Limit of $x_n=sum_{k=np+1}^{nq}frac{1}{k}$ using Riemann sum












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I am trying to find the limit of the following sequence using Riemann sum:
$$x_n=sum_{k=np+1}^{nq}frac{1}{k}qquad p,qinmathbb{N}quad p<q$$
I have tried to develope the expression:
$$frac{1}{np+1}+frac{1}{np+2}+...+frac{1}{nq}=frac{1}{n}(frac{1}{p+frac{1}{n}}+frac{1}{p+frac{2}{n}}+...+frac{1}{p+frac{n(q-p)}{n}})=frac{1}{n}sum_{k=1}^{n(q-p)}frac{1}{p+frac{k}{n}}$$
But I need an expression like $frac{1}{n}sum_{k=1}^{n}f(frac{k}{n})$, from $k=1$ to $n$, not to $n(q-p)$, so I can calculate the limit as $int_0^1f(x)dx.$



Could you give me some hints? Thanks!






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    I am trying to find the limit of the following sequence using Riemann sum:
    $$x_n=sum_{k=np+1}^{nq}frac{1}{k}qquad p,qinmathbb{N}quad p<q$$
    I have tried to develope the expression:
    $$frac{1}{np+1}+frac{1}{np+2}+...+frac{1}{nq}=frac{1}{n}(frac{1}{p+frac{1}{n}}+frac{1}{p+frac{2}{n}}+...+frac{1}{p+frac{n(q-p)}{n}})=frac{1}{n}sum_{k=1}^{n(q-p)}frac{1}{p+frac{k}{n}}$$
    But I need an expression like $frac{1}{n}sum_{k=1}^{n}f(frac{k}{n})$, from $k=1$ to $n$, not to $n(q-p)$, so I can calculate the limit as $int_0^1f(x)dx.$



    Could you give me some hints? Thanks!






    sequences-and-series limits riemann-sum







    share|cite|improve this question

























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      I am trying to find the limit of the following sequence using Riemann sum:
      $$x_n=sum_{k=np+1}^{nq}frac{1}{k}qquad p,qinmathbb{N}quad p<q$$
      I have tried to develope the expression:
      $$frac{1}{np+1}+frac{1}{np+2}+...+frac{1}{nq}=frac{1}{n}(frac{1}{p+frac{1}{n}}+frac{1}{p+frac{2}{n}}+...+frac{1}{p+frac{n(q-p)}{n}})=frac{1}{n}sum_{k=1}^{n(q-p)}frac{1}{p+frac{k}{n}}$$
      But I need an expression like $frac{1}{n}sum_{k=1}^{n}f(frac{k}{n})$, from $k=1$ to $n$, not to $n(q-p)$, so I can calculate the limit as $int_0^1f(x)dx.$



      Could you give me some hints? Thanks!






      sequences-and-series limits riemann-sum







      share|cite|improve this question













      I am trying to find the limit of the following sequence using Riemann sum:
      $$x_n=sum_{k=np+1}^{nq}frac{1}{k}qquad p,qinmathbb{N}quad p<q$$
      I have tried to develope the expression:
      $$frac{1}{np+1}+frac{1}{np+2}+...+frac{1}{nq}=frac{1}{n}(frac{1}{p+frac{1}{n}}+frac{1}{p+frac{2}{n}}+...+frac{1}{p+frac{n(q-p)}{n}})=frac{1}{n}sum_{k=1}^{n(q-p)}frac{1}{p+frac{k}{n}}$$
      But I need an expression like $frac{1}{n}sum_{k=1}^{n}f(frac{k}{n})$, from $k=1$ to $n$, not to $n(q-p)$, so I can calculate the limit as $int_0^1f(x)dx.$



      Could you give me some hints? Thanks!






      sequences-and-series limits riemann-sum




      sequences-and-series limits riemann-sum






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      asked Jan 4 at 15:13









      GibbsGibbs

      13110




      13110






















          3 Answers
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          We just have to calculate $int_0^{q-p}$ instead of $int_0^1$. Then, with
          $$f(x)=frac{1}{p+x}$$
          $$int_0^{q-p}frac{1}{p+x}dx=[ln|p+x|]_0^{q-p}=ln|p+q-p|-ln|p|=lnfrac{q}{p}$$
          And we are done.






          share|cite|improve this answer





























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            $$lim_{nto infty}x_{n}{=lim_{nto infty}sum_{k=np}^{nq+1}{1over k}\=lim_{nto infty}sum_{k=np}^{nq+1}{nover k}cdot {1over n}\=lim_{nto infty}sum_{k=np}^{nq+1}{1over{kover n}}cdot {1over n}\=int_p^q{dxover x}\=ln {qover p}}$$






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              0














              Some additional transformation have indeed to be made: you got
              $x_n=frac{1}{n}sum_{k=1}^{n(q-p)}frac{1}{p+frac{k}{n}}$; we can continue by writing
              $$
              x_n=frac{q-p}{nleft(q-pright)}sum_{k=1}^{n(q-p)}frac{1}{p+frac{kleft(q-pright)}{nleft(q-pright)}}= left(q-pright)frac 1{N_n}sum_{k=1}^{N_n}fleft(frac{k}{N_n}right)
              $$
              where $N_n=n (q-p)$ and $fcolon xmapsto 1/left(p+x(q-p)right)$.






              share|cite|improve this answer





















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                3 Answers
                3






                active

                oldest

                votes








                3 Answers
                3






                active

                oldest

                votes









                active

                oldest

                votes






                active

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                0














                We just have to calculate $int_0^{q-p}$ instead of $int_0^1$. Then, with
                $$f(x)=frac{1}{p+x}$$
                $$int_0^{q-p}frac{1}{p+x}dx=[ln|p+x|]_0^{q-p}=ln|p+q-p|-ln|p|=lnfrac{q}{p}$$
                And we are done.






                share|cite|improve this answer


























                  0














                  We just have to calculate $int_0^{q-p}$ instead of $int_0^1$. Then, with
                  $$f(x)=frac{1}{p+x}$$
                  $$int_0^{q-p}frac{1}{p+x}dx=[ln|p+x|]_0^{q-p}=ln|p+q-p|-ln|p|=lnfrac{q}{p}$$
                  And we are done.






                  share|cite|improve this answer
























                    0












                    0








                    0






                    We just have to calculate $int_0^{q-p}$ instead of $int_0^1$. Then, with
                    $$f(x)=frac{1}{p+x}$$
                    $$int_0^{q-p}frac{1}{p+x}dx=[ln|p+x|]_0^{q-p}=ln|p+q-p|-ln|p|=lnfrac{q}{p}$$
                    And we are done.






                    share|cite|improve this answer












                    We just have to calculate $int_0^{q-p}$ instead of $int_0^1$. Then, with
                    $$f(x)=frac{1}{p+x}$$
                    $$int_0^{q-p}frac{1}{p+x}dx=[ln|p+x|]_0^{q-p}=ln|p+q-p|-ln|p|=lnfrac{q}{p}$$
                    And we are done.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 4 at 15:45









                    GibbsGibbs

                    13110




                    13110























                        0














                        $$lim_{nto infty}x_{n}{=lim_{nto infty}sum_{k=np}^{nq+1}{1over k}\=lim_{nto infty}sum_{k=np}^{nq+1}{nover k}cdot {1over n}\=lim_{nto infty}sum_{k=np}^{nq+1}{1over{kover n}}cdot {1over n}\=int_p^q{dxover x}\=ln {qover p}}$$






                        share|cite|improve this answer


























                          0














                          $$lim_{nto infty}x_{n}{=lim_{nto infty}sum_{k=np}^{nq+1}{1over k}\=lim_{nto infty}sum_{k=np}^{nq+1}{nover k}cdot {1over n}\=lim_{nto infty}sum_{k=np}^{nq+1}{1over{kover n}}cdot {1over n}\=int_p^q{dxover x}\=ln {qover p}}$$






                          share|cite|improve this answer
























                            0












                            0








                            0






                            $$lim_{nto infty}x_{n}{=lim_{nto infty}sum_{k=np}^{nq+1}{1over k}\=lim_{nto infty}sum_{k=np}^{nq+1}{nover k}cdot {1over n}\=lim_{nto infty}sum_{k=np}^{nq+1}{1over{kover n}}cdot {1over n}\=int_p^q{dxover x}\=ln {qover p}}$$






                            share|cite|improve this answer












                            $$lim_{nto infty}x_{n}{=lim_{nto infty}sum_{k=np}^{nq+1}{1over k}\=lim_{nto infty}sum_{k=np}^{nq+1}{nover k}cdot {1over n}\=lim_{nto infty}sum_{k=np}^{nq+1}{1over{kover n}}cdot {1over n}\=int_p^q{dxover x}\=ln {qover p}}$$







                            share|cite|improve this answer












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                            answered Jan 5 at 6:57









                            Mostafa AyazMostafa Ayaz

                            14.4k3937




                            14.4k3937























                                0














                                Some additional transformation have indeed to be made: you got
                                $x_n=frac{1}{n}sum_{k=1}^{n(q-p)}frac{1}{p+frac{k}{n}}$; we can continue by writing
                                $$
                                x_n=frac{q-p}{nleft(q-pright)}sum_{k=1}^{n(q-p)}frac{1}{p+frac{kleft(q-pright)}{nleft(q-pright)}}= left(q-pright)frac 1{N_n}sum_{k=1}^{N_n}fleft(frac{k}{N_n}right)
                                $$
                                where $N_n=n (q-p)$ and $fcolon xmapsto 1/left(p+x(q-p)right)$.






                                share|cite|improve this answer


























                                  0














                                  Some additional transformation have indeed to be made: you got
                                  $x_n=frac{1}{n}sum_{k=1}^{n(q-p)}frac{1}{p+frac{k}{n}}$; we can continue by writing
                                  $$
                                  x_n=frac{q-p}{nleft(q-pright)}sum_{k=1}^{n(q-p)}frac{1}{p+frac{kleft(q-pright)}{nleft(q-pright)}}= left(q-pright)frac 1{N_n}sum_{k=1}^{N_n}fleft(frac{k}{N_n}right)
                                  $$
                                  where $N_n=n (q-p)$ and $fcolon xmapsto 1/left(p+x(q-p)right)$.






                                  share|cite|improve this answer
























                                    0












                                    0








                                    0






                                    Some additional transformation have indeed to be made: you got
                                    $x_n=frac{1}{n}sum_{k=1}^{n(q-p)}frac{1}{p+frac{k}{n}}$; we can continue by writing
                                    $$
                                    x_n=frac{q-p}{nleft(q-pright)}sum_{k=1}^{n(q-p)}frac{1}{p+frac{kleft(q-pright)}{nleft(q-pright)}}= left(q-pright)frac 1{N_n}sum_{k=1}^{N_n}fleft(frac{k}{N_n}right)
                                    $$
                                    where $N_n=n (q-p)$ and $fcolon xmapsto 1/left(p+x(q-p)right)$.






                                    share|cite|improve this answer












                                    Some additional transformation have indeed to be made: you got
                                    $x_n=frac{1}{n}sum_{k=1}^{n(q-p)}frac{1}{p+frac{k}{n}}$; we can continue by writing
                                    $$
                                    x_n=frac{q-p}{nleft(q-pright)}sum_{k=1}^{n(q-p)}frac{1}{p+frac{kleft(q-pright)}{nleft(q-pright)}}= left(q-pright)frac 1{N_n}sum_{k=1}^{N_n}fleft(frac{k}{N_n}right)
                                    $$
                                    where $N_n=n (q-p)$ and $fcolon xmapsto 1/left(p+x(q-p)right)$.







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Jan 5 at 13:16









                                    Davide GiraudoDavide Giraudo

                                    125k16150261




                                    125k16150261






























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