Expressing Riemann sums as integrals












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$$L_2=lim_{n→∞}sum_{k=1}^nfrac{(k-cos^2(k))^4}{n^5}.$$
My teacher said that when brackets at the numarator is expanded the limit of sum except $dfrac{k^4}{n^5}$ equals $0$, so the Riemann sum becomes $limlimits_{n→∞}sumlimits_{k=1}^ndfrac{k^4}{n^5}$. I don't understand this point. Please explain this point. Sorry for my bad English.






limits definite-integrals







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    $$L_2=lim_{n→∞}sum_{k=1}^nfrac{(k-cos^2(k))^4}{n^5}.$$
    My teacher said that when brackets at the numarator is expanded the limit of sum except $dfrac{k^4}{n^5}$ equals $0$, so the Riemann sum becomes $limlimits_{n→∞}sumlimits_{k=1}^ndfrac{k^4}{n^5}$. I don't understand this point. Please explain this point. Sorry for my bad English.






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      $begingroup$


      $$L_2=lim_{n→∞}sum_{k=1}^nfrac{(k-cos^2(k))^4}{n^5}.$$
      My teacher said that when brackets at the numarator is expanded the limit of sum except $dfrac{k^4}{n^5}$ equals $0$, so the Riemann sum becomes $limlimits_{n→∞}sumlimits_{k=1}^ndfrac{k^4}{n^5}$. I don't understand this point. Please explain this point. Sorry for my bad English.






      limits definite-integrals







      share|cite|improve this question











      $endgroup$




      $$L_2=lim_{n→∞}sum_{k=1}^nfrac{(k-cos^2(k))^4}{n^5}.$$
      My teacher said that when brackets at the numarator is expanded the limit of sum except $dfrac{k^4}{n^5}$ equals $0$, so the Riemann sum becomes $limlimits_{n→∞}sumlimits_{k=1}^ndfrac{k^4}{n^5}$. I don't understand this point. Please explain this point. Sorry for my bad English.






      limits definite-integrals




      limits definite-integrals






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      edited Dec 21 '18 at 8:11









      Saad

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      asked Dec 21 '18 at 7:38









      1u2i3o41u2i3o4

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          $begingroup$

          $sum_{k=1}^{n} k^{j} leq int_1^{n} x^{j} dx =frac {n^{j+1} -1} {j+1}$ and $frac {n^{j+1} -1} {n^{5}} to 0$ as $ n to infty$ if $j <4$. Hence, when you expand $(k-cos^{2}(k))^{4}$ all terms except $k^{4}$ give limit $0$. [Note that $cos^{2} k leq 1$].






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            $begingroup$

            $$sumfrac{(k-cos^2 k)^4}{n^5}=sum left(frac{k^4-4k^3cos ^2left(kright)+6k^2cos ^4left(kright)-4kcos ^6left(kright)+cos ^8left(kright)}{n^5}right)$$Every term is involving $cos k$ except the term $k^4$. Applying summation and use algebra of limits together with the fact $cos k$ is bounded by $1$ to get your result






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              $begingroup$

              Since
              $$
              sum_{k=1}^nk^3=frac{n^2(n+1)^2}4
              $$
              we get
              $$
              begin{align}
              lim_{ntoinfty}sum_{k=1}^nfrac{left(k-cos^2(k)right)^4}{n^5}
              &=lim_{ntoinfty}sum_{k=1}^nleft[frac{k^4}{n^5}+O!left(frac{k^3}{n^5}right)right]\
              &=lim_{ntoinfty}left[sum_{k=1}^nfrac{k^4}{n^4}frac1n+O!left(frac1nright)right]\
              &=int_0^1x^4,mathrm{d}x+0\[6pt]
              &=frac15
              end{align}
              $$






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                3 Answers
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                3 Answers
                3






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                3












                $begingroup$

                $sum_{k=1}^{n} k^{j} leq int_1^{n} x^{j} dx =frac {n^{j+1} -1} {j+1}$ and $frac {n^{j+1} -1} {n^{5}} to 0$ as $ n to infty$ if $j <4$. Hence, when you expand $(k-cos^{2}(k))^{4}$ all terms except $k^{4}$ give limit $0$. [Note that $cos^{2} k leq 1$].






                share|cite|improve this answer









                $endgroup$


















                  3












                  $begingroup$

                  $sum_{k=1}^{n} k^{j} leq int_1^{n} x^{j} dx =frac {n^{j+1} -1} {j+1}$ and $frac {n^{j+1} -1} {n^{5}} to 0$ as $ n to infty$ if $j <4$. Hence, when you expand $(k-cos^{2}(k))^{4}$ all terms except $k^{4}$ give limit $0$. [Note that $cos^{2} k leq 1$].






                  share|cite|improve this answer









                  $endgroup$
















                    3












                    3








                    3





                    $begingroup$

                    $sum_{k=1}^{n} k^{j} leq int_1^{n} x^{j} dx =frac {n^{j+1} -1} {j+1}$ and $frac {n^{j+1} -1} {n^{5}} to 0$ as $ n to infty$ if $j <4$. Hence, when you expand $(k-cos^{2}(k))^{4}$ all terms except $k^{4}$ give limit $0$. [Note that $cos^{2} k leq 1$].






                    share|cite|improve this answer









                    $endgroup$



                    $sum_{k=1}^{n} k^{j} leq int_1^{n} x^{j} dx =frac {n^{j+1} -1} {j+1}$ and $frac {n^{j+1} -1} {n^{5}} to 0$ as $ n to infty$ if $j <4$. Hence, when you expand $(k-cos^{2}(k))^{4}$ all terms except $k^{4}$ give limit $0$. [Note that $cos^{2} k leq 1$].







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 21 '18 at 7:50









                    Kavi Rama MurthyKavi Rama Murthy

                    60.6k42161




                    60.6k42161























                        0












                        $begingroup$

                        $$sumfrac{(k-cos^2 k)^4}{n^5}=sum left(frac{k^4-4k^3cos ^2left(kright)+6k^2cos ^4left(kright)-4kcos ^6left(kright)+cos ^8left(kright)}{n^5}right)$$Every term is involving $cos k$ except the term $k^4$. Applying summation and use algebra of limits together with the fact $cos k$ is bounded by $1$ to get your result






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          $$sumfrac{(k-cos^2 k)^4}{n^5}=sum left(frac{k^4-4k^3cos ^2left(kright)+6k^2cos ^4left(kright)-4kcos ^6left(kright)+cos ^8left(kright)}{n^5}right)$$Every term is involving $cos k$ except the term $k^4$. Applying summation and use algebra of limits together with the fact $cos k$ is bounded by $1$ to get your result






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            $$sumfrac{(k-cos^2 k)^4}{n^5}=sum left(frac{k^4-4k^3cos ^2left(kright)+6k^2cos ^4left(kright)-4kcos ^6left(kright)+cos ^8left(kright)}{n^5}right)$$Every term is involving $cos k$ except the term $k^4$. Applying summation and use algebra of limits together with the fact $cos k$ is bounded by $1$ to get your result






                            share|cite|improve this answer









                            $endgroup$



                            $$sumfrac{(k-cos^2 k)^4}{n^5}=sum left(frac{k^4-4k^3cos ^2left(kright)+6k^2cos ^4left(kright)-4kcos ^6left(kright)+cos ^8left(kright)}{n^5}right)$$Every term is involving $cos k$ except the term $k^4$. Applying summation and use algebra of limits together with the fact $cos k$ is bounded by $1$ to get your result







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 21 '18 at 7:58









                            Chinnapparaj RChinnapparaj R

                            5,5072928




                            5,5072928























                                0












                                $begingroup$

                                Since
                                $$
                                sum_{k=1}^nk^3=frac{n^2(n+1)^2}4
                                $$
                                we get
                                $$
                                begin{align}
                                lim_{ntoinfty}sum_{k=1}^nfrac{left(k-cos^2(k)right)^4}{n^5}
                                &=lim_{ntoinfty}sum_{k=1}^nleft[frac{k^4}{n^5}+O!left(frac{k^3}{n^5}right)right]\
                                &=lim_{ntoinfty}left[sum_{k=1}^nfrac{k^4}{n^4}frac1n+O!left(frac1nright)right]\
                                &=int_0^1x^4,mathrm{d}x+0\[6pt]
                                &=frac15
                                end{align}
                                $$






                                share|cite|improve this answer









                                $endgroup$


















                                  0












                                  $begingroup$

                                  Since
                                  $$
                                  sum_{k=1}^nk^3=frac{n^2(n+1)^2}4
                                  $$
                                  we get
                                  $$
                                  begin{align}
                                  lim_{ntoinfty}sum_{k=1}^nfrac{left(k-cos^2(k)right)^4}{n^5}
                                  &=lim_{ntoinfty}sum_{k=1}^nleft[frac{k^4}{n^5}+O!left(frac{k^3}{n^5}right)right]\
                                  &=lim_{ntoinfty}left[sum_{k=1}^nfrac{k^4}{n^4}frac1n+O!left(frac1nright)right]\
                                  &=int_0^1x^4,mathrm{d}x+0\[6pt]
                                  &=frac15
                                  end{align}
                                  $$






                                  share|cite|improve this answer









                                  $endgroup$
















                                    0












                                    0








                                    0





                                    $begingroup$

                                    Since
                                    $$
                                    sum_{k=1}^nk^3=frac{n^2(n+1)^2}4
                                    $$
                                    we get
                                    $$
                                    begin{align}
                                    lim_{ntoinfty}sum_{k=1}^nfrac{left(k-cos^2(k)right)^4}{n^5}
                                    &=lim_{ntoinfty}sum_{k=1}^nleft[frac{k^4}{n^5}+O!left(frac{k^3}{n^5}right)right]\
                                    &=lim_{ntoinfty}left[sum_{k=1}^nfrac{k^4}{n^4}frac1n+O!left(frac1nright)right]\
                                    &=int_0^1x^4,mathrm{d}x+0\[6pt]
                                    &=frac15
                                    end{align}
                                    $$






                                    share|cite|improve this answer









                                    $endgroup$



                                    Since
                                    $$
                                    sum_{k=1}^nk^3=frac{n^2(n+1)^2}4
                                    $$
                                    we get
                                    $$
                                    begin{align}
                                    lim_{ntoinfty}sum_{k=1}^nfrac{left(k-cos^2(k)right)^4}{n^5}
                                    &=lim_{ntoinfty}sum_{k=1}^nleft[frac{k^4}{n^5}+O!left(frac{k^3}{n^5}right)right]\
                                    &=lim_{ntoinfty}left[sum_{k=1}^nfrac{k^4}{n^4}frac1n+O!left(frac1nright)right]\
                                    &=int_0^1x^4,mathrm{d}x+0\[6pt]
                                    &=frac15
                                    end{align}
                                    $$







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Dec 21 '18 at 9:15









                                    robjohn♦robjohn

                                    268k27308633




                                    268k27308633






























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