634























              3












              $begingroup$

              You should look at the global situation rather than focusing on the specific example.



              The sum of a convergent sequence and a divergent one diverges. If this wasn't the case the difference of the sum and the convergent sequence would converge. That can't be as this is equal to the divergent sequence.






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                You should look at the global situation rather than focusing on the specific example.



                The sum of a convergent sequence and a divergent one diverges. If this wasn't the case the difference of the sum and the convergent sequence would converge. That can't be as this is equal to the divergent sequence.






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  You should look at the global situation rather than focusing on the specific example.



                  The sum of a convergent sequence and a divergent one diverges. If this wasn't the case the difference of the sum and the convergent sequence would converge. That can't be as this is equal to the divergent sequence.






                  share|cite|improve this answer









                  $endgroup$



                  You should look at the global situation rather than focusing on the specific example.



                  The sum of a convergent sequence and a divergent one diverges. If this wasn't the case the difference of the sum and the convergent sequence would converge. That can't be as this is equal to the divergent sequence.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 5 '18 at 19:54









                  mathcounterexamples.netmathcounterexamples.net

                  25.6k21953




                  25.6k21953























                      2












                      $begingroup$

                      Let consider




                      • n odd $implies a_n := (-1)^n + frac{1}{n^2 +1} to -1$


                      • n even $implies a_n := (-1)^n + frac{1}{n^2 +1} to 1$



                      what can we conclude, recalling that for a convergent sequence all the subsequence need to converge to the same limit?






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        We conclude is that $a_n$ cannot be convergent! Thank you.
                        $endgroup$
                        – Rikk
                        Dec 5 '18 at 20:12










                      • $begingroup$
                        @Rikk Yes of course that the general way to prove that. You are welcome! Bye
                        $endgroup$
                        – gimusi
                        Dec 5 '18 at 20:14
















                      2












                      $begingroup$

                      Let consider




                      • n odd $implies a_n := (-1)^n + frac{1}{n^2 +1} to -1$


                      • n even $implies a_n := (-1)^n + frac{1}{n^2 +1} to 1$



                      what can we conclude, recalling that for a convergent sequence all the subsequence need to converge to the same limit?






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        We conclude is that $a_n$ cannot be convergent! Thank you.
                        $endgroup$
                        – Rikk
                        Dec 5 '18 at 20:12










                      • $begingroup$
                        @Rikk Yes of course that the general way to prove that. You are welcome! Bye
                        $endgroup$
                        – gimusi
                        Dec 5 '18 at 20:14














                      2












                      2








                      2





                      $begingroup$

                      Let consider




                      • n odd $implies a_n := (-1)^n + frac{1}{n^2 +1} to -1$


                      • n even $implies a_n := (-1)^n + frac{1}{n^2 +1} to 1$



                      what can we conclude, recalling that for a convergent sequence all the subsequence need to converge to the same limit?






                      share|cite|improve this answer











                      $endgroup$



                      Let consider




                      • n odd $implies a_n := (-1)^n + frac{1}{n^2 +1} to -1$


                      • n even $implies a_n := (-1)^n + frac{1}{n^2 +1} to 1$



                      what can we conclude, recalling that for a convergent sequence all the subsequence need to converge to the same limit?







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Dec 5 '18 at 19:57

























                      answered Dec 5 '18 at 19:49









                      gimusigimusi

                      1




                      1












                      • $begingroup$
                        We conclude is that $a_n$ cannot be convergent! Thank you.
                        $endgroup$
                        – Rikk
                        Dec 5 '18 at 20:12










                      • $begingroup$
                        @Rikk Yes of course that the general way to prove that. You are welcome! Bye
                        $endgroup$
                        – gimusi
                        Dec 5 '18 at 20:14


















                      • $begingroup$
                        We conclude is that $a_n$ cannot be convergent! Thank you.
                        $endgroup$
                        – Rikk
                        Dec 5 '18 at 20:12










                      • $begingroup$
                        @Rikk Yes of course that the general way to prove that. You are welcome! Bye
                        $endgroup$
                        – gimusi
                        Dec 5 '18 at 20:14
















                      $begingroup$
                      We conclude is that $a_n$ cannot be convergent! Thank you.
                      $endgroup$
                      – Rikk
                      Dec 5 '18 at 20:12




                      $begingroup$
                      We conclude is that $a_n$ cannot be convergent! Thank you.
                      $endgroup$
                      – Rikk
                      Dec 5 '18 at 20:12












                      $begingroup$
                      @Rikk Yes of course that the general way to prove that. You are welcome! Bye
                      $endgroup$
                      – gimusi
                      Dec 5 '18 at 20:14




                      $begingroup$
                      @Rikk Yes of course that the general way to prove that. You are welcome! Bye
                      $endgroup$
                      – gimusi
                      Dec 5 '18 at 20:14


















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