Convergence of Sum of Sequences











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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.










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  • 1




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















up vote
4
down vote

favorite












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.










share|cite|improve this question







New contributor




Rikk is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
















  • 1




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













up vote
4
down vote

favorite









up vote
4
down vote

favorite











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.










share|cite|improve this question







New contributor




Rikk is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











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






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asked Dec 5 at 19:43









Rikk

424




424




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New contributor





Rikk is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Rikk is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 1




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














  • 1




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








1




1




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




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










3 Answers
3






active

oldest

votes

















up vote
2
down vote



accepted










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




























    up vote
    3
    down vote













    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




























      up vote
      2
      down vote













      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























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










      • @Rikk Yes of course that the general way to prove that. You are welcome! Bye
        – gimusi
        Dec 5 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








      up vote
      2
      down vote



      accepted










      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

























        up vote
        2
        down vote



        accepted










        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























          up vote
          2
          down vote



          accepted







          up vote
          2
          down vote



          accepted






          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












          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 at 19:50









          hamam_Abdallah

          37.3k21634




          37.3k21634






















              up vote
              3
              down vote













              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

























                up vote
                3
                down vote













                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























                  up vote
                  3
                  down vote










                  up vote
                  3
                  down vote









                  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












                  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 at 19:54









                  mathcounterexamples.net

                  23.8k21753




                  23.8k21753






















                      up vote
                      2
                      down vote













                      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























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










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















                      up vote
                      2
                      down vote













                      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























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










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













                      up vote
                      2
                      down vote










                      up vote
                      2
                      down vote









                      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














                      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 at 19:57

























                      answered Dec 5 at 19:49









                      gimusi

                      91k74495




                      91k74495












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










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


















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










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
















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




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












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




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










                      Rikk is a new contributor. Be nice, and check out our Code of Conduct.










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