How to show $lim_{ntoinfty}sum_{k=1}^n(1+(k)/n)^3(1/n)=15/4$?












0














How would I show $$lim_{ntoinfty}sum_{k=1}^n(1+(k)/n)^3(1/n)=15/4?$$



My attempt is using the Riemann Sum technique. We know $(1+(k)/n)^2=f(zeta_k)$ and $(1/n)=Delta x$. So the definite integral goes from $a$ to $b$ where $b-a=1$.



I think I should find the formula for $$sum_{k=1}^nk^3.$$ This would be $1+8+27+cdots+n^3$... but then I get stuck.










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  • Ask Faulhaber...
    – Yves Daoust
    Dec 3 '18 at 14:30






  • 1




    As a Riemannian sum, $int_0^1(1+x)^3dx=15/4$.
    – Yves Daoust
    Dec 3 '18 at 14:31








  • 1




    This is odd, you are asked to find the limit of some sums, thus you turn to integrals (good idea), but then, instead of evaluating the relevant integral (a trivial task), you suddenly make a U-turn to come back to sums... where you declare you are stuck.
    – Did
    Dec 3 '18 at 14:52
















0














How would I show $$lim_{ntoinfty}sum_{k=1}^n(1+(k)/n)^3(1/n)=15/4?$$



My attempt is using the Riemann Sum technique. We know $(1+(k)/n)^2=f(zeta_k)$ and $(1/n)=Delta x$. So the definite integral goes from $a$ to $b$ where $b-a=1$.



I think I should find the formula for $$sum_{k=1}^nk^3.$$ This would be $1+8+27+cdots+n^3$... but then I get stuck.










share|cite|improve this question






















  • Ask Faulhaber...
    – Yves Daoust
    Dec 3 '18 at 14:30






  • 1




    As a Riemannian sum, $int_0^1(1+x)^3dx=15/4$.
    – Yves Daoust
    Dec 3 '18 at 14:31








  • 1




    This is odd, you are asked to find the limit of some sums, thus you turn to integrals (good idea), but then, instead of evaluating the relevant integral (a trivial task), you suddenly make a U-turn to come back to sums... where you declare you are stuck.
    – Did
    Dec 3 '18 at 14:52














0












0








0







How would I show $$lim_{ntoinfty}sum_{k=1}^n(1+(k)/n)^3(1/n)=15/4?$$



My attempt is using the Riemann Sum technique. We know $(1+(k)/n)^2=f(zeta_k)$ and $(1/n)=Delta x$. So the definite integral goes from $a$ to $b$ where $b-a=1$.



I think I should find the formula for $$sum_{k=1}^nk^3.$$ This would be $1+8+27+cdots+n^3$... but then I get stuck.










share|cite|improve this question













How would I show $$lim_{ntoinfty}sum_{k=1}^n(1+(k)/n)^3(1/n)=15/4?$$



My attempt is using the Riemann Sum technique. We know $(1+(k)/n)^2=f(zeta_k)$ and $(1/n)=Delta x$. So the definite integral goes from $a$ to $b$ where $b-a=1$.



I think I should find the formula for $$sum_{k=1}^nk^3.$$ This would be $1+8+27+cdots+n^3$... but then I get stuck.







real-analysis definite-integrals riemann-integration riemann-sum






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asked Dec 3 '18 at 14:29









kaisa

1019




1019












  • Ask Faulhaber...
    – Yves Daoust
    Dec 3 '18 at 14:30






  • 1




    As a Riemannian sum, $int_0^1(1+x)^3dx=15/4$.
    – Yves Daoust
    Dec 3 '18 at 14:31








  • 1




    This is odd, you are asked to find the limit of some sums, thus you turn to integrals (good idea), but then, instead of evaluating the relevant integral (a trivial task), you suddenly make a U-turn to come back to sums... where you declare you are stuck.
    – Did
    Dec 3 '18 at 14:52


















  • Ask Faulhaber...
    – Yves Daoust
    Dec 3 '18 at 14:30






  • 1




    As a Riemannian sum, $int_0^1(1+x)^3dx=15/4$.
    – Yves Daoust
    Dec 3 '18 at 14:31








  • 1




    This is odd, you are asked to find the limit of some sums, thus you turn to integrals (good idea), but then, instead of evaluating the relevant integral (a trivial task), you suddenly make a U-turn to come back to sums... where you declare you are stuck.
    – Did
    Dec 3 '18 at 14:52
















Ask Faulhaber...
– Yves Daoust
Dec 3 '18 at 14:30




Ask Faulhaber...
– Yves Daoust
Dec 3 '18 at 14:30




1




1




As a Riemannian sum, $int_0^1(1+x)^3dx=15/4$.
– Yves Daoust
Dec 3 '18 at 14:31






As a Riemannian sum, $int_0^1(1+x)^3dx=15/4$.
– Yves Daoust
Dec 3 '18 at 14:31






1




1




This is odd, you are asked to find the limit of some sums, thus you turn to integrals (good idea), but then, instead of evaluating the relevant integral (a trivial task), you suddenly make a U-turn to come back to sums... where you declare you are stuck.
– Did
Dec 3 '18 at 14:52




This is odd, you are asked to find the limit of some sums, thus you turn to integrals (good idea), but then, instead of evaluating the relevant integral (a trivial task), you suddenly make a U-turn to come back to sums... where you declare you are stuck.
– Did
Dec 3 '18 at 14:52










1 Answer
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Hint:



$$S_n:=sum_{k=1}^n k^3$$ must be a quartic polynomial in $n$ because $S_{n}-S_{n-1}=n^3$ is a cubic polynomial.



Now



$$S_n-S_{n-1}=an^4+bn^3+cn^2+cdots-a(n-1)^4-b(n-1)^3-c(n-1)^2-cdots
\=4an^3+bn^3-bn^3+cdots$$
and lower order terms. You conclude that



$$S_nsimfrac{n^4}4$$ and this is enough for you to evaluate the limit.






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    1 Answer
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    active

    oldest

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    1 Answer
    1






    active

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    active

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    active

    oldest

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    -1














    Hint:



    $$S_n:=sum_{k=1}^n k^3$$ must be a quartic polynomial in $n$ because $S_{n}-S_{n-1}=n^3$ is a cubic polynomial.



    Now



    $$S_n-S_{n-1}=an^4+bn^3+cn^2+cdots-a(n-1)^4-b(n-1)^3-c(n-1)^2-cdots
    \=4an^3+bn^3-bn^3+cdots$$
    and lower order terms. You conclude that



    $$S_nsimfrac{n^4}4$$ and this is enough for you to evaluate the limit.






    share|cite|improve this answer


























      -1














      Hint:



      $$S_n:=sum_{k=1}^n k^3$$ must be a quartic polynomial in $n$ because $S_{n}-S_{n-1}=n^3$ is a cubic polynomial.



      Now



      $$S_n-S_{n-1}=an^4+bn^3+cn^2+cdots-a(n-1)^4-b(n-1)^3-c(n-1)^2-cdots
      \=4an^3+bn^3-bn^3+cdots$$
      and lower order terms. You conclude that



      $$S_nsimfrac{n^4}4$$ and this is enough for you to evaluate the limit.






      share|cite|improve this answer
























        -1












        -1








        -1






        Hint:



        $$S_n:=sum_{k=1}^n k^3$$ must be a quartic polynomial in $n$ because $S_{n}-S_{n-1}=n^3$ is a cubic polynomial.



        Now



        $$S_n-S_{n-1}=an^4+bn^3+cn^2+cdots-a(n-1)^4-b(n-1)^3-c(n-1)^2-cdots
        \=4an^3+bn^3-bn^3+cdots$$
        and lower order terms. You conclude that



        $$S_nsimfrac{n^4}4$$ and this is enough for you to evaluate the limit.






        share|cite|improve this answer












        Hint:



        $$S_n:=sum_{k=1}^n k^3$$ must be a quartic polynomial in $n$ because $S_{n}-S_{n-1}=n^3$ is a cubic polynomial.



        Now



        $$S_n-S_{n-1}=an^4+bn^3+cn^2+cdots-a(n-1)^4-b(n-1)^3-c(n-1)^2-cdots
        \=4an^3+bn^3-bn^3+cdots$$
        and lower order terms. You conclude that



        $$S_nsimfrac{n^4}4$$ and this is enough for you to evaluate the limit.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 3 '18 at 14:36









        Yves Daoust

        124k671221




        124k671221






























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