How to evaluate $lim_{n to infty} sum_{k=1}^n frac{n+k}{n^3+k}$?











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How to evaluate the following limit?
$$lim_{n to infty} sum_{k=1}^n frac{n+k}{n^3+k}$$




I think that every term of the sum is greater than the first one and smaller than the last one and then from the squeeze theorem the limit is $0$. However, I can't prove that inequality.






calculus sequences-and-series limits







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  • 1




    What are you asking?
    – Will M.
    Nov 19 at 18:31






  • 1




    The sum is less than $n frac{n +n}{n^3+1}$
    – RRL
    Nov 19 at 18:33

















up vote
0
down vote

favorite













How to evaluate the following limit?
$$lim_{n to infty} sum_{k=1}^n frac{n+k}{n^3+k}$$




I think that every term of the sum is greater than the first one and smaller than the last one and then from the squeeze theorem the limit is $0$. However, I can't prove that inequality.






calculus sequences-and-series limits







share|cite|improve this question




















  • 1




    What are you asking?
    – Will M.
    Nov 19 at 18:31






  • 1




    The sum is less than $n frac{n +n}{n^3+1}$
    – RRL
    Nov 19 at 18:33















up vote
0
down vote

favorite









up vote
0
down vote

favorite












How to evaluate the following limit?
$$lim_{n to infty} sum_{k=1}^n frac{n+k}{n^3+k}$$




I think that every term of the sum is greater than the first one and smaller than the last one and then from the squeeze theorem the limit is $0$. However, I can't prove that inequality.






calculus sequences-and-series limits







share|cite|improve this question
















How to evaluate the following limit?
$$lim_{n to infty} sum_{k=1}^n frac{n+k}{n^3+k}$$




I think that every term of the sum is greater than the first one and smaller than the last one and then from the squeeze theorem the limit is $0$. However, I can't prove that inequality.






calculus sequences-and-series limits




calculus sequences-and-series limits






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 days ago









user587192

1,21810




1,21810










asked Nov 19 at 18:27









user69503

486




486








  • 1




    What are you asking?
    – Will M.
    Nov 19 at 18:31






  • 1




    The sum is less than $n frac{n +n}{n^3+1}$
    – RRL
    Nov 19 at 18:33
















  • 1




    What are you asking?
    – Will M.
    Nov 19 at 18:31






  • 1




    The sum is less than $n frac{n +n}{n^3+1}$
    – RRL
    Nov 19 at 18:33










1




1




What are you asking?
– Will M.
Nov 19 at 18:31




What are you asking?
– Will M.
Nov 19 at 18:31




1




1




The sum is less than $n frac{n +n}{n^3+1}$
– RRL
Nov 19 at 18:33






The sum is less than $n frac{n +n}{n^3+1}$
– RRL
Nov 19 at 18:33












2 Answers
2






active

oldest

votes

















up vote
5
down vote



accepted










This should work:
$$sum_{k=1}^nfrac{n+k}{n^3+k}leqsum_{k=1}^nfrac{n+k}{n^3}leqsum_{k=1}^nfrac{n+n}{n^3}=frac{2n^2}{n^3}=frac{2}{n}to0$$



Also, every term is $geq 0$ so $0$ is also a lower bound.






share|cite|improve this answer




























    up vote
    1
    down vote













    HINT:



    Note that $n+1le n+kle 2n$ and $n^3+1le n^3 +kle n^3+n$.






    share|cite|improve this answer





















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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      5
      down vote



      accepted










      This should work:
      $$sum_{k=1}^nfrac{n+k}{n^3+k}leqsum_{k=1}^nfrac{n+k}{n^3}leqsum_{k=1}^nfrac{n+n}{n^3}=frac{2n^2}{n^3}=frac{2}{n}to0$$



      Also, every term is $geq 0$ so $0$ is also a lower bound.






      share|cite|improve this answer

























        up vote
        5
        down vote



        accepted










        This should work:
        $$sum_{k=1}^nfrac{n+k}{n^3+k}leqsum_{k=1}^nfrac{n+k}{n^3}leqsum_{k=1}^nfrac{n+n}{n^3}=frac{2n^2}{n^3}=frac{2}{n}to0$$



        Also, every term is $geq 0$ so $0$ is also a lower bound.






        share|cite|improve this answer























          up vote
          5
          down vote



          accepted







          up vote
          5
          down vote



          accepted






          This should work:
          $$sum_{k=1}^nfrac{n+k}{n^3+k}leqsum_{k=1}^nfrac{n+k}{n^3}leqsum_{k=1}^nfrac{n+n}{n^3}=frac{2n^2}{n^3}=frac{2}{n}to0$$



          Also, every term is $geq 0$ so $0$ is also a lower bound.






          share|cite|improve this answer












          This should work:
          $$sum_{k=1}^nfrac{n+k}{n^3+k}leqsum_{k=1}^nfrac{n+k}{n^3}leqsum_{k=1}^nfrac{n+n}{n^3}=frac{2n^2}{n^3}=frac{2}{n}to0$$



          Also, every term is $geq 0$ so $0$ is also a lower bound.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 19 at 18:33









          Marco

          1909




          1909






















              up vote
              1
              down vote













              HINT:



              Note that $n+1le n+kle 2n$ and $n^3+1le n^3 +kle n^3+n$.






              share|cite|improve this answer

























                up vote
                1
                down vote













                HINT:



                Note that $n+1le n+kle 2n$ and $n^3+1le n^3 +kle n^3+n$.






                share|cite|improve this answer























                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  HINT:



                  Note that $n+1le n+kle 2n$ and $n^3+1le n^3 +kle n^3+n$.






                  share|cite|improve this answer












                  HINT:



                  Note that $n+1le n+kle 2n$ and $n^3+1le n^3 +kle n^3+n$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 19 at 18:35









                  Mark Viola

                  129k1273170




                  129k1273170






























                       

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