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












0












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






  • 1




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








  • 1




    $begingroup$
    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.
    $endgroup$
    – Did
    Dec 3 '18 at 14:52
















0












$begingroup$


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







share|cite|improve this question









$endgroup$












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






  • 1




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








  • 1




    $begingroup$
    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.
    $endgroup$
    – Did
    Dec 3 '18 at 14:52














0












0








0





$begingroup$


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







share|cite|improve this question









$endgroup$




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




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 3 '18 at 14:29









kaisakaisa

2019




2019












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






  • 1




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








  • 1




    $begingroup$
    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.
    $endgroup$
    – Did
    Dec 3 '18 at 14:52


















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






  • 1




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








  • 1




    $begingroup$
    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.
    $endgroup$
    – Did
    Dec 3 '18 at 14:52
















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




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




1




1




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






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






1




1




$begingroup$
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.
$endgroup$
– Did
Dec 3 '18 at 14:52




$begingroup$
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.
$endgroup$
– Did
Dec 3 '18 at 14:52










1 Answer
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$begingroup$

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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    -1












    $begingroup$

    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









    $endgroup$


















      -1












      $begingroup$

      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









      $endgroup$
















        -1












        -1








        -1





        $begingroup$

        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









        $endgroup$



        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 DaoustYves Daoust

        128k675227




        128k675227






























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