Limit of sum as definite integral












2














I don't understand why $$displaystyle sum_{k=1}^n dfrac{n}{n^2+kn+k^2} < lim_{nto infty}sum_{k=1}^n dfrac{n}{n^2+kn+k^2}$$



whereas



$$displaystyle sum_{k=0}^{n-1} dfrac{n}{n^2+kn+k^2} > lim_{nto infty}sum_{k=0}^{n-1} dfrac{n}{n^2+kn+k^2} $$



I know that $$lim_{nto infty}sum_{k=1}^{n} dfrac{n}{n^2+kn+k^2}=dfrac{pi}{3sqrt{3}}$$ $$lim_{nto infty}sum_{k=0}^{n-1} dfrac{n}{n^2+kn+k^2}=dfrac{pi}{3sqrt{3}}$$



I saw somewhere on the internet that $$displaystyle dfrac{1}{n}sum_{k=0}^{n-1} fleft(dfrac{k}{n}right) > int_0^1 f(x)dx > dfrac{1}{n}sum_{k=1}^{n} fleft(dfrac{k}{n}right)$$ Why is this true?






integration definite-integrals riemann-sum







share|cite|improve this question
























  • Your last inequality should be reversed!! See my answer below.
    – Robert Z
    Dec 12 '18 at 8:34










  • Refer also to Perfect understanding of Riemann Sums and Understanding Riemann sums.
    – gimusi
    Dec 12 '18 at 9:11










  • Also see this answer.
    – rtybase
    Dec 12 '18 at 9:13


















2














I don't understand why $$displaystyle sum_{k=1}^n dfrac{n}{n^2+kn+k^2} < lim_{nto infty}sum_{k=1}^n dfrac{n}{n^2+kn+k^2}$$



whereas



$$displaystyle sum_{k=0}^{n-1} dfrac{n}{n^2+kn+k^2} > lim_{nto infty}sum_{k=0}^{n-1} dfrac{n}{n^2+kn+k^2} $$



I know that $$lim_{nto infty}sum_{k=1}^{n} dfrac{n}{n^2+kn+k^2}=dfrac{pi}{3sqrt{3}}$$ $$lim_{nto infty}sum_{k=0}^{n-1} dfrac{n}{n^2+kn+k^2}=dfrac{pi}{3sqrt{3}}$$



I saw somewhere on the internet that $$displaystyle dfrac{1}{n}sum_{k=0}^{n-1} fleft(dfrac{k}{n}right) > int_0^1 f(x)dx > dfrac{1}{n}sum_{k=1}^{n} fleft(dfrac{k}{n}right)$$ Why is this true?






integration definite-integrals riemann-sum







share|cite|improve this question
























  • Your last inequality should be reversed!! See my answer below.
    – Robert Z
    Dec 12 '18 at 8:34










  • Refer also to Perfect understanding of Riemann Sums and Understanding Riemann sums.
    – gimusi
    Dec 12 '18 at 9:11










  • Also see this answer.
    – rtybase
    Dec 12 '18 at 9:13
















2












2








2







I don't understand why $$displaystyle sum_{k=1}^n dfrac{n}{n^2+kn+k^2} < lim_{nto infty}sum_{k=1}^n dfrac{n}{n^2+kn+k^2}$$



whereas



$$displaystyle sum_{k=0}^{n-1} dfrac{n}{n^2+kn+k^2} > lim_{nto infty}sum_{k=0}^{n-1} dfrac{n}{n^2+kn+k^2} $$



I know that $$lim_{nto infty}sum_{k=1}^{n} dfrac{n}{n^2+kn+k^2}=dfrac{pi}{3sqrt{3}}$$ $$lim_{nto infty}sum_{k=0}^{n-1} dfrac{n}{n^2+kn+k^2}=dfrac{pi}{3sqrt{3}}$$



I saw somewhere on the internet that $$displaystyle dfrac{1}{n}sum_{k=0}^{n-1} fleft(dfrac{k}{n}right) > int_0^1 f(x)dx > dfrac{1}{n}sum_{k=1}^{n} fleft(dfrac{k}{n}right)$$ Why is this true?






integration definite-integrals riemann-sum







share|cite|improve this question















I don't understand why $$displaystyle sum_{k=1}^n dfrac{n}{n^2+kn+k^2} < lim_{nto infty}sum_{k=1}^n dfrac{n}{n^2+kn+k^2}$$



whereas



$$displaystyle sum_{k=0}^{n-1} dfrac{n}{n^2+kn+k^2} > lim_{nto infty}sum_{k=0}^{n-1} dfrac{n}{n^2+kn+k^2} $$



I know that $$lim_{nto infty}sum_{k=1}^{n} dfrac{n}{n^2+kn+k^2}=dfrac{pi}{3sqrt{3}}$$ $$lim_{nto infty}sum_{k=0}^{n-1} dfrac{n}{n^2+kn+k^2}=dfrac{pi}{3sqrt{3}}$$



I saw somewhere on the internet that $$displaystyle dfrac{1}{n}sum_{k=0}^{n-1} fleft(dfrac{k}{n}right) > int_0^1 f(x)dx > dfrac{1}{n}sum_{k=1}^{n} fleft(dfrac{k}{n}right)$$ Why is this true?






integration definite-integrals riemann-sum




integration definite-integrals riemann-sum






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 12 '18 at 9:11









rtybase

10.4k21433




10.4k21433










asked Dec 12 '18 at 7:46









Loop BackLoop Back

28711




28711












  • Your last inequality should be reversed!! See my answer below.
    – Robert Z
    Dec 12 '18 at 8:34










  • Refer also to Perfect understanding of Riemann Sums and Understanding Riemann sums.
    – gimusi
    Dec 12 '18 at 9:11










  • Also see this answer.
    – rtybase
    Dec 12 '18 at 9:13




















  • Your last inequality should be reversed!! See my answer below.
    – Robert Z
    Dec 12 '18 at 8:34










  • Refer also to Perfect understanding of Riemann Sums and Understanding Riemann sums.
    – gimusi
    Dec 12 '18 at 9:11










  • Also see this answer.
    – rtybase
    Dec 12 '18 at 9:13


















Your last inequality should be reversed!! See my answer below.
– Robert Z
Dec 12 '18 at 8:34




Your last inequality should be reversed!! See my answer below.
– Robert Z
Dec 12 '18 at 8:34












Refer also to Perfect understanding of Riemann Sums and Understanding Riemann sums.
– gimusi
Dec 12 '18 at 9:11




Refer also to Perfect understanding of Riemann Sums and Understanding Riemann sums.
– gimusi
Dec 12 '18 at 9:11












Also see this answer.
– rtybase
Dec 12 '18 at 9:13






Also see this answer.
– rtybase
Dec 12 '18 at 9:13












2 Answers
2






active

oldest

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2














Consider the function
$$f(x)=frac{1}{1+x+x^2}$$
and note that the your inequality holds because $f$ is strictly decreasing.



Indeed for $ngeq 1$, $kgeq 0$, and $xin [frac{k}{n},frac{k+1}{n}],$
$$f(frac{k}{n})> f(x)> f(frac{k+1}{n}).$$
By integrating over the interval $[frac{k}{n},frac{k+1}{n}]$, we get
$$frac{f(frac{k}{n})}{n}=int_{frac{k}{n}}^{frac{k+1}{n}}f(frac{k}{n})dx> int_{frac{k}{n}}^{frac{k+1}{n}}f(x)dx> int_{frac{k}{n}}^{frac{k+1}{n}}f(frac{k+1}{n})dx=frac{f(frac{k+1}{n}) }{n}.$$
Finally we take the sum for $k=0,dots,n-1$,
$$frac{1}{n}sum_{k=0}^{n-1}f(frac{k}{n})>int_0^1 f(x),dx
>frac{1}{n}sum_{k=0}^{n-1}f(frac{k+1}{n}).$$
Note that the last sum on the right is equal to
$$frac{1}{n}sum_{k=1}^{n}f(frac{k}{n})=
frac{1}{n}sum_{k=0}^{n-1}f(frac{k}{n})+frac{f(1)-f(0)}{n}.$$



P.S. Once we have the double inequality, we may conclude that
$$lim_{nto infty}sum_{k=0}^{n-1} dfrac{n}{n^2+kn+k^2}=lim_{nto infty}sum_{k=1}^{n} dfrac{n}{n^2+kn+k^2}=int_0^1frac{dx}{1+x+x^2}=dfrac{pi}{3sqrt{3}}.$$






share|cite|improve this answer























  • Taking n=1 last inequality proves to be correct
    – Loop Back
    Dec 12 '18 at 8:24










  • Are you talking about my "last inequality" or yours?
    – Robert Z
    Dec 12 '18 at 8:34










  • Yes...............
    – Loop Back
    Dec 12 '18 at 8:39










  • @LoopBack Yes what?
    – Robert Z
    Dec 12 '18 at 8:42












  • My last inequality is correct for n=1
    – Loop Back
    Dec 12 '18 at 8:43



















2














In this case $f(x)=frac{1}{1+x+x^2}$ is decreasing on $[0,,1]$, so the leftmost expression is the sum of areas of width-$1/n$ rectangles that overestimate the integral, while the rightmost expression is an analogous underestimation with rectangles.






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









    2














    Consider the function
    $$f(x)=frac{1}{1+x+x^2}$$
    and note that the your inequality holds because $f$ is strictly decreasing.



    Indeed for $ngeq 1$, $kgeq 0$, and $xin [frac{k}{n},frac{k+1}{n}],$
    $$f(frac{k}{n})> f(x)> f(frac{k+1}{n}).$$
    By integrating over the interval $[frac{k}{n},frac{k+1}{n}]$, we get
    $$frac{f(frac{k}{n})}{n}=int_{frac{k}{n}}^{frac{k+1}{n}}f(frac{k}{n})dx> int_{frac{k}{n}}^{frac{k+1}{n}}f(x)dx> int_{frac{k}{n}}^{frac{k+1}{n}}f(frac{k+1}{n})dx=frac{f(frac{k+1}{n}) }{n}.$$
    Finally we take the sum for $k=0,dots,n-1$,
    $$frac{1}{n}sum_{k=0}^{n-1}f(frac{k}{n})>int_0^1 f(x),dx
    >frac{1}{n}sum_{k=0}^{n-1}f(frac{k+1}{n}).$$
    Note that the last sum on the right is equal to
    $$frac{1}{n}sum_{k=1}^{n}f(frac{k}{n})=
    frac{1}{n}sum_{k=0}^{n-1}f(frac{k}{n})+frac{f(1)-f(0)}{n}.$$



    P.S. Once we have the double inequality, we may conclude that
    $$lim_{nto infty}sum_{k=0}^{n-1} dfrac{n}{n^2+kn+k^2}=lim_{nto infty}sum_{k=1}^{n} dfrac{n}{n^2+kn+k^2}=int_0^1frac{dx}{1+x+x^2}=dfrac{pi}{3sqrt{3}}.$$






    share|cite|improve this answer























    • Taking n=1 last inequality proves to be correct
      – Loop Back
      Dec 12 '18 at 8:24










    • Are you talking about my "last inequality" or yours?
      – Robert Z
      Dec 12 '18 at 8:34










    • Yes...............
      – Loop Back
      Dec 12 '18 at 8:39










    • @LoopBack Yes what?
      – Robert Z
      Dec 12 '18 at 8:42












    • My last inequality is correct for n=1
      – Loop Back
      Dec 12 '18 at 8:43
















    2














    Consider the function
    $$f(x)=frac{1}{1+x+x^2}$$
    and note that the your inequality holds because $f$ is strictly decreasing.



    Indeed for $ngeq 1$, $kgeq 0$, and $xin [frac{k}{n},frac{k+1}{n}],$
    $$f(frac{k}{n})> f(x)> f(frac{k+1}{n}).$$
    By integrating over the interval $[frac{k}{n},frac{k+1}{n}]$, we get
    $$frac{f(frac{k}{n})}{n}=int_{frac{k}{n}}^{frac{k+1}{n}}f(frac{k}{n})dx> int_{frac{k}{n}}^{frac{k+1}{n}}f(x)dx> int_{frac{k}{n}}^{frac{k+1}{n}}f(frac{k+1}{n})dx=frac{f(frac{k+1}{n}) }{n}.$$
    Finally we take the sum for $k=0,dots,n-1$,
    $$frac{1}{n}sum_{k=0}^{n-1}f(frac{k}{n})>int_0^1 f(x),dx
    >frac{1}{n}sum_{k=0}^{n-1}f(frac{k+1}{n}).$$
    Note that the last sum on the right is equal to
    $$frac{1}{n}sum_{k=1}^{n}f(frac{k}{n})=
    frac{1}{n}sum_{k=0}^{n-1}f(frac{k}{n})+frac{f(1)-f(0)}{n}.$$



    P.S. Once we have the double inequality, we may conclude that
    $$lim_{nto infty}sum_{k=0}^{n-1} dfrac{n}{n^2+kn+k^2}=lim_{nto infty}sum_{k=1}^{n} dfrac{n}{n^2+kn+k^2}=int_0^1frac{dx}{1+x+x^2}=dfrac{pi}{3sqrt{3}}.$$






    share|cite|improve this answer























    • Taking n=1 last inequality proves to be correct
      – Loop Back
      Dec 12 '18 at 8:24










    • Are you talking about my "last inequality" or yours?
      – Robert Z
      Dec 12 '18 at 8:34










    • Yes...............
      – Loop Back
      Dec 12 '18 at 8:39










    • @LoopBack Yes what?
      – Robert Z
      Dec 12 '18 at 8:42












    • My last inequality is correct for n=1
      – Loop Back
      Dec 12 '18 at 8:43














    2












    2








    2






    Consider the function
    $$f(x)=frac{1}{1+x+x^2}$$
    and note that the your inequality holds because $f$ is strictly decreasing.



    Indeed for $ngeq 1$, $kgeq 0$, and $xin [frac{k}{n},frac{k+1}{n}],$
    $$f(frac{k}{n})> f(x)> f(frac{k+1}{n}).$$
    By integrating over the interval $[frac{k}{n},frac{k+1}{n}]$, we get
    $$frac{f(frac{k}{n})}{n}=int_{frac{k}{n}}^{frac{k+1}{n}}f(frac{k}{n})dx> int_{frac{k}{n}}^{frac{k+1}{n}}f(x)dx> int_{frac{k}{n}}^{frac{k+1}{n}}f(frac{k+1}{n})dx=frac{f(frac{k+1}{n}) }{n}.$$
    Finally we take the sum for $k=0,dots,n-1$,
    $$frac{1}{n}sum_{k=0}^{n-1}f(frac{k}{n})>int_0^1 f(x),dx
    >frac{1}{n}sum_{k=0}^{n-1}f(frac{k+1}{n}).$$
    Note that the last sum on the right is equal to
    $$frac{1}{n}sum_{k=1}^{n}f(frac{k}{n})=
    frac{1}{n}sum_{k=0}^{n-1}f(frac{k}{n})+frac{f(1)-f(0)}{n}.$$



    P.S. Once we have the double inequality, we may conclude that
    $$lim_{nto infty}sum_{k=0}^{n-1} dfrac{n}{n^2+kn+k^2}=lim_{nto infty}sum_{k=1}^{n} dfrac{n}{n^2+kn+k^2}=int_0^1frac{dx}{1+x+x^2}=dfrac{pi}{3sqrt{3}}.$$






    share|cite|improve this answer














    Consider the function
    $$f(x)=frac{1}{1+x+x^2}$$
    and note that the your inequality holds because $f$ is strictly decreasing.



    Indeed for $ngeq 1$, $kgeq 0$, and $xin [frac{k}{n},frac{k+1}{n}],$
    $$f(frac{k}{n})> f(x)> f(frac{k+1}{n}).$$
    By integrating over the interval $[frac{k}{n},frac{k+1}{n}]$, we get
    $$frac{f(frac{k}{n})}{n}=int_{frac{k}{n}}^{frac{k+1}{n}}f(frac{k}{n})dx> int_{frac{k}{n}}^{frac{k+1}{n}}f(x)dx> int_{frac{k}{n}}^{frac{k+1}{n}}f(frac{k+1}{n})dx=frac{f(frac{k+1}{n}) }{n}.$$
    Finally we take the sum for $k=0,dots,n-1$,
    $$frac{1}{n}sum_{k=0}^{n-1}f(frac{k}{n})>int_0^1 f(x),dx
    >frac{1}{n}sum_{k=0}^{n-1}f(frac{k+1}{n}).$$
    Note that the last sum on the right is equal to
    $$frac{1}{n}sum_{k=1}^{n}f(frac{k}{n})=
    frac{1}{n}sum_{k=0}^{n-1}f(frac{k}{n})+frac{f(1)-f(0)}{n}.$$



    P.S. Once we have the double inequality, we may conclude that
    $$lim_{nto infty}sum_{k=0}^{n-1} dfrac{n}{n^2+kn+k^2}=lim_{nto infty}sum_{k=1}^{n} dfrac{n}{n^2+kn+k^2}=int_0^1frac{dx}{1+x+x^2}=dfrac{pi}{3sqrt{3}}.$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 12 '18 at 9:13

























    answered Dec 12 '18 at 7:56









    Robert ZRobert Z

    93.8k1061132




    93.8k1061132












    • Taking n=1 last inequality proves to be correct
      – Loop Back
      Dec 12 '18 at 8:24










    • Are you talking about my "last inequality" or yours?
      – Robert Z
      Dec 12 '18 at 8:34










    • Yes...............
      – Loop Back
      Dec 12 '18 at 8:39










    • @LoopBack Yes what?
      – Robert Z
      Dec 12 '18 at 8:42












    • My last inequality is correct for n=1
      – Loop Back
      Dec 12 '18 at 8:43


















    • Taking n=1 last inequality proves to be correct
      – Loop Back
      Dec 12 '18 at 8:24










    • Are you talking about my "last inequality" or yours?
      – Robert Z
      Dec 12 '18 at 8:34










    • Yes...............
      – Loop Back
      Dec 12 '18 at 8:39










    • @LoopBack Yes what?
      – Robert Z
      Dec 12 '18 at 8:42












    • My last inequality is correct for n=1
      – Loop Back
      Dec 12 '18 at 8:43
















    Taking n=1 last inequality proves to be correct
    – Loop Back
    Dec 12 '18 at 8:24




    Taking n=1 last inequality proves to be correct
    – Loop Back
    Dec 12 '18 at 8:24












    Are you talking about my "last inequality" or yours?
    – Robert Z
    Dec 12 '18 at 8:34




    Are you talking about my "last inequality" or yours?
    – Robert Z
    Dec 12 '18 at 8:34












    Yes...............
    – Loop Back
    Dec 12 '18 at 8:39




    Yes...............
    – Loop Back
    Dec 12 '18 at 8:39












    @LoopBack Yes what?
    – Robert Z
    Dec 12 '18 at 8:42






    @LoopBack Yes what?
    – Robert Z
    Dec 12 '18 at 8:42














    My last inequality is correct for n=1
    – Loop Back
    Dec 12 '18 at 8:43




    My last inequality is correct for n=1
    – Loop Back
    Dec 12 '18 at 8:43











    2














    In this case $f(x)=frac{1}{1+x+x^2}$ is decreasing on $[0,,1]$, so the leftmost expression is the sum of areas of width-$1/n$ rectangles that overestimate the integral, while the rightmost expression is an analogous underestimation with rectangles.






    share|cite|improve this answer


























      2














      In this case $f(x)=frac{1}{1+x+x^2}$ is decreasing on $[0,,1]$, so the leftmost expression is the sum of areas of width-$1/n$ rectangles that overestimate the integral, while the rightmost expression is an analogous underestimation with rectangles.






      share|cite|improve this answer
























        2












        2








        2






        In this case $f(x)=frac{1}{1+x+x^2}$ is decreasing on $[0,,1]$, so the leftmost expression is the sum of areas of width-$1/n$ rectangles that overestimate the integral, while the rightmost expression is an analogous underestimation with rectangles.






        share|cite|improve this answer












        In this case $f(x)=frac{1}{1+x+x^2}$ is decreasing on $[0,,1]$, so the leftmost expression is the sum of areas of width-$1/n$ rectangles that overestimate the integral, while the rightmost expression is an analogous underestimation with rectangles.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 12 '18 at 7:57









        J.G.J.G.

        23.3k22137




        23.3k22137






























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            Cabo Verde

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              Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
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            Gyllenstierna

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