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 Nov 19 at 21:03









user587192

1,424112




1,424112










asked Nov 19 at 18:27









user69503

546




546








  • 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





















  • Typo: $n + n = 2n$.
    – Paul Frost
    Dec 2 at 10:25


















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





















    • Typo: $n + n = 2n$.
      – Paul Frost
      Dec 2 at 10:25















    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





















    • Typo: $n + n = 2n$.
      – Paul Frost
      Dec 2 at 10:25













    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

    31019




    31019












    • Typo: $n + n = 2n$.
      – Paul Frost
      Dec 2 at 10:25


















    • Typo: $n + n = 2n$.
      – Paul Frost
      Dec 2 at 10:25
















    Typo: $n + n = 2n$.
    – Paul Frost
    Dec 2 at 10:25




    Typo: $n + n = 2n$.
    – Paul Frost
    Dec 2 at 10:25










    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

        130k1273170




        130k1273170






























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